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```Instructor’s Solution Manual
Introduction to Electrodynamics
Fourth Edition
David J. Griffiths
2014
2
Contents
1 Vector Analysis
4
2 Electrostatics
26
3 Potential
53
4 Electric Fields in Matter
92
5 Magnetostatics
110
6 Magnetic Fields in Matter
133
7 Electrodynamics
145
8 Conservation Laws
168
9 Electromagnetic Waves
185
10 Potentials and Fields
210
231
12 Electrodynamics and Relativity
262
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
3
Preface
Although I wrote these solutions, much of the typesetting was done by Jonah Gollub, Christopher Lee, and
James Terwilliger (any mistakes are, of course, entirely their fault). Chris also did many of the figures, and I
would like to thank him particularly for all his help. If you find errors, please let me know ([email protected]).
David Griffiths
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
3
CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS
4
3
CHAPTER 1. VECTOR ANALYSIS
3
CHAPTER 1. VECTOR ANALYSIS
Chapter 1
Chapter 1
Chapter 1
1
VectorChapter
Analysis
Vector Analysis
Vector
Analysis
Vector
Analysis
Problem 1.1
✒
✣
Problem 1.1
}} }
✒
✣
|C| sin θ2
C
+
C
C
✒
✣
B
(a) From
the diagram,
|B1.1
+ C| cos θ3 = |B| cos θ1 + |C| cos θ2 . Multiply by |A|.
Problem
Problem 1.1
|A||B + C| cos θ(a)
|A||B|
θ1 + |A||C|
cos
θ2θ.3 = |B| cos θ1 + |C| cos θ2 .
thecos
diagram,
|B + C|
cos
3 =From
From the diagram,
|B
+
C|
cos
✓=
|B| cos
✓1θ ++|C|
cos ✓2cos
. Multiply
by |A|.
3 =
|A||B
+
C|
cos
θ
|A||B|
cos
θ2θ.2 .
So:(a)A·(B
+ C)
=
A·B
+
A·C.
(Dot
product
(a) From the diagram, |B
|B|
cos|A||C|
θ1 + |C|
cos
3 + C| cos θ3 = is
1 distributive)
|C| sin
B
B
C
+
C
}
C
+
|A||B + C| cos
✓3 +
=A·(B
|A||B|
✓=1 A·B
+ |A||C|
cos
✓(Dot
2 . cosproduct
|C| sin θ2
|A||B
C|
cos+
θ3cos
= |A||B|
cos+θ1A·C.
+
|A||C|
θ2 .
So:
C)
is distributive)
θ2
So: A·(B
C)
=
A·B
+
A·C.
(Dot
product
is
Similarly:
|B ++C|
sin
θ
=
|B|
sin
θ
+
|C|
sin
θ
Mulitply
by |A| n̂.
2 . distributive)
So: A·(B3 + C) = A·B 1+ A·C. (Dot product
is distributive)
θ3 ✯
θ2
θ3 = |B|sin
sinθθ1n̂.
+ |C| sin θ2 . Mulitply by |A| n̂.B
|A||B + C| sin θ3 n̂Similarly:
= |A||B||B
sin+θ1C|n̂sin
+ |A||C|
θ3 ✯ |B| sin θ1
2
θ2
Similarly:
|B
+
C|
sin
θ
=
|B|
sin
θ
+
|C|
sin
θ
.
Mulitply
by
|A|
n̂.
Similarly: |B
+|A||B
C| sin+
✓C|
=
|B|
sin
✓
+
|C|
sin
✓
.
Mulitply
by
|A|
n̂.
3
1
2
3 sin θ3 n̂ =1 |A||B| sin θ21 n̂ + |A||C| sin θ2 n̂.
θ1 θ3 ✯
B
|B| sin
✲
If n̂ is|A||B
the unit
vector
pointing
out
of the
it✓ follows
that
\$|B| sin
|A||B
C|the
sin θunit
=vector
|A||B|
sin
θpage,
|A||C|
n̂.
θ1
3 n̂sin
1 n̂ +
θ1
+ C| sin
✓3 n̂
n̂+is
=
|A||B|
✓1 n̂ +pointing
|A||C|
sin
n̂.sin
2of
If
out
theθ2page,
it follows that! "#θ1B \$ !! "#
A
"# θ2 \$ !✲ "# \$ ✲
A×(B
C)the
= unit
(A×B)
(A×C).
(Cross
product
is
distributive)
|B|
cos
θ1 \$ ! |C|"#
cos
If n̂ vector
is the+unit
vector out
pointing
out
of theit page,
it follows
that
A
!
"#
\$
If +
n̂ is
pointing
of
the
page,
follows
that
A×(B + C) = (A×B) + (A×C). (Cross product is distributive)
|B| cos θ1
|C|
A cos θ2
A×(B + C) = (A×B) + (A×C). (Cross product is distributive)
|B| cos θ1
|C| cos θ2
A⇥(B
+ C)case,
= (A⇥B)
+ E.
(A⇥C).
product
is distributive)
(b) For the
general
see G.
Hay’s (Cross
Vector
and
Tensor
Analysis,
Chapter
1, Section 71,(dot
product)
and
general
E.
Hay’s
Vector
and Analysis,
Tensor
Analysis,
Section
7 (dot
(b)(b)
ForFor
the the
general
case,case,
see G.see
E. G.
Hay’s
Vector
and Tensor
Chapter 1,Chapter
Section 7 (dot
product)
and product)
Section
(cross
product)
Section
8 (cross
(b) For8 the
general
case,
see
G.
E. product)
Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and
Section
8 (cross
product)
Section
8
(cross
product)
Problem 1.2 Problem
Problem
1.2 1.2
Problem 1.2
C
CC ✻
TheThe
triple
cross-product
is not isinassociative.
general
associative.
For example,
triple
cross-product
not
in general
associative.
For example,
The triple cross-product
is not
in general
For
example,
✻
✻
The triple cross-product
is
not
in
general
associative.
For
example,
suppose
A
=
B
and
C
is
perpendicular
to
A,
as
in
the
diagram.
suppose
= B and C istoperpendicular
to diagram.
A, as in the diagram.
suppose A = B and
C isAperpendicular
A, as in the
✲
out-of-the-page,
andasA×(B×C)
points down,
suppose A Then
= Then
B (B×C)
and(B×C)
C points
is perpendicular
to A,
in
theA×(B×C)
diagram.
✲AA==BB ✲ A = B
points
out-of-the-page,
and
points down,
Then (B×C) points
A×(B×C)
down,
andpoints
hasout-of-the-page,
magnitude
ABC. and
But
(A×B)
= 0, sopoints
(A×B)×C
= 0 ̸=
❂
Then (B⇥C)
out-of-the-page,
and
A⇥(B⇥C)
points
down,
and
has magnitude
ABC. But (A×B)
= 0, so=(A×B)×C
=B×C
0 ̸=
❂
and has
ABC.
But
(A×B)
A×(B×C).
❂
❄
A×(B×C)
and magnitude
has magnitude
ABC.
But
(A⇥B)==0,0,so
so (A×B)×C
(A⇥B)⇥C =
00 6≠=
B×C ❄
A×(B×C).
A×(B×C)
B×C ❄
A×(B×C).
A×(B×C)
A⇥(B⇥C).
}
}
z
Problem 1.3
z✻✻
Problem
1.3 1.3 Problem 1.3
Problem
z✻
√
p
A = +1 x̂ + 1√
ŷ −p1 ẑ; A = 3; B√= 1 x̂ + 1 ŷ + 1 ẑ;
√
A =x̂ +1
ẑ;=A =3;1 ŷB
3; B
ŷ+√
+
B
= 1 ŷ3.
✣B
=1 =
1 x̂1Ax̂
+=+1 1ŷ3;
ẑ;ẑ;1B
3.+ 1 ẑ;
A = +1
+ 1x̂ŷ+−1 ŷ1Aẑ;=1A+1
ẑ;
B131√
=
x̂ =
+
3
cos
θ
⇒
cos
θ.
A·B = +1 +x̂1+
− 1 =−
1=
ABpcos
θ
=
p
B
√ √ 11
√ √
✣ θ
✣
✲By
& AB
A·B
+ 11A·B
11−1
==
1%+1
=
cos
✓=◦=1 3=3AB
3cos
cos
✓θ⇒
)
cos
✓ 3=
=cos
..θ ⇒ cos θ.
3
1
−
1
cos
=
A·B =
+1=++1
1−
=
=
AB
θ
=
3
θ
cos
θ
1 +cos
3
θ
θ
θ = cos
3
3 ≈ 70.5288
✲y
❲
✲y
%
&
% & 1 θ=
A
✰
cos◦−1 31 ≈ 70.5288◦
✓
=
70.5288
−1cos1 1 3 ⇡
x
❲
θ = cos
❲
3 ≈ 70.5288
A
✰
A
✰
x
Problem 1.4
x
Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
Problem 1.4
Problem 1.4
we might pick the base (A) and the left side (B):
The cross-product
of any two vectors in the plane will give a vector perpendicular to the plane. For example,
The cross-product The
of any
two vectorsofin
the
plane
willingive
a vector
to the plane.
example,
cross-product
any
two
vectors
the plane
willperpendicular
give a vector perpendicular
to For
the plane.
For exam
we might pick the base (A) and the left side (B):
c base
we Pearson
might
pick
base
(A)
and
the NJ.
leftAllside
(B):
⃝2005
Upper
River,
rights
reserved. This material is
we might pick the
(A)Education,
and the
theInc.,
left
side
(B):
A=
protected under all copyright laws as they currently exist. No portion of this material may be
1 x̂reproduced,
+ 2 ŷ +in0 any
ẑ; B
=or by
1 x̂any+means,
0 ŷ +without
3 ẑ. permission in writing from the publisher.
form
A = −1 x̂ + 2 ŷ⃝2005
+ 0 ẑ;Pearson
B = Education,
−1 x̂ + 0Inc.,
ŷ +Upper
c
This material is
protected under all copyright laws as they currently exist. No portion of this material may be
c
2012
Pearson
Education,
Inc.,
Upper
River,
NJ.
All rights
reserved. This material is
reproduced, in any form or by any means, without permission in writing from the
publisher.
protected
under
laws as they
currently
exist.
c
⃝2005
River,
NJ.all
All
rights reserved.
This
material
is No portion of this material may be
in anyNo
form
or by any
means,
without
permission
in writing from the publisher.
protected under all copyright laws as theyreproduced,
currently exist.
portion
of this
material
may
be
reproduced, in any form or by any means, without permission in writing from the publisher.
5
CHAPTER 1. VECTOR ANALYSIS
x̂ ŷ ẑ
1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ.
1 0 3
This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its
length:
p
A⇥B
6
3
2
|A⇥B| = 36 + 9 + 4 = 7.
n̂ = |A
⇥B| = 7 x̂ + 7 ŷ + 7 ẑ .
A⇥B =
Problem 1.5
A⇥(B⇥C) =
(By Cz
x̂
Ax
Bz Cy ) (Bz Cx
ŷ
Ay
Bx Cz ) (Bx Cy
ẑ
Az
By Cx )
= x̂[Ay (Bx Cy By Cx ) Az (Bz Cx Bx Cz )] + ŷ() + ẑ()
(I’ll just check the x-component; the others go the same way)
= x̂(Ay Bx Cy Ay By Cx Az Bz Cx + Az Bx Cz ) + ŷ() + ẑ().
C(A·B) = [Bx (Ax Cx + Ay Cy + Az Cz ) Cx (Ax Bx + Ay By + Az Bz )] x̂ + () ŷ + () ẑ
= x̂(Ay Bx Cy + Az Bx Cz Ay By Cx Az Bz Cx ) + ŷ() + ẑ(). They agree.
B(A·C)
Problem 1.6
A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C) C(A·B)+C(A·B) A(C·B)+A(B·C) B(C·A) = 0.
So: A⇥(B⇥C) (A⇥B)⇥C = B⇥(C⇥A) = A(B·C) C(A·B).
If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or
one is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.)
Conclusion: A⇥(B⇥C) = (A⇥B)⇥C () either A is parallel to C, or B is perpendicular to A and C.
Problem 1.7
r
r
= (4 x̂ + 6 ŷ + 8 ẑ)
p
= 4+4+1= 3
r̂
=
r
r
=
2
3 x̂
2
3 ŷ
(2 x̂ + 8 ŷ + 7 ẑ) = 2 x̂
2 ŷ + ẑ
+ 13 ẑ
Problem 1.8
(a) Āy B̄y + Āz B̄z = (cos Ay + sin Az )(cos By + sin Bz ) + ( sin Ay + cos Az )( sin By + cos Bz )
= cos2 Ay By + sin cos (Ay Bz + Az By ) + sin2 Az Bz + sin2 Ay By sin cos (Ay Bz + Az By ) +
2
cos Az Bz
= (cos2 + sin2 )Ay By + (sin2 + cos2 )Az Bz = Ay By + Az Bz . X
(b) (Ax )2 + (Ay )2 + (Az )2 = ⌃3i=1 Ai Ai = ⌃3i=1 ⌃3j=1 Rij Aj ⌃3k=1 Rik Ak = ⌃j,k (⌃i Rij Rik ) Aj Ak .
⇢
1 if j = k
This equals A2x + A2y + A2z provided ⌃3i=1 Rij Rik =
0 if j 6= k
Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but also
necessary. For suppose A = (1, 0, 0). Then ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 , and this must equal 1 (since we
2
2
2
want Ax +Ay +Az = 1). Likewise, ⌃3i=1 Ri2 Ri2 = ⌃3i=1 Ri3 Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0).
Then we want 2 = ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 + ⌃i Ri2 Ri2 + ⌃i Ri1 Ri2 + ⌃i Ri2 Ri1 . But we already
know that the first two sums are both 1; the third and fourth are equal, so ⌃i Ri1 Ri2 = ⌃i Ri2 Ri1 = 0, and so
on for other unequal combinations of j, k. X In matrix notation: R̃R = 1, where R̃ is the transpose of R.
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
6
CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS
5
CHAPTER 1. VECTOR ANALYSIS
5
Problem 1.9
y✻
y
z′ ✻
✻
y✻
y
❃
z′ ✻
✻
✿
✿ ❃
down the
axis:
✲ x LookingLooking
down
the axis:
✒ ❄
✲x
Looking down the axis:
✒ ❄
′
■
&y
′
■z ✰
&
✠
y
✰
&
z
′
✰
z
x
✠
x &x
✰′
z
x
A 120 rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az ,
Ay = Ax , Az = Ay .
0
1
0 01
R = @1 0 0A
0 10
Problem 1.10
(a) No change. (Ax = Ax , Ay = Ay , Az = Az )
(b) A !
A, in the sense (Ax =
Ax , Ay =
Ay , Az =
Az )
(c) (A⇥B) ! ( A)⇥( B) = (A⇥B). That is, if C = A⇥B, C ! C . No minus sign, in contrast to
behavior of an “ordinary” vector, as given by (b). If A and B are pseudovectors, then (A⇥B) ! (A)⇥(B) =
(A⇥B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vector
and a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector.
Angular momentum (L = r⇥p) and torque (N = r⇥F) are pseudovectors.
(d) A·(B⇥C) ! ( A)·(( B)⇥( C)) = A·(B⇥C). So, if a = A·(B⇥C), then a !
changes sign under inversion of coordinates.
Problem 1.11
a; a pseudoscalar
(a)rf = 2x x̂ + 3y 2 ŷ + 4z 3 ẑ
(b)rf = 2xy 3 z 4 x̂ + 3x2 y 2 z 4 ŷ + 4x2 y 3 z 3 ẑ
(c)rf = ex sin y ln z x̂ + ex cos y ln z ŷ + ex sin y(1/z) ẑ
Problem 1.12
(a) rh = 10[(2y 6x 18) x̂ + (2x 8y + 28) ŷ]. rh = 0 at summit, so
2y 6x 18 = 0
2y 18 24y + 84 = 0.
2x 8y + 28 = 0 =) 6x 24y + 84 = 0
22y = 66 =) y = 3 =) 2x 24 + 28 = 0 =) x = 2.
Top is 3 miles north, 2 miles west, of South Hadley.
(b) Putting in x = 2, y = 3:
h = 10( 12 12 36 + 36 + 84 + 12) = 720 ft.
(c) Putting in x = 1, y = 1: rh = 10[(2 6 18) x̂ + (2
p
|rh| = 220 2 ⇡ 311 ft/mile; direction: northwest.
8 + 28) ŷ] = 10( 22 x̂ + 22 ŷ) = 220( x̂ + ŷ).
protected under all copyright laws as they currently exist. No portion of this material may be
in any
form
or All
by rights
any means,
without
c
⃝2005
Pearson Education, Inc.,reproduced,
River,
NJ.
reserved.
Thispermission
material isin writing from the publisher.
c
⃝2005
Pearson
Education,
Inc.,
NJ.
All rights
reserved.
This material
is
protected
under
laws River,
as they
currently
exist.
No portion
of this material
may be
protected under
laws
as they
exist.
No portion
of this
may the
be publisher.
reproduced,
in any
form
or bycurrently
any means,
without
permission
in material
writing from
reproduced, in any form or by any means, without permission in writing from the publisher.
7
CHAPTER 1. VECTOR ANALYSIS
Problem 1.13
r
= (x
x0 ) x̂ + (y
y 0 ) ŷ + (z
r
z 0 ) ẑ;
=
p
(x
x0 )2 + (y
y 0 )2 + (z
z 0 )2 .
(a) r( r 2 ) =
@
@x [(x
@
@
x0 )2 +(y y 0 )2 +(z z 0 )2 ] x̂+ @y
() ŷ + @z
() ẑ = 2(x x0 ) x̂+2(y y 0 ) ŷ +2(z z 0 ) ẑ = 2
(b) r( r1 ) =
@
@x [(x
x0 )2 + (y
(c)
1
2 ()
=
=
()
@
@x (
r
n
3
2
3
2
2(x x ) x̂
[(x x0 ) x̂ + (y
0
)=n
r
r
n 1@
@x
1
2 ()
y 0 )2 + (z
3
2
z 0 )2 ]
1
2 ()
2(y y ) ŷ
y ) ŷ + (z z 0 ) ẑ] =
0
0
=n
r
r 2r
n 1 1 1
(2
x)
1
2
x̂ +
3
2
@
@y ()
0
1
2
ŷ +
2(z z ) ẑ
(1/ r 3 ) r =
=n
r
n 1
r̂
Problem 1.14
x,
@
@z ()
1
2
r
.
ẑ
(1/ r 2 ) r̂ .
so r( r n ) = n
r
n 1
r̂
y = +y cos + z sin ; multiply by sin : y sin = +y sin cos + z sin2 .
z = y sin + z cos ; multiply by cos : z cos = y sin cos + z cos2 .
Add: y sin + z cos = z(sin2 + cos2 ) = z. Likewise, y cos
z sin = y.
@y
@y
@z
@z
So @y = cos ; @z = sin ; @y = sin ; @z = cos . Therefore
)
@f @y
@f @z
(rf )y = @f
@y = @y @y + @z @y = + cos (rf )y + sin (rf )z
So rf transforms as a vector.
@f @y
@f @z
(rf )z = @f
sin (rf )y + cos (rf )z
@z = @y @z + @z @z =
qed
Problem 1.15
(a)r·va =
@
2
@x (x )
+
@
2
@y (3xz )
(b)r·vb =
@
@x (xy)
+
@
@y (2yz)
(c)r·vc =
@
2
@x (y )
+
@
@y (2xy
Problem 1.16
+
+
@
@z (
2xz) = 2x + 0
@
@z (3xz)
+ z2) +
2x = 0.
= y + 2z + 3x.
@
@z (2yz)
= 0 + (2x) + (2y) = 2(x + y)
h
@
@
@
@
r·v = @x
( rx3 ) + @y
( ry3 ) + @z
( rz3 ) = @x
x(x2 + y 2 + z 2 )
h
i
h
i
3
3
@
@
+ @y
y(x2 + y 2 + z 2 ) 2 + @z
z(x2 + y 2 + z 2 ) 2
3
5
= () 2 + x( 3/2)() 2 2x + ()
5
+ z( 3/2)() 2 2z = 3r 3 3r
3
2
5
3
2
i
3
+ y( 3/2)() 2 2y + () 2
5 2
(x + y 2 + z 2 ) = 3r 3 3r
3
= 0.
This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the
origin. How, then, can r·v = 0? The answer is that r·v = 0 everywhere except at the origin, but at the
origin our calculation is no good, since r = 0, and the expression for v blows up. In fact, r·v is infinite at
that one point, and zero elsewhere, as we shall see in Sect. 1.5.
Problem 1.17
v y = cos vy + sin vz ; v z = ⇣sin vy + cos v⌘z .
⇣
⌘
@v y
@vy
@vy @y
@vy @z
@vz
@vz @z
z @y
cos + @v
sin . Use result in Prob. 1.14:
@y = @y cos + @y sin =
@y @y + @z @y
@y @y + @z @y
⇣
⌘
⇣
⌘
@vy
@vy
@vz
@vz
= @y cos + @z sin
cos + @y cos + @z sin
sin .
⇣
⌘
⇣
⌘
@vy
@vy @y
@vy @z
@v z
@vz
@vz @y
@vz @z
=
sin
+
cos
=
+
sin
+
+
cos
@z
@z
@y @z
@z @z
⇣@z
⌘ @y @z ⇣ @z @z
⌘
@vy
@vy
@vz
@vz
=
sin +
cos . So
@y sin + @z cos
@y sin + @z cos
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
8
@v y
@y
+
@v z
@z
=
@vy
@y
cos2
@vy
@z
+
CHAPTER 1. VECTOR ANALYSIS
sin cos +
sin cos +
@vz
@y
=
@vy
@y
cos
+ sin
+
2
2
@vz
@y
sin cos +
@vz
@z
cos
sin2
@vz
@z
@vz
@z
sin2
+
@vy
@y
@vy
@z
sin2
sin cos
2
+ cos2
=
@vy
@y
+
@vz
@z .
X
Problem 1.18
(a) r⇥va =
x̂
ŷ
@
@x
2
@
@y
x̂
ŷ
ẑ
@
@x
@
@y
@
@z
x 3xz
(b) r⇥vb =
ẑ
@
@z
= x̂(0
6xz) + ŷ(0 + 2z) + ẑ(3z 2
0) =
6xz x̂ + 2z ŷ + 3z 2 ẑ.
2xz
2
= x̂(0
2y) + ŷ(0
3z) + ẑ(0
x) =
2y x̂
3z ŷ
x ẑ.
xy 2yz 3xz
(c) r⇥vc =
x̂
ŷ
@
@x
2
@
@y
ẑ
@
@z
= x̂(2z
2z) + ŷ(0
0) + ẑ(2y
2y) = 0.
y (2xy + z ) 2yz
2
Problem 1.19
As we go from point A to point B (9 o’clock to 10 o’clock), x
increases, y increases, vx increases, and vy decreases, so @vx /@y >
0, while @vy /@y < 0. On the circle, vz = 0, and there is no
dependence on z, so Eq. 1.41 says
✓
◆
@vy
@vx
r ⇥ v = ẑ
@x
@y
y
v
v
v
B
A
x
points in the negative z direction (into the page), as the right
hand rule would suggest. (Pick any other nearby points on the
circle and you will come to the same conclusion.) [I’m sorry, but I
cannot remember who suggested this cute illustration.]
z
v
Problem 1.20
v = y x̂ + x ŷ; or v = yz x̂ + xz ŷ + xy ẑ; or v = (3x2 z
or v = (sin x)(cosh y) x̂ (cos x)(sinh y) ŷ; etc.
Problem 1.21
(i) r(f g) =
@(f g)
@x
=f
⇣
x̂ +
@g
@x
(iv) r·(A⇥B) =
=
=
x̂ +
@(f g)
@y
@g
@y
ŷ +
ŷ +
(v) r⇥ (f A) =
⇣
⇣
⌘
⇣
⌘
⇣
⌘
@g
@g
@f
@g
@f
ẑ = f @x
+ g @f
x̂
+
f
+
g
ŷ
+
f
+
g
ẑ
@x
@y
@z
@z
⌘
⇣
⌘ @y
@f
@f
ẑ + g @f
qed
@x x̂ + @y ŷ + @z ẑ = f (rg) + g(rf ).
@g
@z
Az By ) +
Ay
@(f Az )
@y
3xz 2 ) ẑ;
@(f g)
@z
@
@
Ax Bz ) + @z
(Ax By
@y (Az Bx
@A
@B
@Bz
@Az
@Bx
y
y
z
Ay @x + Bz @x
Az @x
By @x + Az @y + Bx @A
@y
@By
@A
x
x
+Ax @z + By @A
Ay @B
Bx @zy
@z
@z
⇣
⌘
⇣
⌘
@A
@A
@Az
@Ax
y
z
x
Bx @A
+ By @A
+ Bz @xy
@y
@z
@z
@x
@y
@
@x
(Ay Bz
z 3 ) x̂ + 3 ŷ + (x3
@Bx
@z
@(f Ay )
@z
@Bz
@x
⌘
x̂ +
⇣
Az
⇣
@(f Ax )
@z
@By
@x
⌘
Ay Bx )
z
Ax @B
@y
Ax
⇣
@Bz
@y
x
Bz @A
@y
@By
@z
= B· (r⇥A) A· (r⇥B).
⌘
⇣
⌘
@(f Ay )
@(f Az )
@(f Ax )
ŷ
+
ẑ
@x
@x
@y
@Bx
@y
⌘
qed
protected under all copyright laws as they currently exist. No portion of this material may be
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9
CHAPTER 1. VECTOR ANALYSIS
⇣
⌘
⇣
⌘
@A
@f
@f
@Ax
z
z
= f @A
f @zy Ay @f
f @A
Az @f
@y + Az @y
@z x̂ + f @z + Ax @z
@x
@x ŷ
⇣
⌘
@A
x
+ f @xy + Ay @f
f @A
Ax @f
@x
@y
@y ẑ
h⇣
⌘
⇣
⌘ i
@Ay
@Ay
@Ax
@Az
@Ax
z
= f @A
x̂
+
ŷ
+
ẑ
@y
@z
@x
h⇣ @z
⌘ @x⇣
⌘@y ⇣
⌘ i
@f
@f
@f
@f
@f
Ay @f
A
x̂
+
A
A
ŷ
+
A
A
z
x
z
x
y
@z
@y
@x
@z
@y
@x ẑ
= f (r⇥A) A⇥ (rf ). qed
Problem 1.22
⇣
⌘
⇣
⌘
@By
@By
@By
@Bx
@Bx
x
(a) (A·r) B = Ax @B
+
A
+
A
x̂
+
A
+
A
+
A
ŷ
y
z
x
y
z
@x
@y
@z
@y
@z
⇣
⌘@x
@Bz
@Bz
@Bz
+ Ax @x + Ay @y + Az @z ẑ.
x x̂+y ŷ+z ẑ
=p
. Let’s just do the x component.
x2 +y 2 +z 2
⇣
⌘
@
@
@
p 2 x2 2
[(r̂·r)r̂]x = p1 x @x
+ y @y
+ z @z
x +y +z
n h
i
h
i
h
1
1
p 3 2y + zx
= 1r x p1 + x( 12 ) (p1 )3 2x + yx
2(
)
(b) r̂ =
r
r
=
1
r
x
r
1
r3
x3 + xy 2 + xz 2
=
1
r
x
r
x
r3
1
1
p 3 2z
2(
)
x2 + y 2 + z 2
=
1
r
x
r
io
x
r
= 0.
Same goes for the other components. Hence: (r̂·r) r̂ = 0 .
⇣
⌘
@
@
@
(c) (va ·r) vb = x2 @x
+ 3xz 2 @y
2xz @z
(xy x̂ + 2yz ŷ + 3xz ẑ)
= x2 (y x̂ + 0 ŷ + 3z ẑ) + 3xz 2 (x x̂ + 2z ŷ + 0 ẑ) 2xz (0 x̂ + 2y ŷ + 3x ẑ)
= x2 y + 3x2 z 2 x̂ + 6xz 3 4xyz ŷ + 3x2 z 6x2 z ẑ
= x2 y + 3z 2 x̂ + 2xz 3z 2
2y ŷ
3x2 z ẑ
Problem 1.23
(ii) [r(A·B)]x =
@
@x (Ax Bx
+ Ay By + Az Bz ) =
[A⇥(r⇥B)]x = Ay (r⇥B)z
@Ax
@x Bx
Az (r⇥B)y = Ay
x
+ Ax @B
@x +
@By
@x
@Bx
@y
@Ay
@x By
+ Ay
@Bx
@z
Az
@By
@x
+
@Az
@x Bz
z
+ Az @B
@x
@Bz
@x
@A
@Ax
@Az
x
[B⇥(r⇥A)]x = By @xy
Bz @A
@y
@z
@x
@Bx
@Bx
@
@
@
x
[(A·r)B]x = Ax @x
+ Ay @y
+ Az @z
Bx = Ax @B
@x + Ay @y + Az @z
@Ax
@Ax
@Ax
[(B·r)A]x = Bx @x + By @y + Bz @z
So [A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A]x
@B
@Ay
@Bz
@Az
x
x
x
x
= Ay @xy Ay @B
Az @B
By @A
Bz @A
@y
@z + Az @x + By @x
@y
@z + Bz @x
@Bx
@Bx
@Ax
@Ax
@Ax
x
+Ax @B
@x + Ay @y + Az @z + Bx @x + By @y + Bz @z
@Bx
x
= Bx @A
@x + Ax @x + By
/
/
/
@Ax
@y
+Bz
+
+ Az
= [r(A·B)]x (same for y and z)
@Ax
@z
z
+ @A
@x
@Ay
@x
(vi) [r⇥(A⇥B)]x =
=
@Ax
@z
/ + Ay @[email protected]
@B
/ + @[email protected] + @[email protected]/
@z
x
+ @A
@y
x
y
z
/
@Bx
@y
/
x
+ @B
@y
x
@
@
@
Ay Bx )
@y (A⇥B)z
@z (A⇥B)y = @y (Ax By
@By
@Ay
@Ax
@Bx
@Az
Ay @y
@y By + Ax @y
@y Bx
@z Bx
[(B·r)A (A·r)B + A(r·B) B(r·A)]x
@Ax
@Ax
x
x
x
= Bx @A
Ax @B
Ay @B
@x + By @y + Bz @z
@x
@y
x
Az @B
@z + Ax
@Bx
@x
@
Ax Bz )
@z (Az Bx
@Bx
@Ax
z
Az @z + @z Bz + Ax @B
@z
+
@By
@y
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
z
+ @B
@z
Bx
@Ax
@x
+
@Ay
@y
z
+ @A
@z
10
x
= By @A
@y + Ax
/
@Bx
@x
/
x
+ @B
@x +
CHAPTER 1. VECTOR ANALYSIS
@By
@y
+
@Bz
@z
x
Bz @A
@z
+ Bx
/
/
@Ax
@x
@Ay
@y
@Ax
@x
@Az
@z
x
x
+ Ay @B
+ Az @B
+
@y
@z
= [r⇥(A⇥B)]x (same for y and z)
Problem 1.24
r(f /g) =
@
@
@
@x (f /g) x̂ + @y (f /g) ŷ + @z (f /g) ẑ
@f
@g
@f
@g
g
f
g @x f @x
g @f f @g
x̂ + @y g2 @y ŷ + @z g2 @z ẑ
2
g
h ⇣
⌘
⇣
@f
@f
@f
@g
@g
1
f @x
x̂ + @y
ŷ
g 2 g @x x̂ + @y ŷ + @z ẑ
=
=
r·(A/g) =
=
=
+
@
@
@
@x (Ax /g) + @y (Ay /g) + @z (Az /g)
@A
y
@g
@g
x
g
Ay
g @A
Ax @x
g @Az A @g
@x
+ @y g2 @y + @z g2 z @x
h g⇣2
⌘ ⇣
@Ay
@g
@g
@Ax
@Az
1
Ax @x
+ Ay @y
g2 g
@x + @y + @z
[r⇥(A/g)]x =
=
=
=
@
@
@y (Az /g)
@z (Ay /g)
@Az
@A
@g
g @y Az @y
g @zy Ay @g
@z
h g⇣2
⌘g2 ⇣
@Ay
@g
@Az
1
Az @y
g2 g
@y
@z
g(r⇥A)x +(A⇥rg)x
(same for
g2
Ay @g
@z
@g
@z ẑ
⌘i
=
+ Az @g
@z
grf f rg
.
g2
⌘i
=
qed
gr·A A·rg
.
g2
qed
⌘i
y and z). qed
Problem 1.25
(a) A⇥B =
x̂ ŷ ẑ
x 2y 3z = x̂(6xz) + ŷ(9zy) + ẑ( 2x2
3y 2x 0
@
@
r·(A⇥B) = @x
(6xz) + @y
(9zy) +
⇣
⌘
@
@
r⇥A = x̂ @y (3z) @z (2y) + ŷ
⇣
⌘
@
@
r⇥B = x̂ @y
(0) @z
( 2x) + ŷ
?
r·(A⇥B) = B·(r⇥A)
(b) A·B = 3xy
4xy =
6y 2 )
@
2
6y 2 )
@z ( 2x
@
@
@z (x)
@x (3z)
@
@z (3y)
A·(r⇥B) = 0
= 6z + 9z + 0 = 15z
⇣
⌘
@
@
+ ẑ @x
(2y) @y
(x) = 0; B·(r⇥A) = 0
⇣
⌘
@
@
@
(0)
+
ẑ
(
2x)
(3y)
= 5 ẑ; A·(r⇥B) =
@x
@x
@y
15z
( 15z) = 15z. X
@
@
xy ; r(A·B) = r( xy) = x̂ @x
( xy) + ŷ @y
( xy) =
y x̂
x ŷ
x̂ ŷ ẑ
A⇥(r⇥B) = x 2y 3z = x̂( 10y) + ŷ(5x); B⇥(r⇥A) = 0
0 0 5
⇣
⌘
@
@
@
(A·r)B = x @x
+ 2y @y
+ 3z @z
(3y x̂ 2x ŷ) = x̂(6y) + ŷ( 2x)
⇣
⌘
@
@
(B·r)A = 3y @x
2x @y
(x x̂ + 2y ŷ + 3z ẑ) = x̂(3y) + ŷ( 4x)
A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A
= 10y x̂ + 5x ŷ + 6y x̂ 2x ŷ + 3y x̂ 4x ŷ = y x̂ x ŷ = r·(A·B). X
⇣
⌘
⇣
@
@
@
@
@
(c) r⇥(A⇥B) = x̂ @y
( 2x2 6y 2 ) @z
(9zy) + ŷ @z
(6xz) @x
( 2x2 6y 2 ) + ẑ @x
(9zy)
= x̂( 12y 9y) + ŷ(6x + 4x) + ẑ(0) = 21y x̂ + 10x ŷ
r·A =
@
@x (x)
+
@
@y (2y)
+
@
@z (3z)
= 1 + 2 + 3 = 6; r·B =
@
@x (3y)
+
@
@y (
⌘
@
@y (6xz)
2x) = 0
protected under all copyright laws as they currently exist. No portion of this material may be
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11
CHAPTER 1. VECTOR ANALYSIS
(B·r)A (A·r)B + A(r·B)
= r⇥(A⇥B). X
B(r·A) = 3y x̂
4x ŷ
6y x̂ + 2x ŷ
18y x̂ + 12x ŷ =
21y x̂ + 10x ŷ
Problem 1.26
(a)
@ 2 Ta
@x2
= 2;
(b)
@ 2 Tb
@x2
=
(c)
@ 2 Tc
@x2
= 25Tc ; @@yT2c =
(d)
@ 2 vx
@x2
@ 2 vy
@x2
@ 2 vz
@x2
@ 2 Ta
@y 2
@ 2 Tb
@y 2
=
=
@ 2 Ta
@z 2
@ 2 Tb
@z 2
= 0 ) r2 Ta = 2.
=
Tb ) r2 Tb =
2
3Tb =
3 sin x sin y sin z.
2
9Tc ) r2 Tc = 0.
9
2
2
= 2 ; @@yv2x = @@zv2x = 0 ) r2 vx = 2 >
=
@2v
@2v
r2 v = 2 x̂ + 6x ŷ.
= @y2y = 0 ; @z2y = 6x ) r2 vy = 6x
>
;
@ 2 vz
@ 2 vz
2
= @y2 = @z2 = 0 ) r vz = 0
Problem 1.27
r·(r⇥v) =
⇣ 2
@ vz
= @x
@y
@
@x
⇣
16Tc ; @@zT2c =
@vy
@z
@vz
@y
@ 2 vz
@y @x
⌘
+
From Prob. 1.18: r⇥va =
Problem 1.28
⇣
⌘
@ 2 vx
@y @z
@
@
@
@x @y @z
@t @t @t
@x @y @z
@vx
@z
@
@y
@ 2 vx
@z @y
⌘
⇣
@vy
@vz
@
@x + @z
@x
⇣ 2
⌘
@ vy
@ 2 vy
+ @z @x @x @z
@vx
@y
= x̂
@2t
@y @z
@2t
@z @y
+ ŷ
@2t
@z @x
⌘
= 0, by equality of cross-derivatives.
6xz x̂+2z ŷ+3z 2 ẑ ) r·(r⇥va ) =
x̂ ŷ ẑ
r⇥(rt) =
+
@2t
@x @z
@
@x (
@
@
6xz)+ @y
(2z)+ @z
(3z 2 ) =
+ ẑ
@2t
@x @y
6z+6z = 0.
@2t
@y @x
= 0, by equality of cross-derivatives.
In Prob. 1.11(b), rf = 2xy 3 z 4 x̂ + 3x2 y 2 z 4 ŷ + 4x2 y 3 z 3 ẑ, so
x̂
ŷ
ẑ
@
@
@
r⇥(rf ) =
@x
@y
@z
3 4
2 2 4
2 3 3
2xy z 3x y z 4x y z
= x̂(3 · 4x2 y 2 z 3 4 · 3x2 y 2 z 3 ) + ŷ(4 · 2xy 3 z 3 2 · 4xy 3 z 3 ) + ẑ(2 · 3xy 2 z 4
Problem 1.29
(a) (0, 0, 0)
(1, 0, 0)
(1, 1, 0)
R
Total:
(b) (0, 0, 0)
(0, 0, 1)
(0, 1, 1)
R
Total:
3 · 2xy 2 z 4 ) = 0. X
R
R1
! (1, 0, 0). x : 0 ! 1, y = z = 0; dl = dx x̂; v · dl = x2 dx; v · dl R= 0 x2 dx = (x3 /3)|10 = 1/3.
! (1, 1, 0). x = 1, y : 0 ! 1, z = 0; dl = dy ŷ; v · dl = 2yz dy = 0; v · dl = 0.
R
R1
! (1, 1, 1). x = y = 1, z : 0 ! 1; dl = dz ẑ; v · dl = y 2 dz = dz; v · dl = 0 dz = z|10 = 1.
v · dl = (1/3) + 0 + 1 = 4/3.
R
! (0, 0, 1). x = y = 0, z : 0 ! 1; dl = dz ẑ; v · dl = y 2 dz = 0; v · dl = 0.
R
R1
! (0, 1, 1). x = 0, y : 0 ! 1, z = 1; dl = dy ŷ; v · dl = 2yz dy = 2y dy; v · dl = 0 2y dy = y 2 |10 = 1.
R
R1
! (1, 1, 1). x : 0 ! 1, y = z = 1; dl = dx x̂; v · dl = x2 dx; v · dl = 0 x2 dx = (x3 /3)|10 = 1/3.
v · dl = 0 + 1 + (1/3) = 4/3.
(c) x = y = z : 0 ! 1; dx = dy = dz; v · dl = x2 dx + 2yz dy + y 2 dz = x2 dx + 2x2 dx + x2 dx = 4x2 dx;
R
R1
v · dl = 0 4x2 dx = (4x3 /3)|10 = 4/3.
H
(d) v · dl = (4/3) (4/3) = 0.
protected under all copyright laws as they currently exist. No portion of this material may be
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12
CHAPTER 1. VECTOR ANALYSIS
Problem 1.30
x, y : 0 ! 1, z = 0; da = dx dy ẑ; v · da = y(z 2
3) dx dy =
2
R
3y dx dy; v · da =
3
R2
0
dx
R2
0
y dy =
3(x|20 )( y2 |20 ) = 3(2)(2) =
12. In Ex. 1.7 we got 20, for the same boundary line (the square in the
xy-plane), so the answer is no: the surface integral does not depend only on the boundary line. The total flux
for the cube is 20 + 12 = 32.
Problem 1.31
R
R
T d⌧ = z 2 dx dy dz. You can do the integrals in any order—here it is simplest to save z for last:
Z ✓Z
◆
Z
z2
dx dy dz.
R (1 y z)
The sloping surface is x+y +z = 1, so the x integral is 0
dx = 1 y z. For a given z, y ranges from 0 to
R (1 z)
(1 z)
1 z, so the y integral is 0
(1 y z) dy = [(1 z)y (y 2 /2)]|0
= (1 z)2 [(1 z)2 /2] = (1 z)2 /2 =
R
R 1 z2
2
4
3
1 2 1
z
z4
z5 1
2
(1/2) z + (z /2). Finally, the z integral is 0 z ( 2 z + 2 ) dz = 0 ( 2
z 3 + z2 ) dz = ( z6
4 + 10 )|0 =
1
6
1
4
+
1
10
= 1/60.
Problem 1.32
T (b) = 1 + 4 + 2 = 7; T (a) = 0. ) T (b)
T (a) = 7.
rT = (2x + 4y)x̂ + (4x + 2z 3 )ŷ + (6yz 2 )ẑ; rT ·dl = (2x + 4y)dx + (4x + 2z 3 )dy + (6yz 2 )dz
9
R
R1
1
>
(a) Segment 1: x : 0 ! 1, y = z = dy = dz = 0. rT ·dl = 0 (2x) dx = x2 0 = 1.
=R
R
R1
b
1
rT ·dl = 7. X
Segment 2: y : 0 ! 1, x = 1, z = 0, dx = dz = 0. rT ·dl = 0 (4) dy = 4y|0 = 4.
a
>
R
R1 2
1
;
3
Segment 3: z : 0 ! 1, x = y = 1, dx = dy = 0. rT ·dl = 0 (6z ) dz = 2z 0 = 2.
9
R
R1
>
(b) Segment 1: z : 0 ! 1, x = y = dx = dy = 0. rT ·dl = 0 (0) dz = 0.
>
R
R1
>
1
Segment 2: y : 0 ! 1, x = 0, z = 1, dx = dz = 0. rT ·dl = 0 (2) dy = 2y|0 = 2. = R b
R
R1
rT ·dl = 7. X
a
>
Segment 3: x : 0 ! 1, y = z = 1, dy = dz = 0. rT ·dl = 0 (2x + 4) dx
>
>
1
;
= (x2 + 4x) 0 = 1 + 4 = 5.
(c) x : 0 ! 1, y = x, z = x2 , dy = dx, dz = 2x dx.
rT ·dl = (2x + 4x)dx + (4x + 2x6 )dx + (6xx4 )2x dx = (10x + 14x6 )dx.
Rb
R1
1
rT ·dl = 0 (10x + 14x6 )dx = (5x2 + 2x7 ) 0 = 5 + 2 = 7. X
a
Problem 1.33
r·v = y + 2z + 3x
o
R
R
RR nR 2
(r·v)d⌧ = (y + 2z + 3x) dx dy dz =
(y
+
2z
+
3x)
dx
dy dz
0
⇥
⇤2
(y + 2z)x + 32 x2 0 = 2(y + 2z) + 6
n
o
R R2
=
(2y + 4z + 6)dy dz
0
⇥ 2
⇤2
y + (4z + 6)y 0 = 4 + 2(4z + 6) = 8z + 16
,!
,!
=
R2
0
(8z + 16)dz = (4z 2 + 16z)
2
0
= 16 + 32 = 48.
Numbering the surfaces as in Fig. 1.29:
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13
CHAPTER 1. VECTOR ANALYSIS
R
RR
2
(i) da = dy dz x̂, x = 2. v·da = 2y dyRdz. v·da = 2y dy dz = 2y 2 0 = 8.
(ii) da = dy dz x̂, x = 0. v·da = 0. v·daR= 0.
RR
(iii) da = dx dz ŷ, y = 2. v·da = 4z dx
R dz. v·da = 4z dx dz = 16.
(iv) da = dx dz ŷ, y = 0. v·da = 0. v·da
R = 0.
(v) da = dx dy ẑ, z = 2. v·da = 6x dxRdy. v·da = 24.
(vi)R da = dx dy ẑ, z = 0. v·da = 0. v·da = 0.
) v·da = 8 + 16 + 24 = 48 X
Problem 1.34
r⇥v = x̂(0 2y) + ŷ(0 3z) + ẑ(0 x) = 2y x̂ 3z ŷ x ẑ.
da = dy dz x̂, if we agree that the path integral shall run counterclockwise. So
(r⇥v)·da = 2y dy dz.
o
R
R nR 2 z
(r⇥v)·da =
(
2y)dy
dz
0
z6
2 2 z
2
@
y 0 = (2 z)
⇣
⌘
@
2
R2
3
@
=
(4 4z + z 2 )dz =
4z 2z 2 + z3
0
0
@
8
=
8 8 + 83 =
@
3
@ y
Meanwhile, v·dl = (xy)dx + (2yz)dy + (3zx)dz. There are three segments.
y
,!
=
2
z
z6
@
@@
I (2)
(3)
@@
@
?
@
@ - y
(1)
R
(1) x = z = 0; dx = dz = 0. y : 0 ! 2. v·dl = 0.
(2) x = 0; z = 2 y; dx = 0, dz = dy, y : 2 ! 0. v·dl = 2yz dy.
R
R0
R2
2
v·dl = 2 2y(2 y)dy =
(4y 2y 2 )dy =
2y 2 32 y 3 0 =
8 23 · 8 =
0
R
H
(3) x = y = 0; dx = dy = 0; z : 2 ! 0. v·dl = 0. v·dl = 0. So v·dl = 83 . X
8
3.
Problem 1.35
R
By Corollary 1, (r⇥v)·da should equal 43 . r⇥v = (4z 2
(i) da = dy dz x̂, x = 1; y, z : 0 ! 1. (r⇥v)·da = (4z 2
( 43 z 3
1
0
2x)x̂ + 2z ẑ.
R
R1
2)dy dz; (r⇥v)·da = 0 (4z 2
=
2z) =
2=
R
(ii) da = dx dy ẑ, z = 0; x, y : 0 ! 1. (r⇥v)·da = 0; R (r⇥v)·da = 0.
(iii) da = dx dz ŷ, y = 1; x, z : 0 ! 1. (r⇥v)·da = 0; (r⇥v)·da
= 0.
R
(iv) da = dx dz ŷ, y = 0; x, z : 0 ! 1. (r⇥v)·da = 0; (r⇥v)·da
= 0.
R
(v) da = dx dy ẑ, z = 1; x, y : 0 ! 1. (r⇥v)·da = 2 dx dy; (r⇥v)·da = 2.
R
) (r⇥v)·da = 23 + 2 = 43 . X
4
3
2
3.
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2)dz
14
CHAPTER 1. VECTOR ANALYSIS
Problem 1.36
(a) Use the product rule r⇥(f A) = f (r⇥A) A ⇥ (rf ) :
Z
Z
Z
I
Z
f (r⇥A) · da =
r⇥(f A) · da + [A ⇥ (rf )] · da =
f A · dl + [A ⇥ (rf )] · da. qed
S
S
S
P
S
(I used Stokes’ theorem in the last step.)
(b) Use the product rule r·(A ⇥ B) = B · (r⇥A) A · (r⇥B) :
Z
Z
Z
I
Z
B · (r⇥A)d⌧ =
r·(A ⇥ B) d⌧ +
A · (r⇥B) d⌧ = (A ⇥ B) · da +
A · (r⇥B) d⌧. qed
V
V
V
S
V
(I used the divergence theorem in the last step.)
Problem 1.37 r =
Problem 1.38
p
x2 + y 2 + z 2 ;
✓ = cos
1
✓
p
z
x2 +y 2 +z 2
◆
;
= tan
1
y
x
.
There are many ways to do this one—probably the most illuminating way is to work it out by trigonometry
from Fig. 1.36. The most systematic approach is to study the expression:
r = x x̂ + y ŷ + z ẑ = r sin ✓ cos x̂ + r sin ✓ sin ŷ + r cos ✓ ẑ.
@
If I only vary r slightly, then dr = @r
(r)dr is a short vector pointing in the direction of increase in r. To make
it a unit vector, I must divide by its length. Thus:
r̂ =
@r
@r
@r
@✓
@r
@
= sin ✓ cos x̂ + sin ✓ sin ŷ + cos ✓ ẑ;
= r cos ✓ cos x̂ + r cos ✓ sin ŷ
=
@r
@r
@r
@r
; ✓ˆ =
@r 2
@r
r sin ✓ ẑ;
r sin ✓ sin x̂ + r sin ✓ cos ŷ;
@r 2
@
@r
@✓
@r
@✓
; ˆ=
= sin2 ✓ cos2
@r 2
=
@✓
2
2
@r
@
@r
@
.
+ sin2 ✓ sin2
r2 cos2 ✓ cos2
= r sin ✓ sin
2
+ cos2 ✓ = 1.
+ r2 cos2 ✓ sin2
+ r2 sin ✓ cos2
2
+ r2 sin2 ✓ = r2 .
= r2 sin ✓.
2
r̂ = sin ✓ cos x̂ + sin ✓ sin ŷ + cos ✓ ẑ.
) ✓ˆ = cos ✓ cos x̂ + cos ✓ sin ŷ sin ✓ ẑ.
ˆ = sin x̂ + cos ŷ.
Check: r̂·r̂ = sin2 ✓(cos2 + sin2 ) + cos2 ✓ = sin2 ✓ + cos2 ✓ = 1, X
ˆ ˆ = cos ✓ sin cos + cos ✓ sin cos = 0, X etc.
✓·
sin ✓ r̂ = sin2 ✓ cos x̂ + sin2 ✓ sin ŷ + sin ✓ cos ✓ ẑ.
cos ✓ ✓ˆ = cos2 ✓ cos x̂ + cos2 ✓ sin ŷ sin ✓ cos ✓ ẑ.
(1) sin ✓ r̂ + cos ✓ ✓ˆ = + cos x̂ + sin ŷ;
ˆ = sin x̂ + cos ŷ.
(2)
Multiply (1) by cos , (2) by sin , and subtract:
x̂ = sin ✓ cos r̂ + cos ✓ cos ✓ˆ
sin
ˆ.
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15
CHAPTER 1. VECTOR ANALYSIS
Multiply (1) by sin , (2) by cos , and add:
ŷ = sin ✓ sin r̂ + cos ✓ sin ✓ˆ + cos
ˆ.
cos ✓ r̂ = sin ✓ cos ✓ cos x̂ + sin ✓ cos ✓ sin ŷ + cos2 ✓ ẑ.
sin ✓ ✓ˆ = sin ✓ cos ✓ cos x̂ + sin ✓ cos ✓ sin ŷ sin2 ✓ ẑ.
Subtract these:
ˆ
sin ✓ ✓.
ẑ = cos ✓ r̂
Problem 1.39
@
(a) r·v1 = r12 @r
(r2 r2 ) = r12 4r3 = 4r
⇣ 4⌘
R
R
R
RR
R⇡
R
(r·v1 )d⌧ = (4r)(r2 sin ✓ dr d✓ d ) = (4) 0 r3 dr 0 sin ✓ d✓ 2⇡
(2)(2⇡) = 4⇡R4
0 d = (4)
4
R
R 2
R
R
⇡
2⇡
v1 ·da = (r r̂)·(r2 sin ✓ d✓ d r̂) = r4 0 sin ✓ d✓ 0 d = 4⇡R4 X (Note: at surface of sphere r = R.)
R
@
(b) r·v2 = r12 @r
r2 r12 = 0 ) (r·v2 )d⌧ = 0
R
R 1
R
2
v2 ·da =
r 2 r̂ (r sin ✓ d✓ d r̂) = sin ✓ d✓ d = 4⇡.
They don’t agree!
The point is that this divergence is zero except at the origin, where it blows up, so our
R
calculation of (r·v2 ) is incorrect. The right answer is 4⇡.
Problem 1.40
@
1
@
1
@
r·v = r12 @r
(r2 r cos ✓) + r sin
✓ @✓ (sin ✓ r sin ✓) + r sin ✓ @ (r sin ✓ cos )
1
1
1
= r2 3r2 cos ✓ + r sin ✓ r 2 sin ✓ cos ✓ + r sin ✓ r sin ✓( sin )
= 3 cos ✓ + 2 cos ✓ sin = 5 cos ✓ sin
R
R
RR
R ✓ hR 2⇡
(r·v)d⌧ = (5 cos ✓ sin ) r2 sin ✓ dr d✓ d = 0 r2 dr 02 0 (5 cos ✓
=
=
⇣
R3
3
⌘
(10⇡)
R
⇡
2
0
sin ) d
i
,!2⇡(5 cos ✓)
d✓ sin ✓
sin ✓ cos ✓ d✓
⇡
2
,! sin2 ✓ 0
2
=
1
2
5⇡ 3
3 R .
Two surfaces—one the hemisphere: da = R2 sin ✓ d✓ d r̂; r = R; : 0 ! 2⇡, ✓ : 0 ! ⇡2 .
R
R
R⇡
R 2⇡
v·da = (r cos ✓)R2 sin ✓ d✓ d = R3 02 sin ✓ cos ✓ d✓ 0 d = R3 12 (2⇡) = ⇡R3 .
ˆ = r dr d ✓ˆ (here ✓ = ⇡ ). r : 0 ! R, : 0 ! 2⇡.
other the flat bottom: da = (dr)(r sin ✓ d )(+✓)
2
R
R
R R 2 R 2⇡
R3
v·da
=
(r
sin
✓)(r
dr
d
)
=
r
dr
d
=
2⇡
.
3
0
0
R
Total: v·da = ⇡R3 + 23 ⇡R3 = 53 ⇡R3 . X
Problem 1.41 rt = (cos ✓ + sin ✓ cos )r̂ + ( sin ✓ + cos ✓ cos )✓ˆ +
1
sin
/✓
( sin
/ ✓ sin ) ˆ
r2 t = r·(rt)
@
1
@
1
@
= r12 @r
r2 (cos ✓ + sin ✓ cos ) + r sin
✓ @✓ (sin ✓( sin ✓ + cos ✓ cos )) + r sin ✓ @ ( sin )
2
1
1
1
2
= r2 2r(cos ✓ + sin ✓ cos ) + r sin ✓ ( 2 sin ✓ cos ✓ + cos ✓ cos
sin ✓ cos ) r sin
✓ cos
2
2
1
2
= r sin ✓ [2
sin
✓
cos
✓
+
2
sin
✓
cos
2
sin
✓
cos
✓
+
cos
✓
cos
sin
✓
cos
cos
]
⇥
⇤
2
1
2
= r sin
cos = 0.
✓ (sin ✓ + cos ✓) cos
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16
CHAPTER 1. VECTOR ANALYSIS
) r2 t = 0
Check: r cos ✓ = z, r sin ✓ cos = x ) in Cartesian coordinates t = x + z. Obviously Laplacian is zero.
Rb
Gradient Theorem: a rt·dl = t(b) t(a)
Segment 1: ✓ = ⇡2 , = 0, r : 0 ! 2. dl = dr r̂; rt·dl = (cos ✓ + sin ✓ cos )dr = (0 + 1)dr = dr.
R
R2
rt·dl = 0 dr = 2.
Segment 2: ✓ =
⇡
2,
r = 2,
:0!
⇡
2.
rt·dl = ( sin )(2 d ) =
dl = r sin ✓ d ˆ = 2 d ˆ.
⇡
R
R ⇡2
2 sin d . rt·dl =
2 sin d = 2 cos |02 =
0
2.
Segment 3: r = 2, = ⇡2 ; ✓ : ⇡2 ! 0.
ˆ rt·dl = ( sin ✓ + cos ✓ cos )(2 d✓) = 2 sin ✓ d✓.
dl = r d✓ ✓ˆ = 2 d✓ ✓;
R
R0
0
rt·dl =
⇡ 2 sin ✓ d✓ = 2 cos ✓| ⇡ = 2.
2
2
Rb
Total: a rt·dl = 2 2 + 2 = 2 . Meanwhile, t(b) t(a) = [2(1 + 0)] [0( )] = 2. X
Problem 1.42 From Fig. 1.42, ŝ = cos x̂ + sin ŷ; ˆ =
Multiply first by cos , second by sin , and subtract:
ˆ sin = cos2 x̂ + cos sin ŷ + sin2 x̂
ŝ cos
So x̂ = cos ŝ sin ˆ.
sin x̂ + cos ŷ; ẑ = ẑ
sin cos ŷ = x̂(sin2
+ cos2 ) = x̂.
Multiply first by sin , second by cos , and add:
ŝ sin + ˆ cos = sin cos x̂ + sin2 ŷ sin cos x̂ + cos2 ŷ = ŷ(sin2
So ŷ = sin ŝ + cos ˆ.
ẑ = ẑ.
+ cos2 ) = ŷ.
Problem 1.43
@
@
(a) r·v = 1s @s
s s(2 + sin2 ) + 1s @@ (s sin cos ) + @z
(3z)
2
2
1
1
2
= s 2s(2 + sin ) + s s(cos
sin ) + 3
= 4 + 2 sin2 + cos2
sin2 + 3
2
= 4 + sin + cos2 + 3 = 8.
R
R
R2
R⇡
R5
(b) (r·v)d⌧ = (8)s ds d dz = 8 0 s ds 02 d 0 dz = 8(2) ⇡2 (5) = 40⇡.
Meanwhile, the surface integral has five parts:
R
R2
R⇡
top: z = 5, da = s ds d ẑ; v·da = 3z s ds d = 15s ds d . R v·da = 15 0 s ds 02 d = 15⇡.
bottom: z = 0, da = s ds d ẑ; v·da = 3z s ds d = 0. v·da = 0.
R
back: = ⇡2 , da = ds dz ˆ; v·da = s sin cos ds dz = 0. v·da = 0.
R
left: = 0, da = ds dz ˆ; v·da = s sin cos ds dz = 0. v·da = 0.
front: s = 2, da = s d dz ŝ; v·da = s(2 + sin2 )s d dz = 4(2 + sin2 )d dz.
R
R⇡
R5
v·da = 4 02 (2 + sin2 )d 0 dz = (4)(⇡ + ⇡4 )(5) = 25⇡.
H
So v·da = 15⇡ + 25⇡ = 40⇡. X
⇣
⌘
@
@
@
ˆ
(c) r⇥v = 1s @@ (3z) @z
(s sin cos ) ŝ + @z
s(2 + sin2 )
@s (3z)
⇣
⌘
@
+ 1s @s
(s2 sin cos ) @@ s(2 + sin2 ) ẑ
=
1
s (2s sin
cos
s 2 sin cos ) ẑ = 0.
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17
CHAPTER 1. VECTOR ANALYSIS
Problem 1.44
(a) 3(32 )
2(3)
1 = 27
6
1 = 20.
(b) cos ⇡ = -1.
(c) zero.
(d) ln( 2 + 3) = ln 1 = zero.
Problem 1.45
R2
(a) 2 (2x + 3) 13 (x) dx = 13 (0 + 3) = 1.
(b) By Eq. 1.94, (1
(c)
R1
1
1), so 1 + 3 + 2 = 6.
x) = (x
1 2 1
3
3
9x2 13 (x + 13 ) dx = 9
=
1
3.
(d) 1 (if a > b), 0 (if a < b).
Problem 1.46
⇥ d
⇤
R1
1
(a) 1 f (x) x dx
(x) dx = x f (x) (x)| 1
R1
d
(x f (x)) (x) dx.
1 dx
df
d
dx
dx (x f (x)) = x dx + dx f
The first term is zero, since (x) = 0 at ±1;
⌘
R 1 ⇣ df
So the integral is
x
+
f
(x) dx = 0
dx
1
(b)
So,
R1
d
x dx
(x) =
(x).
f (0) =
= x df + f.
R 1 dx
f (x) (x) dx.
1
f (0) =
qed
1
d✓
f (x) dx
dx = f (x)✓(x)| 1
R1
= f (0) = 1 f (x) (x) dx. So
1
R1
d✓
dx
df
✓(x)dx
1 dx
= f (1)
= (x). qed
R1
0
df
dx dx
= f (1)
(f (1)
f (0))
Problem 1.47
(a) ⇢(r) = q 3 (r
r0 ). Check:
(b) ⇢(r) = q 3 (r
a)
q 3 (r).
(c) Evidently ⇢(r) = A (r
R
R
Q = ⇢ d⌧ = A (r
R
⇢(r)d⌧ = q
R
3
(r
r0 ) d⌧ = q.
X
R). To determine the constant A, we require
R)4⇡r2 dr = A 4⇡R2 .
So A =
Q
4⇡R2 .
⇢(r) =
Q
4⇡R2
(r
R).
Problem 1.48
(a) a2 + a·a + a2 = 3a2 .
R
(b) (r
b)2 513
3
(r) d⌧ =
1 2
125 b
=
1
2
125 (4
+ 32 ) =
1
5.
(c) c2 = 25 + 9 + 4 = 38 > 36 = 62 , so c is outside V, so the integral is zero.
2
2
(d) (e (2 x̂ + 2 ŷ + 2 ẑ)) = (1 x̂ + 0 ŷ + ( 1) ẑ) = 1 + 1 = 2 < (1.5)2 = 2.25, so e is inside V,
and hence the integral is e·(d e) = (3, 2, 1)·( 2, 0, 2) = 6 + 0 + 2 = -4.
Problem 1.49
R
First method: use Eq. 1.99 to write J = e r 4⇡ 3(r) d⌧ = 4⇡e
Second method: integrating by parts (use Eq. 1.59).
0
= 4⇡.
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18
Z
J =
=
Z
V
r̂
· r(e
r2
1
e
r2
= 4⇡
r
e
r
) d⌧ +
4⇡r dr +
2
R
0
r
Problem 1.50 (a) r·F1 =
I
e
r
S
Z
+ 4⇡e
CHAPTER 1. VECTOR ANALYSIS
r
e
R
r̂
· da. But r e
r2
r̂
· r2 sin ✓ d✓ d r̂ = 4⇡
r2
= 4⇡
@
@x (0)
+
R
e
@
@y (0)
+
x̂ ŷ ẑ
r⇥F1 =
@
@
@
@x @y @z
2
0
=
ŷ
0 x
F2 is a gradient; F1 is a curl
U2 =
⇣
⌘
@A
@Az
x
For A1 , we want @zy
= @A
@y
@z
1
2
(b) r·F3 =
= 0;
A1 = 13 x2 ŷ
+e
@
@z
0
x2 = 0 ;
R
r
dr + e
= 4⇡.X
r·F2 =
r
◆
r̂ =
R
Z
e
r
r̂.
sin ✓ d✓ d
Here R = 1, so e
@x
@x
+
@y
@y
+
@z
@z
R
= 0.
=1+1+1= 3
x̂ ŷ ẑ
2xŷ ;
x3 + y 2 + z 2
= 0;
e
@
e
@r
0
+ 4⇡e
@
x2 =
@x
@Az
@x
ZR
✓
=
r
r⇥F2 =
@
@
@
@x @y @z
x
y
= 0
z
would do (F2 = rU2 ).
@Ay
@x
@Ax
@y
= x2 . Ay =
x3
3 ,
Ax = Az = 0 would do it.
(F1 = r⇥A1 ) . (But these are not unique.)
x̂ ŷ ẑ
@
@x (yz)
>
:
@Az
@y
@Ax
@z
@Ay
@x
@
@y (xz)
@Ay
@z
@Az
@x
@Ax
@y
+
@
@z (xy)
@
@
@
@x @y @z
= x̂ (x x) + ŷ (y y) + ẑ (z z) = 0.
yz xz xy
So F3 can be written as the gradient of a scalar (F3 = rU3 ) and as the curl of a vector (F3 = r⇥A3 ). In
fact, U3 = xyz does the job. For the vector potential, we have
8
>
<
+
r⇥F3 =
= yz, which suggests Az = 14 y 2 z + f (x, z); Ay =
= xz, suggesting
Ax = 14 z 2 x + h(x, y); Az =
= xy, so
Ay = 14 x2 y + k(y, z); Ax =
Putting this all together: A3 =
1
4
x z2
y 2 x̂ + y x2
z 2 ŷ + z y 2
x2 ẑ
9
1
2
>
4 yz + g(x, y) =
1
2
4 zx + j(y, z) >
;
1
2
4 xy + l(x, z)
(again, not unique).
Problem 1.51
(d) ) (a): Hr⇥F = r⇥(
rU ) = 0 (Eq. 1.44 – curl of gradient is always zero).
R
(a) ) (c): F · dl = (r⇥F) · da = 0 (Eq. 1.57–Stokes’ theorem).
Rb
Rb
Rb
Ra
H
(c) ) (b): a I F · dl
F · dl = a I F · dl + b II F · dl = F · dl = 0, so
a II
Z
b
a I
F · dl =
Z
b
a II
F · dl.
(b) ) (c): same as (c) ) (b), only in reverse; (c) ) (a): same as (a)) (c).
Problem 1.52
(d) ) (a): Hr·F = r·(r⇥W)
= 0 (Eq 1.46—divergence of curl is always zero).
R
(a) ) (c): F · da = (r·F) d⌧ = 0 (Eq. 1.56—divergence theorem).
protected under all copyright laws as they currently exist. No portion of this material may be
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19
CHAPTER 1. VECTOR ANALYSIS
(c) ) (b):
R
I
F · da
R
II
F · da =
H
F · da = 0, so
Z
Z
F · da =
I
II
F · da.
H
(Note: sign change because for F · da, da is outward, whereas for surface II it is inward.)
(b) ) (c): same as (c) ) (b), in reverse; (c)) (a): same as (a)) (c) .
Problem 1.53
In Prob. 1.15 we found that r·va = 0; in Prob. 1.18 we found that r⇥vc = 0. So
vc can be written as the gradient of a scalar; va can be written as the curl of a vector.
(a) To find t:
(1)
= y 2 ) t = y 2 x + f (y, z)
(2)
@t
@x
@t
@y
= 2xy + z 2
(3)
@t
@z
= 2yz
@f
2
2
@z = 2yz ) f = yz + g(y) ) t = y x
@g
@y = 0. We may as well pick g = 0; then
From (1) & (3) we get
2xy + z (from (2)) )
2
(b) To find W:
@Wy
@z
@Wz
@y
= x2 ;
@Wx
@z
@Wz
@x
= 3z 2 x;
@Wy
@x
+ yz 2 + g(y), so
@t
@y
= 2xy + z 2 +
@g
@y
=
t = xy + yz .
@Wx
@y
2
=
2
2xz.
Pick Wx = 0; then
@Wz
=
@x
@Wy
=
@x
@Wz
@y
W=
Check:
@Wy
@z
=
x2 z ŷ
3xz 2 ) Wz =
2xz ) Wy =
@f
@y
@g
@z
+ x2
3 2 2
2x z
r⇥W =
3 2 2
x z + f (y, z)
2
x2 z + g(y, z).
= x2 )
@f
@y
@g
@z
= 0. May as well pick f = g = 0.
ẑ.
x̂
ŷ
@
@x
@
@y
2
0
x z
ẑ
@
@z
3 2 2
2x z
= x̂ x2 + ŷ 3xz 2 + ẑ ( 2xz).X
You can add any gradient (rt) to W without changing its curl, so this answer is far from unique. Some
other solutions:
W = xz 3 x̂
x2 z ŷ;
W = 2xyz + xz 3 x̂ + x2 y ẑ;
W = xyz x̂
1 2
2x z
ŷ + 12 x2 y
3z 2 ẑ.
protected under all copyright laws as they currently exist. No portion of this material may be
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20
CHAPTER 1. VECTOR ANALYSIS
Problem 1.54
1 @
1
@
r2 r2 cos ✓ +
sin ✓ r2 cos
2
r @r
r sin ✓ @✓
1
1
1
= 2 4r3 cos ✓ +
cos ✓ r2 cos +
r
r sin ✓
r sin ✓
r cos ✓
=
[4 sin ✓ + cos
cos ] = 4r cos ✓.
sin ✓
r·v =
Z
(r·v) d⌧ =
Z
(4r cos ✓)r sin ✓ dr d✓ d = 4
2
0
✓ ◆⇣ ⌘
1
⇡
=
2
2
= R4
ZR
⇡R
4
4
+
1
@
r sin ✓ @
r2 cos ✓ cos
Z⇡/2
Z⇡/2
r dr
cos ✓ sin ✓ d✓
d
3
0
0
.
Surface consists of four parts:
(1) Curved: da = R2 sin ✓ d✓ d r̂; r = R. v · da = R2 cos ✓
Z
.
✓ ◆⇣ ⌘
Z⇡/2
Z⇡/2
1
⇡
⇡R4
v · da = R
cos ✓ sin ✓ d✓
d = R4
=
.
2
2
4
0
r dr d✓ ˆ;
(3) Back: da = r dr d✓ ˆ;
0
= 0. v · da = r2 cos ✓ sin
= ⇡/2. v · da =
Z
v · da =
ZR
0
(r dr d✓) = 0.
r cos ✓ sin
2
Z⇡/2
r dr
cos ✓ d✓ =
3
0
ˆ ✓ = ⇡/2. v · da = r2 cos
(4) Bottom: da = r sin ✓ dr d ✓;
Z
H
R2 sin ✓ d✓ d
4
(2) Left: da =
Total:
r2 cos ✓ sin
v · da = ⇡R4 /4 + 0
1 4
4R
v · da =
+ 14 R4 =
ZR
0
⇡R4
4 .
(r dr d✓) =
✓
R
v · da = 0.
r3 cos ✓ dr d✓.
◆
1 4
R (+1) =
4
1 4
R .
4
(r dr d ) .
Z⇡/2
1
r dr
cos d = R4 .
4
3
0
X
Problem 1.55
x̂ ŷ ẑ
R
@
@
@
r⇥v = @x
a). So
(r⇥v) · da = (b a)⇡R2 .
@y @z = ẑ (b
ay bx 0
v · dl = (ay x̂ + bx ŷ) · (dx x̂ + dy ŷ + dz ẑ) = ay dx + bx dy; x2 + y 2 = R2 ) 2x dx + 2y dy = 0,
so dy = (x/y) dx. So v · dl = ay dx + bx( x/y) dx = y1 ay 2 bx2 dx.
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21
CHAPTER 1. VECTOR ANALYSIS
For the “upper” semicircle, y =
Z
v · dl =
ZR
R
p
R2
aR2 (a + b)x2
p
dx =
R2 x2
⇢
aR2 sin
R
1 2
R (a b) sin
2
1
= ⇡R2 (b a).
2
=
1
a(R2 x2 ) bx2
p
R2 x2
x2 , so v · dl =
(x/R)
=
+R
1
⇣x⌘
(a + b)
R
1 2
R (a
2
dx.
b) sin
1

( 1)
xp 2
R
2
sin
1
x2 +
(+1) =
R2
sin
2
1 2
R (a
2
1
⇣x⌘
R
R
b)
+R
⇣ ⇡
2
⇡⌘
2
And
the same for the lower semicircle (y changes sign, but the limits on the integral are reversed) so
H
v · dl = ⇡R2 (b a). X
Problem 1.56
R
(1) x = z = 0; dx = dz = 0; y : 0 ! 1. v · dl = (yz 2 ) dy = 0;
v · dl = 0.
(2) x = 0; z = 2 2y; dz = 2 dy; y : 1 ! 0. v · dl = (yz 2 ) dy + (3y + z) dz = y(2
Z
v · dl = 2
Z0
(2y 3
4y 2 + y
2) dy = 2
1
✓
4y 3
y2
+
3
2
y4
2
2y
2y)2 dy
◆
0
=
1
(3y + 2
14
.
3
(3) x = y = 0; dx = dy = 0; z : 2 ! 0. v · dl = (3y + z) dz = z dz;
Z
Total:
H
v · dl = 0 +
2=
14
3
v · dl =
Z0
z dz =
z2
2
0
2
=
2.
2
8
3.
H
R
Meanwhile, Stokes’ thereom says v · dl = (r⇥v) · da. Here da = dy dz x̂, so all we need is
@
@
(r⇥v)x = @y
(3y + z) @z
(yz 2 ) = 3 2yz. Therefore
Z
(r⇥v) · da =
=
Z Z
Z
0
=
1
(3
⇥
3(2
2yz) dy dz =
Z
1
0
2y)
y 4 + 83 y 3
2y 12 (2
5y 2 + 6y
Z
2 2y
(3
0
⇤
2y)2 dy =
1
0
=
1+
8
3
2yz) dz dy
Z
1
( 4y 3 + 8y 2
10y + 6) dy
0
5 + 6 = 83 . X
Problem 1.57
Start at the origin.
(1) ✓ =
⇡
2,
= 0; r : 0 ! 1. v · dl = r cos2 ✓ (dr) = 0.
(2) r = 1, ✓ =
⇡
2;
R
v · dl = 0.
: 0 ! ⇡/2. v · dl = (3r)(r sin ✓ d ) = 3 d .
R
v · dl = 3
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⇡/2
R
0
d =
3⇡
2 .
2y)2 dy;
22
(3)
=
⇡
2;
CHAPTER 1. VECTOR ANALYSIS
r sin ✓ = y = 1, so r =
1
sin ✓ ,
dr =
1
sin2 ✓
cos ✓ d✓, ✓ :
⇡
2
! ✓0 ⌘ tan
1
(1/2).
✓
◆
cos2 ✓
cos ✓
cos ✓ sin ✓
v · dl = r cos2 ✓ (dr) (r cos ✓ sin ✓)(r d✓) =
d✓
d✓
sin ✓
sin2 ✓
sin2 ✓
✓ 3
◆
✓
◆
cos ✓ cos ✓
cos ✓ cos2 ✓ + sin2 ✓
cos ✓
=
+
d✓
=
d✓ =
d✓.
sin ✓
sin ✓
sin3 ✓
sin2 ✓
sin3 ✓
Therefore
(4) ✓ = ✓0 ,
Z
=
⇡
2;
Z✓0
v · dl =
cos ✓
1
3 d✓ =
sin ✓
2 sin2 ✓
⇡/2
r:
p
✓0
1
2 · (1/5)
=
⇡/2
1
5
=
2 · (1)
2
1
= 2.
2
5 ! 0. v · dl = r cos2 ✓ (dr) = 45 r dr.
Z
Z0
4
4 r2
v · dl =
r dr =
5p
5 2
5
0
p
4 5
· =
5 2
=
5
2.
Total:
I
v · dl = 0 +
3⇡
+2
2
2=
3⇡
2 .
R
Stokes’ theorem says this should equal (r⇥v) · da


1
@
@
1
1 @
r⇥v =
(sin ✓ 3r)
( r sin ✓ cos ✓) r̂ +
r cos2 ✓
r sin ✓ @✓
@
r sin ✓ @

1 @
@
ˆ
+
( rr cos ✓ sin ✓)
r cos2 ✓
r @r
@✓
1
1
1
=
[3r cos ✓] r̂ + [ 6r] ✓ˆ + [ 2r cos ✓ sin ✓ + 2r cos ✓ sin ✓] ˆ
r sin ✓
r
r
ˆ
= 3 cot ✓ r̂ 6 ✓.
(1) Back face: da =
(2) Bottom: da =
r dr d✓ ˆ; (r⇥v) · da = 0.
R
(r⇥v) · da = 0.
ˆ (r⇥v) · da = 6r sin ✓ dr d . ✓ =
r sin ✓ dr d ✓;
Z
(r⇥v) · da =
Z1
0
(2) Bottom: dz = 0.
x; dz =
⇡
2,
so (r⇥v) · da = 6r dr d
Z⇡/2
1 ⇡
3⇡
6r dr
d =6· · =
. X
2 2
2
0
Problem 1.58
v · dl = y dz.
(1) Left side: z = a
@
(r3r) ✓ˆ
@r
dx; y = 0. Therefore
R
Therefore v · dl = 0.
R
v · dl = 0.
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23
CHAPTER 1. VECTOR ANALYSIS
(3) Back: z = a
1
2 y;
dz =
R
1/2 dy; y : 2a ! 0.
v · dl =
R0
1
2
y
2a
2
1y
2 2
dy =
0
2a
=
4a2
4
R
Meanwhile, r⇥v = x̂, so (r⇥v) · da is the projection of this surface on the x y plane =
1
2
Problem 1.59
= a2 .
· a · 2a = a2 . X
1 @
1
@
1
@
r2 r2 sin ✓ +
sin ✓ 4r2 cos ✓ +
r2 tan ✓
r2 @r
r sin ✓ @✓
r sin ✓ @
1
1
4r
= 2 4r3 sin ✓ +
4r2 cos2 ✓ sin2 ✓ =
sin2 ✓ + cos2 ✓ sin2 ✓
r
r sin ✓
sin ✓
cos2 ✓
= 4r
.
sin ✓
r·v =
Z
Z ✓

Z⇡/6
Z2⇡
✓ sin 2✓
(r·v) d⌧ =
r sin ✓ dr d✓ d = 4r dr
cos2 ✓ d✓ d = R4 (2⇡)
+
2
4
0
0
0
!
p
✓
◆
p
4
⇡
sin 60
⇡R4
3
4
= 2⇡R
+
=
⇡+3
= ⇡R
2⇡ + 3 3 .
12
12
4
6
2
cos2 ✓
4r
sin ✓
◆
ZR
2
⇡/6
3
0
Surface coinsists of two parts:
(1) The ice cream: r = R;
R4 sin2 ✓ d✓ d .
Z
: 0 ! 2⇡; ✓ : 0 ! ⇡/6; da = R2 sin ✓ d✓ d r̂; v·da = R2 sin ✓

Z⇡/6
Z2⇡
1
2
4
4
v·da = R
sin ✓ d✓ d = R (2⇡) ✓
2
0
(2) The cone: ✓ =
0
⇡
6;
R
v · da =
⇡/6
= 2⇡R4
0
: 0 ! 2⇡; r : 0 ! R; da = r sin ✓ d dr ✓ˆ =
Z
Therefore
1
sin 2✓
4
⇡R4
2
⇣
p
✓
3
2 r dr d
⇡
12
R2 sin ✓ d✓ d
1
sin 60
4
ˆ v · da =
✓;
◆
p
⇡R4
=
6
⇡
=
p !
3
3
2
3 r3 dr d
R
2⇡
p
p Z 3 Z
p R4
3 4
v · da = 3 r dr d = 3 ·
· 2⇡ =
⇡R .
4
2
0
⇡
3
p
3
2
+
p ⌘
3 =
0
⇡R4
12
p
2⇡ + 3 3 .
X.
Problem 1.60
H
H
R
R
(a) Corollary 2 says (rT )·dl = 0. Stokes’ theorem says (rT )·dl = [r⇥(rT )]·da. So [r⇥(rT )]·da = 0,
and since this is true for any surface, the integrand must vanish: r⇥(rT ) = 0, confirming Eq. 1.44.
H
H
R
R
(b) Corollary 2 says (r⇥v)·da = 0. Divergence theorem says (r⇥v)·da = r·(r⇥v) d⌧. So r·(r⇥v) d⌧
= 0, and since this is true for any volume, the integrand must vanish: r(r⇥v) = 0, confirming Eq. 1.46.
Problem 1.61
H
R
(a) Divergence theorem: v · da = (r·v) d⌧. Let v = cT , where c is a constant vector. Using product
rule
#5 in front cover:
r·v = r·(cT ) = T (r·c) + c · (rT ). But c is constant so Rr·c = 0. Therefore
we have:
R
R
R
c · (rT ) d⌧ = T c · da. Since c is constant, take it outside the integrals: c · rT d⌧ = c · T da. But c
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24
CHAPTER 1. VECTOR ANALYSIS
is any constant vector—in particular, it could be be x̂, or ŷ, or ẑ—so each component of the integral on left
equals corresponding component on the right, and hence
Z
Z
rT d⌧ = T da.
qed
R
R
(b) Let v ! (v ⇥ c) in divergence theorem. Then r·(v ⇥ c)d⌧ = (v ⇥ c) · da. Product rule #6 )
r·(v ⇥ c) = c · (r⇥v) v · (r⇥c) = c · (r⇥v). (Note: r⇥c = R0, since c is constant.)
R Meanwhile vector
indentity (1) says da · (v ⇥ c) = c · (da ⇥ v) = c · (v ⇥ da). Thus c · (r⇥v) d⌧ =
c · (v ⇥ da). Take c
outside, and again let c be x̂, ŷ, ẑ then:
Z
Z
(r⇥v) d⌧ =
v ⇥ da. qed
R
R
(c) Let v = T rU in divergence theorem: r·(T rU ) d⌧ = T rU · da. Product rule #(5) ) r·(T rU ) =
T r·(rU ) + (rU ) · (rT ) = T r2 U + (rU ) · (rT ). Therefore
Z
Z
2
T r U + (rU ) · (rT ) d⌧ = (T rU ) · da. qed
R
R
(d) Rewrite (c) with T \$ U :
U r2 T + (rT ) · (rU ) d⌧ = (U rT ) · da. Subtract this from (c), noting
that the (rU ) · (rT ) terms cancel:
Z
Z
T r2 U U r2 T d⌧ = (T rU U rT ) · da. qed
R
H
(e) Stokes’ theorem: (r⇥v) · da = v · dl. Let v = cTR. By Product Rule H#(7): r⇥(cT ) = T (r⇥c)
c ⇥ (rT ) = c ⇥ (rT ) (since c is constant). Therefore,
(c ⇥ (rT
R )) · da = T c ·Hdl. Use vector indentity
#1 to rewrite the first term (c ⇥ (rT )) · da = c · (rT ⇥ da). So
c · (rT ⇥ da) = c · T dl. Pull c outside,
and let c ! x̂, ŷ, and ẑ to prove:
Z
I
rT ⇥ da =
T dl. qed
Problem 1.62
(a) da = R2 sin ✓ d✓ d r̂. Let the surface be the northern hemisphere. The x̂ and ŷ components clearly integrate
to zero, and the ẑ component of r̂ is cos ✓, so
Z
Z ⇡/2
sin2 ✓ ⇡/2
2
2
a = R sin ✓ cos ✓ d✓ d ẑ = 2⇡R ẑ
sin ✓ cos ✓ d✓ = 2⇡R2 ẑ
= ⇡R2 ẑ.
2 0
0
H
(b) Let T = 1 in Prob. 1.61(a). Then rT = 0, so da = 0.
qed
(c)
This
follows
from
(b).
For
suppose
a
=
6
a
;
then
if
you
put
them together to make a closed surface,
1
2
H
da = a1 a2 6= 0.
(d) For one such triangle, da = 12 (r ⇥ dl) (since r ⇥ dl is the area ofH the parallelogram, and the direction is
perpendicular to the surface), so for the entire conical surface, a = 12 r ⇥ dl.
(e) Let T = c · r, and use product rule #4: rT = r(c · r) = c ⇥ (r⇥r) + (c · r)r. But r⇥r = 0, and
@
@
@
(c · r)r = (cx @x
+ cy @y
+ cz @z
)(x x̂ + y ŷ + z ẑ) = cx x̂ + cy ŷ + cz ẑ = c. So Prob. 1.61(e) says
I
I
Z
Z
Z
T dl = (c · r) dl =
(rT ) ⇥ da =
c ⇥ da = c ⇥ da = c ⇥ a = a ⇥ c.
qed
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25
CHAPTER 1. VECTOR ANALYSIS
Problem 1.63
(1)
1 @
r·v = 2
r @r
✓
1
r ·
r
2
◆
=
1 @
(r) =
r2 @r
1
r2 .
For a sphere of radius R:
9
R
R 1
R
2
v · da =
>
R r̂ · R sin ✓ d✓ d r̂ = R sin
=
! ✓ d✓ d = 4⇡R.
So divergence
R
R
R 1
R
R
2
(r·v) d⌧ =
r
sin
✓
dr
d✓
d
=
dr
sin
✓
d✓
d
=
4⇡R.
>
2
r
; theorem checks.
0
Evidently there is no delta function at the origin.
r⇥ (rn r̂) =
(except for n =
1 @
1 @
1
r2 rn = 2
rn+2 = 2 (n + 2)rn+1 = (n + 2)rn
r2 @r
r @r
r
1
2, for which we already know (Eq. 1.99) that the divergence is 4⇡ 3(r)).
(2) Geometrically, it should be zero. Likewise, the curl in the spherical coordinates obviously gives zero.
To be certain there is no lurking delta function here, we integrate over a sphere of radius R, using
R
H
?
Prob. 1.61(b): If r⇥(rn r̂) = 0, then (r⇥v) d⌧ = 0 =
v ⇥ da. But v = rn r̂ and da =
2
R sin ✓ d✓ d r̂ are both in the r̂ directions, so v ⇥ da = 0. X
Problem 1.64
(a) Since the argument is not a function of angle, Eq. 1.73 says
 ✓
◆

2r
r3
1 1 d
1
1 d
2
D =
r
=
4⇡ r2 dr
2 (r2 + ✏2 )3/2
4⇡r2 dr (r2 + ✏2 )3/2

2
3
1
3r
3
r 2r
1
3r2
=
=
r2 + ✏2
4⇡r2 (r2 + ✏2 )3/2
2 (r2 + ✏2 )5/3
4⇡r2 (r2 + ✏2 )5/2
(b) Setting r ! 0:
D(0, ✏) =
r2 =
3✏2
.X
4⇡(r2 + ✏2 )5/2
3✏2
3
=
,
5
4⇡✏
4⇡✏3
which goes to infinity as ✏ ! 0. X
(c) From (a) it is clear that D(r, 0) = 0 for r 6= 0. X
(d)
✓
◆
Z
Z 1
r2
1
2
2
2
D(r, ✏) 4⇡r dr = 3✏
dr = 3✏
= 1. X
3✏2
(r2 + ✏2 )5/2
0
(I looked up the integral.) Note that (b), (c), and (d) are the defining conditions for
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
3
(r).
26
CHAPTER 2. ELECTROSTATICS
Chapter 2
Electrostatics
Problem 2.1
(a) Zero.
1 qQ
, where r is the distance from center to each numeral. F points toward the missing q.
4⇡✏0 r2
Explanation: by superposition, this is equivalent to (a), with an extra q at 6 o’clock—since the force of all
twelve is zero, the net force is that of q only.
(b) F =
(c) Zero.
1 qQ
, pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) as
4⇡✏0 r2
a cancellation in pairs of opposite charges (1 o’clock against 7 o’clock; 2 against 8, etc.), with one unpaired q
doing the job, then you’ll need a di↵erent explanation for (d).
(d)
Problem 2.2
This time the “vertical” components cancel, leaving
E=
E=
q
1
4⇡✏0 2
r
2
r
sin ✓ x̂, or
1
qd
4⇡✏0 z 2 + d
2
2 3/2
x̂.
s
q
-E
AAU
A
✓ A
z A
A
As- x
q
1 qd
From far away, (z
d), the field goes like E ⇡ 4⇡✏
3 ẑ, which, as we shall see, is the field of a dipole. (If we
0 z
set d ! 0, we get E = 0, as is appropriate; to the extent that this configuration looks like a single point charge
from far away, the net charge is zero, so E ! 0.)
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 2.3
CHAPTER
2. ELECTROSTATICS
27
1
Problem 2.3
Problem 2.3
z
z ✻
❑✻
❑
θ
η
z dq = λdx
✴L
!x "# \$✲x
θ
r
z
dq = λdx
% L λ dx
1%
2
Ez =1 4πϵ
=2 z 2 +2 x2 ; cos θ z= ηz )
L
2 cos θ; 2(η
R
λ
ηcos
0 dxdx
0
2
2 θ = η) z
EzE = =4πϵ01 0 L
θ;
(η
=
z
2
η
r =+z 2x+; xcos
; cos ✓ = r )
z
4⇡✏0 0 r 2 cos ✓; (
%
L
1 %
1
=1 4πϵ0 λzLR L0 (z12 +x2 )3/2 dx
1λz
1
= =4πϵ
dxdx
2 3/2 '(
z0 &(z2 +x
0
2 ) 2 )3/2 (L
4⇡✏
x '(L
10 & h0 1(z√+x
1 λ√ L
.
=1 4πϵ0 λz1 √z2 x z2 +x(2i L
( =1 4πϵ
L zL
2 +L2
1λ 0√zp
. .
= =4πϵ01λz z z%2 12 zp2 +xx2 ( =0 =4πϵ
2 +L2
0 z z%
z
L
2
2
2
2
4⇡✏0 1%
z λ dx
4⇡✏
0
1
x
dx
z +x θ =
+L
%0λR zx dx
0 −
E = −1 4πϵ0LR L
3/2
λ0 dxdx
1 4πϵ
η 2 sin
(x2x+z
✴L
1
1λ 0
dx2 )
ExEx=x =− 4πϵ
θ=
− 4πϵ
2 sin
2 +z
2 )3/2
&
'(
&
'
✲
sin
✓
=
η
0
0
0
(x
2
L
2
2 )3/2
4⇡✏10 & 0 r
4⇡✏0 1 & (x +z
! "# \$
(
'(
1
1
1' i
L
x
iL
h z−√
= −1 4πϵ0 λh √− √
=
−
.
λ
(
(
1
1
1
1
2
2
2
2
x
4πϵ
x1 +z( =
z1 +L.
1 √p
= =− 4πϵ01λ − xp2 +z
− 4πϵ01λ 0 z −
0=
2
2 +L2
.
z
2
2
2
2
4⇡✏0
z
z +L
)*
+
* 4⇡✏0
+ x, +z 0 0
λ✓
z +◆ *✓
L + ◆,
1 )*
−1 + √z z
x̂ + √L L
ẑ .
E =1 1 λ
√p z 2 + L2x̂ +
√p z 2 + L2ẑ ẑ. .
EE
= = 4πϵ
0 z −1 +
1
+
x̂
+
2
2
2
2
4πϵ
z z+
L L2
z +
L L2
0 z z
2+
2+
4⇡✏
0
1 λL
For z ≫ L you
expect it to
look
like a pointz charge
q = λL: E →1 4πϵ
λL z 2 ẑ. It checks, for with z ≫ L the x̂
1 20 ẑ.
L It checks, for with z ≫ L the x̂
For
z≫
L Lyou
expect
it ittotolook
like
a
point
charge
q
=
λL:
E
→
4πϵ
For
z
you
expect
look
like
a
point
charge
q
=
L:
E
!
L the x̂
1 λL
0 z
2 ẑ. It checks, for with z
4⇡✏
term → 0, and the ẑ term →1 4πϵ
0 z
λ 0Lz z ẑ.
1
L
ẑ.
term
→
0,
and
the
ẑ
term
→
term ! 0, and the ẑ term !4πϵ4⇡✏
ẑ.
0 z z
0 z z
Problem 2.4
Problem
2.4
-q 2 . /. a /2
Problem
2.4 2.1, with L → a and z →
z +a 22 2 (distance from center of edge to P ), field of one edge is:
From Ex.
2
z→
from
center
of of
edge
toto
P ),
field
of of
one
is:is:
z 2z+
From
Ex.
2.1,
with
LL
→!a2 aand
2 +2 a (distance
From
Ex.
2.2,
with
and
z
!
(distance
from
center
edge
P ),
field
oneedge
edge
2
2
1
λa
λa-a
.
E =1 1 -0q 2 .a2.
E1E=1= 4πϵ
a2q 2
a2
+
z
+
z
+
2
2
2
1 4πϵ0
a 4 2
a 24 a 24
4⇡✏0 z 2z+
2 +4 a2 z z+
2 +4 a+
4a
4
4 + 4
z
There are 4 sides, and we want vertical components only, so multiply by 4 cos θ =q4 q
z 2 a2 :
There
are
4
sides,
and
we
want
vertical
components
only,
so
multiply
by
4
cos
θ
=
4
z 2+ :4
z
q
There are 4 sides, and we want vertical components only, so multiply by 4 cos ✓ = 4 z2 + a a2 :
4
z2 +
4
4λaz
1
4λaz
E =1 1 .
ẑ.
/
4
az
2
2
EE
= = 4πϵ
. 0 z 2 a+2 /aq2 z 2 a+2 aẑ.
ẑ.
4πϵ
0 z2 +
4
2
4⇡✏
0 z 2 +4 a2 z z+
2 +2 a2
4
Problem 2.5
Problem 2.5
Problem 2.5
❑
❑
θ
z
θ
z
r
r
2
0%
1
1 θ o ẑ.
10%nR λdl
cos
“Horizontal” components cancel, leaving: E =1 4πϵ
2
1 0 λdl ηcos
θ ✓ẑ. ẑ.
“Horizontal”
components
cancel,
leaving:
= =4πϵ4⇡✏
“Horizontal”
components
cancel,
leaving:E E
2
% dl2 cos
0 0
% η Rrdl = 2πr. So
Here,2 η2 2 =2 r2 2 +2 z2 2 , cos θ z= zηz (both constants), while
Here,
z z, cos
θ ✓==η (both
constants),
while
So So
Here, ηr ==r r++
, cos
constants),
whiledl =dl2πr.
= 2⇡r.
r (both
1
λ(2πr)z
(2⇡r)z
E =1 1 λ(2πr)z
ẑ.
3/2
EE
= = 4⇡✏
ẑ. ẑ.
2 3/2
4πϵ00(r(r22++zz3/2
4πϵ0 (r2 + z 2 )2 ))
r
Problem 2.6
Break it into rings of radius r, and thickness dr, and use Prob. 2.5 to express the field of each ring. Total
charge of a ring is · 2⇡r · dr = · 2⇡r, so = dr is the “line charge” of each ring.
Ering
1 ( dr)2⇡rz
1
=
; Edisk =
2⇡ z
3/2
2
2
4⇡✏0 (r + z )
4⇡✏0
1

1
Z
0
R
r
(r2
1
c
⃝2005
River,
p This material
Edisk
= NJ. All2⇡rights
z reserved.
ẑ. is
2 + z 2 may be
protected under all copyright laws as they currently exist.
4⇡✏0No portionz of thisRmaterial
reproduced, in any form or by any means, without permission in writing from the publisher.
c protected
⃝2005
Pearson
Education,
Inc., laws
Upper
River, NJ.
All No
rights
reserved.
This
material
under
they currently
exist.
portion
of this
material
mayis be
protected
underinall
as means,
they currently
No portion
of this
material
may be
reproduced,
any
form or laws
by any
withoutexist.
permission
in writing
from
the publisher.
reproduced, in any form or by any means, without permission in writing from the publisher.
3/2
+ z2)
dr.
1
CHAPTER 2. ELECTROSTATICS
28
For R
z the second term ! 0, so Eplane =
For z
R,
and E
⇣
⌘
2
p 1
= z1 1 + R
z2
R2 +z 2
1 2⇡R2
1 Q
= 4⇡✏
2z 2 = 4⇡✏0 z 2 ,
0
Contents
1/2
⇡
1
z
⇣
1
4⇡✏0 2⇡
1
1 R2
2 z2
where Q = ⇡R2 .
ẑ =
⌘
2✏0
ẑ.
, so [ ] ⇡
1
z
1
z
+
1 R2
2 z3
=
R2
2z 3 ,
X
Problem 2.7
E is clearly in the z direction. From the diagram,
dq = da = R2 sin ✓ d✓ d ,
r 2 = R2 + z 2 2Rz cos ✓,
cos = z Rrcos ✓ .
z
✻
z
ψ
r
So
θ
R
✲ y
φ
Z
R
1
R2 sin ✓ d✓ d (z R cos ✓)
Ez =
.
d = 2⇡.
4⇡✏0
(R2 + z 2 2Rz cos ✓)3/2
❂
⇢
Z ⇡
x
1
(z R cos ✓) sin ✓
✓ = 0 ) u = +1
=
(2⇡R2 )
d✓.
Let
u
=
cos
✓;
du
=
sin
✓
d✓;
.
2 + z2
3/2
✓
=⇡)u= 1
4⇡✏0
(R
2Rz
cos
✓)
0
Z
1 1 !2 σR21sin θ dθ dφ(z
z Ru
"
− R cos θ)du. Integral
=
can be done by partial fractions—or look it up.
3/2 .
dφ = 2π.
Ez4⇡✏
= 0 (2⇡R ) 21 (R22 + z 2 2Rzu)3/2
4πϵ0
(R
+
z
−
2Rz
cos
θ)
 !
⇢
1
#
\$
1 1 2
1 π zu(z −RR cos θ) sin θ 1 2⇡R2
(z R) ( z R)
θ = 0 ⇒ u = +1
2
p
= = (2⇡R
)
=
.
dθ. z 2 Let |zu = R|
cos θ; du
=R|
− sin θ dθ;
.
2 22Rzu
3/2 0
4⇡✏04πϵ0 (2πR σ)
z 2 0R2(R
|z +
+2z+
θ = π ⇒ u = −1
z − 2Rz cos
1 θ)4⇡✏
! 1
1
z − Ru
=
(2πR2 σ)
Integral can
1 be
q done by partial fractions—or look it up.
q
1 2 4⇡R2
1 du.
2
4πϵ0the sphere),−1E(R
For z > R (outside
= 3/2
z0 −z2Rzu)
2
z =+
4⇡✏
4⇡✏0 z 2 , so E = 4⇡✏ z 2 ẑ.
0
&1
%
#
\$
1 2πR2 σ (z − R) (−z − R)
1
1
zu − R
2
For z < R (inside),
E(2πR
so E
= 0.
√
=
=
−
.
z = 0,σ)
4πϵ0
z 2 R2 + z 2 − 2Rzu −1 4πϵ0 z 2
|z − R|
|z + R|
Problem 2.8
According to Prob. 2.7, all shells interior to the
point (i.e. at smaller 1r) contribute
as though their charge
q
1 q
1 4πR2 σ
ẑ.
= 4πϵ
=
z > R (outside
thecenter,
sphere),
Ez all
= exterior
2 , so E
wereFor
concentrated
at the
while
contribute
nothing.
Therefore:
4πϵ0
z 2 shells
2
0 z
4πϵ z
0
For z < R (inside), Ez = 0, so E = 0.
E(r) =
1 Qint
r̂,
4⇡✏0 r2
Problem 2.8
to Prob.
2.7,interior
all shells
to Outside
the pointthe
(i.e.
at smaller
r) charge
contribute
as though
where QAccording
charge
to interior
the point.
sphere,
all the
is interior,
so their charge
int is the total
were concentrated at the center, while all exterior shells contribute nothing. Therefore:
1 Q
E=
1 r̂.Qint
r̂,
E(r)4⇡✏
=0 r 2
4πϵ0 r2
Inside
the Q
sphere,
only that fraction of the total which is interior to the point counts:
where
int is the total charge interior to the point. Outside the sphere, all the charge is interior, so
Qint =
4
3
3 ⇡r
Q
4
3
3 ⇡R
=
r3
1 r3 1
1 Q
Q, so EE== 1 Q r̂.
Q 2 r̂ =
r.
3
3
3
R
4⇡✏
R
r
4⇡✏
2
0
0 R
4πϵ r
0
Inside 2.9
the sphere, only that fraction of the total which is interior to the point counts:
Problem
@
(a) ⇢ = ✏0 r·E = ✏0 r12 @r
r2 · kr3 = ✏0 r12 k(5r4 ) = 5✏0 kr2 .
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
2012 Pearson
Upper from
reproduced, in any form or by anycmeans,
withoutEducation,
permissionInc.,
in writing
theRiver,
publisher.
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
4
πr43 3 r3 3
1 r3 31
1 Q
r so E =
1 Qr r̂ 1=
1 r.Q
Qint = 43 33πr
Q = 3 Q,
3
2
3
Qint =
Q
=
Q,
so
E
=
Q
r̂
=
r.
R
4πϵ
R
r
4πϵ
0
0R
3
3
2
3
3 πR4 πR
R r3
4πϵ10 Rr3 r 1
4πϵ10 R3Q
4
3 πr3
Qint = 43 3 Q = 3 Q, so E =
Q r̂ =
r.
Problem 2.9
R
4πϵ0 R3 r2
4πϵ0 R3
!
Problem 2.9 1 ∂ 2 ! 3 " 3"πR1
2
(a) ρ = ϵ0 ∇·E = ϵ0 r2 ∂r1 r∂ · kr
= ϵ k(5r4 ) = 5ϵ kr .
(a) ρ = ϵ0 ∇·E = ϵ0 r2 ∂r r2 · kr3 0=r2ϵ0 r12 k(5r4 ) = 0 5ϵ0 kr2 .
Problem 2.9
!#2
"
1 ∂
1 3
4 2 ) = 4πϵ
(b) By
law: =Qϵenc
)(4πR
(a)Gauss’s
ρ = ϵ0 ∇·E
· #kr3 ==ϵ0ϵ(kR
) = 5ϵ kr2 . kR5 .
0 r 2=∂rϵ0 r E·da
0 r 2 k(5r
(b) By Gauss’s law: Qenc = ϵ0\$ E·da =
ϵ0 (kR3 )(4πR2 )0 = 04πϵ0 kR5 .\$
\$
R
CHAPTER
2. ELECTROSTATICS
2
\$ = R (5ϵ
\$ R0 kr2 )(4πr
By
direct integration:
Qenc = ρ#dτ
dr) = 20πϵ0 k 5 0 r\$4Rdr 429
= 4πϵ kR5 .!
0
3kr2 )(4πr
2 2 dr) = 20πϵ
integration:
ρ
dτ
=
(5ϵ
k 0 r dr = 04πϵ0 kR5 .!
0
0
(b)ByBydirect
Gauss’s
law: QencQenc
= ϵ=
E·da
=
ϵ
(kR
)(4πR
)
=
4πϵ
kR
.
0
0
0
\$
\$
\$R 4
Problem 2.10
2
2
5
Problem
2.10integration: QencH = ρ dτ = R (5ϵ
By
direct
kr
)(4πr
dr)
=
20πϵ
k
r dr
= 4πϵ
.!the surface
5
0
0
0 kRup
3
2
0
0
Think
this cube
as one
of
the charge.
of0the
24
which
make
(b)Think
ByofGauss’s
= ✏80ofsurrounding
= ✏0 (kR
)(4⇡R
)Each
= Each
4⇡✏
kR
. squares
enc one
of thislaw:
cubeQas
8E·da
surrounding
the
charge.
of
the
24
squares
which
make
up the surface
Rother
RR 4
R
Problem
2.10
of this
larger
cube
gets the same
fluxRas
every
one,
2 so: 2
5
By larger
direct
integration:
⇢ d⌧
(5✏
kr
)(4⇡r
dr)
=
20⇡✏
k
r
dr
=
4⇡✏
kR
.X
of this
cube
gets theQsame
as=every
other
one,
so:
enc = flux
0
0
0
0
0 squares which make up the surface
Think of this cube as one of 8 surrounding
the charge. Each of the 24
Problem
2.10cube gets the same flux as
of this larger
every
other
one,
so:
%
%
% the 1charge.% Each of the 24 squares which make up the surface
Think of this cube as one of 8 surrounding
1 E·da.
E·da =
=one, so:% E·da.
of this larger cube gets the same flux as every
other
24
%E·da
24
one
whole
1 Z
Z
face one
largewhole
face E·da = 1 large E·da.
cube
E·da = 24cube E·da.
one
%face
24 whole
large
%
one
q whole
q
.cube
The latter is ϵ10 q, 1by Gauss’s law. Therefore faceE·da =
large
= 0cube .
The latter is ϵ0 q, by Gauss’s law. Therefore %E·da24ϵ
24ϵ
one
q0
face one
Z
.
The latter is ϵ10 q, by Gauss’s law. Therefore
face E·da = q
1
The latter is ✏10 q, by Gauss’s law. Therefore oneE·da = 24ϵ0.
24✏0
face
one
Problem 2.11
face
Problem 2.11
1
#
# = E(4πr2 ) =2 1 Qenc
Gaussian surface: Inside: E·da
Problem 2.11
1 = 0 ⇒ E = 0.
ϵ
0
Gaussian surface: Inside: E·da = E(4πr ) = ϵ0 Qenc = 0 ⇒ E = 0.
Problem✙2.11
✙
(As in Prob. 2.7.)
2
#
(As in Prob. 2.7.)
✲
2
2= 0.
1
2 1 Qenc = 0σR
⇒ σR
Gaussiansurface:
surface:Outside:
Inside: E(4πr
E·da2=
E(4πr
E
)=
r̂.
r ✲ ✲ Gaussian
1 ) = ϵ)0 ⇒
2 E=
2 ϵ0 (σ4πR
✲
2
Gaussian surface: Outside: E(4πr ) = ϵ0 (σ4πR ) ⇒ Eϵ=
r̂.
r
r ✙
0
ϵ0σR
r2 2
(As in Prob. 2.7.)
✲
1
2
2
✲
Gaussian surface: Outside: E(4πr ) = ϵ0 (σ4πR ) ⇒ E =
r̂.
r
ϵ0 r 2
Contents
Contents
Gaussian surface
Problem 2.12
Problem 2.12
Problem 2.12
Problem 2.12
}}
}
1 4 1 34
#H = E · 4πr2 = 221 Qenc
E·da
πr ρ. 3 So
E·da =
=E
E ·· 4πr
4⇡r ϵ0=
= ✏110Q
Q=
encϵ0=3 1 4 ⇡r3 ⇢. So
E·da
ϵ0 enc = ϵ✏00 33πr ρ. So
#
Gaussian surface
21 1
E·da = E ·E4πr
= ρrr̂.
1Qenc = ϵ10 43 πr3 ρ. So
r ✙
ϵ1
=E =
0
✒
r✒ ✙
⇢rr̂.
E
=
3ϵ0 3✏0ρrr̂.
Gaussian surface
3ϵ01
r ✙
Q
= 1 ρrr̂.
✒
Q in Prob. 2.8).
Problem
3E=
πR42⇡R
ρ, 2E
r̂ Q(as
Since
QtotQ=tot43 =
31
3ϵ
Since
⇢, E4πϵ
E
r̂ in
(asProb.
in Prob.
R 2.13
0 R14⇡✏
0
ρ,
==
r̂3 (as
2.8).2.8).
Since
Qtot = 433πR
0 3R
4πϵ0 R
❄R
%
❄
Gaussian surface
Q
4
E 2· 2πs · l = 1ϵ100 Q
= ϵ1in
λl.
So 2.8).
Prob.
Since QE·da
tot = =
3 r̂ (as
R
3 πR ρ, E = 4πϵ
Renc
0
Problem
2.13
Problem
2.13
Problem
2.13
✠
❄
Problem
2.13
✻
s
# %H
Gaussian
surface
Gaussian
surface
λ= E · 2⇡s · l1= 1 1 Qenc1= 1 1 l. So
#E·da
E·da
Gaussian surface
Problem 2.13
E·da
=E
·E2πs
·l=
=
So So
ŝ· 2πs
E
=
(same
2.1).
==
· l =asQEx.
= λl.✏10λl.
✏10Qenc
enc
! "# \$
E·da
2πϵ0 sE · 2πs · lϵ0= ϵ0ϵ0 Qencϵ0= ϵ0ϵ0 λl. So
✠
✠
✻
sl ✻
s✻
#
✠
Gaussian
surface
s
1
E·da
Eŝ · (same
2πsas· lEx.
= Eq.
Qenc
= ϵ10 λl. So
E
=λ λλ
as
2.9).
ϵ2.1).
0
E=
ŝ=s(same
ŝ
E
=
(same
as
Ex.
2.1).
2⇡✏
E
=
(same
as
Ex.
2.1).
ŝ
0
✠
& ! '( "# ) ✻
2πϵ2πϵ
\$
0s 0s
& '( s )
2πϵ0λs
l l
l
E=
ŝ (same as Ex. 2.1).
Problem 2.14
Problem
2.14
2πϵ0 s
Problem
Problem2.14
2.14& '( )
l
H
R
R This2 material is
Problem 2.14
c
⃝2005
Pearson Education,
1 River, NJ.
=E
· 4⇡r
= currently
QencRiver,
=
⇢ portion
d⌧
= ✏1of
(kr̄)(r̄ This
sin ✓material
dr̄ d✓ dis )
c
⃝2005
Pearson
Education,
Inc.,
All rights
reserved.
protected
under
all E·da
laws
as Upper
they
exist.
No
✏0 NJ.
0
0 this material may be
Rasr they
protected
currently
Nowriting
portion
of this
may be
4
Problem 2.14
reproduced,
in under
any form
or by1anylaws
means,
without
permission
in
from
the material
publisher.
4⇡k rexist.
⇡k
3
4
= or
kInc.,
4⇡
= ✏0River,
rinrights
. writing
in anyEducation,
form
any Upper
means,
without
permission
from the
publisher.
Gaussian reproduced,
surface
c
⃝2005
Pearson
NJ.✏0All
reserved.
This
material is
✏0 by
4 =
0
protected under all copyright laws as they currently exist. No portion of this material may be
r ✙
✒
reproduced, in any form
means, without permission in writing from the publisher.
1 or by any
2
)
E
=
⇡kr
r̂.
Gaussian surface
4⇡✏0
r ✙
✒
#
Problem 2.15
c 2012
Education,
(i) QPearson
so E =Inc.,
enc = 0,
protected under all copyright laws as they currently exist. No portion of this material may be
Problem
% 2.15
& permission in& writing from the publisher.
reproduced,
in any form or by any means, without
(ii) E·da = E(4πr2 ) = ϵ10 Qenc = ϵ10 ρ dτ = ϵ10 r̄k2 r̄2 sin θ dr̄ dθ dφ
(i) Qenc = 0, so E = 0.
'
(
&r
k r−a
4πk
4πk
&r̂.k 2
=% ϵ0 a dr̄ = ϵ02 (r −1a) ∴ E =1 &
2 1
|E|
(ii) E·da
= E(4πr ) = ϵ0 Qenc = ϵ0 ϵ0ρ dτ r=
dφ
ϵ0
r̄ 2 r̄ sin θ dr̄ dθ ✻
'
&
k r−a
&
4πk 2r
4πk4πkb(r − a)4πk
∴ E = so 2
(iii)=E(4πr
==
ϵ0
a) dr̄
ϵ0 ϵ0a dr̄ = ϵ0 (b − a),
ϵ0
r
'
(
k b−a
& r̂.
E= 2
b
2
dr̄ =
(iii) E(4πr
ϵ0 ) =r4πk
ϵ0
a
'
(
4πk
ϵ0 (b
− a), so
(
r̂.
|E|
✻
✲
Gaussian surface
r ✙
✒
111
Contents
30
CHAPTER 2. ELECTROSTATICS
Problem 2.15
Qenc = 0, so E = 0.
Problem(i)2.15
&
&
(i) Qenc =%0, so E = 0. 2
) = ϵ10 QencR= ϵ10 ρ dτR= ϵ10 r̄k2 r̄2 sin θ dr̄ dθ dφ
H (ii) E·da = E(4πr
1
1
2
2
'1 k (
Problem
2.13
(ii) E·da
= E(4⇡r
✏0 Qenc = ✏0 ⇢ d⌧ =
& r ) =4πk
k ✏0◆r −r̄2ar̄ sin ✓ dr̄ d✓ d
4πk
✓
Problem
r̂.
dr̄ = ϵ0 (r − a) ∴ kE =r a 2
=
R r ϵ0 a2.16
|E|
ϵ0
rr̂.
4⇡k
✻
= 4⇡k
dr̄
=
(r
a)
)
E
=
✏0
✏0
a
2
✏0
r
&
4πk
4πk b
dr̄
=
(b
−
a),
so
(iii) E(4πr2 ) =
R
ϵ0
a
b ϵ0
' a dr̄ (
(iii) E(4⇡r2 ) = 4⇡k
a), so
= 4⇡k
✏0
✏0 (b
k b−a
!
E✓=b a ◆ 2
r̂.
Problem
Problem
2.13
✛
E·da = E · 2πs · l = ϵ10 Qenc = ϵ10 ρπs2 l;
Problem2.13
2.13k(i)
Gaussian surface
ϵ0
r
E=
r̂.
2
✲
Problem
2.16
Problem
Problem 2.16
2.16
✏0
r
ρs
l
r
b
ŝ. a
E=
2ϵ0
Problem
2.16 2.16
Problem
H
✛
E·da
= E · 2⇡s · l = 1 Q = ✏10 ⇢⇡s2 l;
Gaussian
c
⃝2005
Pearson Education, Inc., Upper
NJ. All!
reserved. This material is✏0 enc
!rights
!surface
222
laws
they currently
exist. No portion
of
may
(i)
Q
l;
E·da
E
2πs
s asGaussian
(i)
Q
=
ρπs
l;l;
✛
E·da
=
EE·!·⇢s
2πs
=
(i)
Qenc
= ϵϵ110ϵ010ρπs
ρπs
✛
E·da=
=this
·material
2πs···lll=
= ϵϵ110ϵ010be
(i) protected(ii)
1
1
enc
2
enc=
surface
Gaussian
surface
Gaussian
surface
E·da
=E
· 2πs
·l =
E=
ŝ.publisher.
reproduced, in any form or by any means, without permission in writing
from the
ϵ0 Qenc = ϵ0 ρπa l;
ρs
ρs
ρs
2✏
0
lll
2
ŝ.
E
ŝ.
E
=
ŝ.
E=
=
2ϵ
2ϵ
2ϵ000 E = ρa ŝ.
2ϵ0 s
l
Contents
Contents
(ii)
(ii)
(ii)
(ii)
✻
✻
✻
sss
(iii)
lll
✛
H
✛
✛ ✻Gaussian
Gaussian
surface
Gaussiansurface
surface
E·da = E · 2⇡s ·1l = ✏10 Qenc = ✏10 ⇢⇡a2 l;
✛ Gaussian!!!surface
s
222
E·da
·!··2πs
l;
E·da
=
EE⇢a
2πs
=
Q
=
ρπa
l;l;
E·da=
=E
2πs
= ϵϵ110ϵ010Q
Qenc
= ϵϵ110ϵ010ρπa
ρπa
2 ···lll=
enc
enc=
1
Q
E·da
=
E
·
2πs
·
l
=
E = 222 ŝ.
ϵ0 enc = 0;
ρa
ρa
ρa
2✏0 s
ŝ.
E
ŝ.ŝ.E = 0.
E
=
E=
=
2ϵ
2ϵ
2ϵ000sss
l
✛
✛
✛ Gaussian
Gaussian
surface
Gaussiansurface
surface
✻
✻
✻
sss
!!!H
1
=
E·da
···ll·l=
E·da=
·2πs
2⇡s
l==ϵϵ110ϵ01✏Q
Q
=
0;
E·da
=
EEE···2πs
2πs
=
QQ
==0;
0;0;
E·da
==E
enc
enc
enc
enc
00
E
=
0.
E
=
0.
EE
==
0.0.
(iii)
(iii)
(iii)
(iii)
Contents
|E| ✻
lll
Problem 2.17
✲
|E|
|E|
|E|a✻
On the x z plane E = 0 by symmetry. Set up
Gaussian “pillbox” with
and the
a one face
b in this plane
s
✻
✻
other at y.
c2.17 Pearson
Upper
River, NJ.
Allup
rights
reserved. This
material with
is
Problem⃝2005
On theEducation,
x z planeInc.,
E=
symmetry.
Set
a Gaussian
“pillbox”
one face in this plane
protected under all copyright laws as they currently exist. No portion of this material may be
and the other
at
y.
reproduced, in any form or by any means, without permission in writing from the publisher.
Gaussian pillbox
R
✲
✲
1
1
✲
E·da = aa
E
a · A = ✏0bbbQenc = ✏0ssAy⇢;
s
⇢
E=
y ŷisisis(for |y| < d).
cc⃝2005
⃝2005
Pearson
material
⃝2005
Pearson
Education,
Inc.,
Upper
River,
NJ.
All
rights
reserved.
This
material
c
PearsonEducation,
Education,Inc.,
Inc.,Upper
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
This
material
0maybe
protected
✲
protected
under
all
laws
as
they
currently
exist.
No
portion
of
this
material
may
be
protectedunder
underall
lawsas
asthey
theycurrently
currentlyexist.
exist.No
Noportion
portionof
ofthis
thismaterial
material✏may
be
y ✛ laws
❂
reproduced,
reproduced,
in
any
form
or
by
any
means,
without
permission
in
writing
from
the
publisher.
reproduced,in
inany
anyform
formor
orby
byany
anymeans,
means,without
withoutpermission
permissionin
inwriting
writingfrom
fromthe
thepublisher.
publisher.
Qenc =
1
⇒ E=
ρ
d ŷ c(for
2012yPearson
ϵ0
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
E✻
ρd
ϵ0
−d
d
✲
y
Problem 2.17
On the x z plane E = 0 by symmetry. Set up a Gaussian “pillbox” with one face in this plane and the
other at y.
!
Gaussian pillbox
❂
CHAPTER 2. ELECTROSTATICS
E·da = E · A = ϵ10 Qenc = ϵ10 Ayρ;
ρ
E = y ŷ (for |y| < d).
ϵ0
31
✲ y✛
ρ
Qenc =1 ϵ10 Adρ ⇒ E =⇢ d ŷ (for y > d).
Qenc = ✏0 Ad⇢ ) E = ϵd0 ŷ (for y > d).
✏0
E✻
ρd
ϵ0
−d
d
✲
y
Problem 2.18
Problem 2.18
From Prob. 2.12, the field inside the positive sphere is E+ = 3✏⇢0ρr+ , where r+ is the vector from the positive
From Prob. 2.12, the field inside the positive sphere is E+ = 0 r+ , where r⇢+ is the vector from the positive
center to the point in question. Likewise, the field of the negative3ϵsphere
is 3✏0ρr . So the total field is
center to the point in question. Likewise, the field of the negative sphere is −
3ϵ0 r− . So the total field is
⇢
EE== ρ(r(r
r r) )
r
+ −
+
−
✛ r−*r
3✏3ϵ
0
✯−
0
r+r+ ✕
dd
⇢ρ
r
But
But(see
(seediagram)
diagram)r+r+ −r r−==d.d.SoSoEE==3✏0 d.d.
+
+
3ϵ0
Problem
Problem2.19
2.19
✓
◆
Z"
Z" #
\$ %&
r̂η̂
11
11
η̂r̂
r⇥E
=
r⇥
⇢
d⌧
=
r⇥
(sinceρ⇢depends
dependson
onr′r,0 ,not
notr)r)
∇× r 22 ρ dτ =
∇× r2 2 ρ ⇢dτd⌧ (since
∇×E =
4⇡✏
4⇡✏
0 0
4πϵ
η
4πϵ00
η
✓\$ %
◆
r̂η̂
==0 0 (since
0, from
from Prob.
Prob. 1.62).
1.63).
(sincer⇥
∇× 22 =
rη = 0,
Problem
Problem2.20
2.20
'
'
' x̂ ŷ ŷ ẑ ẑ '
x̂
'@ ∂ @ ∂ @ ∂ '
'
' =k k[x̂(0
(1)
==k [email protected]
(1)r⇥E
∇×E
[x̂(0 −2y)
2y)++ŷ(0
ŷ(0 −3z)
3z)++ẑ(0
ẑ(0 −x)]
x)]6≠=0,0,
1 1
' ∂x @y∂y @z∂z =
'
' xy2yz
xy
2yz3zx
3zx'
sosoEE
ananimpossible
impossibleelectrostatic
electrostaticfield.
field.
1 1is is
'
'
' x̂
x̂
ŷ ŷ
ẑ ẑ ''
'@ ∂
@∂
@∂ '
'
[x̂(2z −2z)
2z)++ŷ(0
ŷ(0 −0)0)++ẑ(2y
ẑ(2y −2y)]
2y)]==0,0,
(2)r⇥E
∇×E
(2)
==k [email protected]
[x̂(2z
2 2
' =k k
' ∂x @y∂y @z∂z =
2
2
'
2y 2xy + 2z 2yz '
y 2xy + z 2yz
possibleelectrostatic
electrostaticfield.
field.
sosoEE
a apossible
2 2is is
z
Let’s go by the indicated path:
E·dl = (y 2 dx + (2xy +
6
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
any form or by any means, without permission in writing from the publisher.
2
r(x0 , y0 , z0 )
zreproduced,
)dy + 2yzindz)k
6
Step I: y = z = 0; dy = dz = 0. E·dl = ky 2 dx = 0.
Step II: x = x0 , y : 0 ! y0 , z = 0. dx = dz = 0.
E·dl
= k(2xy + Rz 2 )dy = 2kx0 y dy.
R
y
E·dl
= 2kx0 0 0 y dy = kx0 y02 .
II
Step III : x = x0 , y = y0 , z : 0 ! z0 ; dx = dy = 0.
E·dl = 2kyz dz = 2ky0 z dz.
x
I⇢⇢
⇢
=
⇢⇢
=
⇢
II
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
III
-
-y
32
R
III
E·dl = 2y0 k
V (x0 , y0 , z0 ) =
(x0 ,y
R0 ,z0 )
R z0
CHAPTER 2. ELECTROSTATICS
z dz = ky0 z02 .
0
E·dl =
k(x0 y02 + y0 z02 ), or V (x, y, z) =
0
k(xy 2 + yz 2 ).
@
@
@
rV =k[ @x
(xy 2 +yz 2 ) x̂+ @y
(xy 2 +yz 2 ) ŷ+ @z
(xy 2 +yz 2 ) ẑ]=k[y 2 x̂+(2xy+z 2 ) ŷ+2yz ẑ]=E. X
Check :
Problem 2.21
Rr
V (r) =
E·dl.
1
8
<Outside the sphere (r > R) : E =
:
Inside the sphere (r < R) :
Rr ⇣
So for r > R: V (r) =
1
RR⇣
and for r < R: V (r) =
=
1 q
4⇡✏0 r̄ 2
⌘
1 q
4⇡✏0 r̄ 2
1
✓
q 1
4⇡✏0 2R
When r > R, rV =
q
@
4⇡✏0 @r
When r < R, rV =
q
1 @
4⇡✏0 2R @r
V(r)
1
4⇡✏0 q
⌘
Rr ⇣
dr̄
◆
2
r
R2
3
r̂ =
⇣
3
1
r
dr̄ =
R
.
q
1
4⇡✏0 r 2 r̂,
⌘
2
r
R2
r̂ =
1
r̄
E=
r
1
so E =
q
1
4⇡✏0 2R
⌘
q
1
4⇡✏0 R3 rr̂.
q 1
,
4⇡✏0 r
=
q
1
4⇡✏0 R3 r̄
1 q
4⇡✏0 r 2 r̂.
dr̄ =
rV =
2r
R2
q
4⇡✏0
h
1
R
q
1
4⇡✏0 r 2 r̂.
1
R3
r 2 R2
2
⌘i
X
q
r
4⇡✏0 R3 r̂;
r̂ =
⇣
so E =
rV =
q
1
4⇡✏0 R3 rr̂.X
1.6
1.4
1.2
(In the figure, r is in units of R, and V (r) is in units of q/4⇡✏0 R.)
1
0.8
0.6
0.4
0.2
0.5
1
1.5
2
2.5
3
r
Problem 2.22
1 2
E = 4⇡✏
ŝ (Prob. 2.13). In this case we cannot set the reference point at 1, since the charge itself
0 s
extends to 1. Let’s set it at s = a. Then
⇣s⌘
Rs⇣ 1 2 ⌘
1
V (s) =
ds̄
=
2
ln
.
4⇡✏0 s̄
a
4⇡✏0
a
(In this form it is clear why a = 1 would be no good—likewise the other “natural” point, a = 0.)
rV =
1
4⇡✏0 2
@
@s
ln
Problem 2.23
R0
V (0) =
E·dl =
1
=
k
✏0
1
a
b
ln
a
b
s
a
ŝ =
Rb
1
4⇡✏0 2
k (b a)
1 ✏0 r 2
1+
a
b
1
s ŝ
dr
=
Ra
b
✓ ◆
k
b
=
ln
.
✏0
a
E. X
k (r a)
✏0 r 2
Problem 2.24
Using Eq. 2.22 and the fields from Prob. 2.16:
Rb
Ra
Rb
V (b) V (0) =
E·dl =
E·dl
E·dl =
0
0
a
⇢
2✏0
dr
Ra
0
R0
a
s ds
(0)dr =
⇢a2
2✏0
Rb
k (b a)
✏0
b
k
✏0
ln
a
b
+a
1
a
1
ds
a s
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
1
b
Contents
33
CHAPTER 2. ELECTROSTATICS
=
⇣
⇢
2✏0
Problem
Problem2.25
2.25
⌘
s2
2
a
0
+
⇢a2
2✏0
b
ln s|a =
⇢a2
4✏0
✓
1 + 2 ln
✓ ◆◆
b
.
a
11
2q
2q
q
..
!
4⇡✏
4πϵ00 z 22+ "dd #22
z + 22
p
R
L
L
1
√z 2 + x2 ) %L
\$ LL p 2λdx
(b) V = 4⇡✏
10
dx2 = 4⇡✏λ0 ln(x +
2 + x2 )% L
z +x
√
(b) V = 4πϵ
z
=
ln(x
+
4πϵ0
−L z 2 +x2
−L
0
"
#
!
p
p
'
&
√
2
2
L + z 2+ L 2
z2 +
( L+
) L2
√
λ
L
+
z
+
L
p
=
ln
= λ ln L+ z2 +L2
.
= 4⇡✏0 ln L + √z 2 + L2
.
= 2⇡✏
2πϵ00 ln
z z
4πϵ0
−L + z 2 + L2
(a)
(a) VV ==
)
%R
√
\$ R σ 2πr dr
σ (* ⇣2
Z
1
2p
2 )% = R
2−z . ⌘
R√
p
(c) V =
r
+
z
R
=
2πσ
(
+
z
2⇡r
dr
1
2
2
4πϵ0
0
pr +z
(c) V =
=
2⇡ ( r2 +0z 2 ) 2ϵ
=0
R2 + z 2 z .
4⇡✏0 0
4⇡✏0
2✏0
0
r2 + z 2
∂V
In each case, by symmetry ∂V
∴ E = − ∂V
∂y = ∂x = 0.
∂z ẑ.
@V
In each case, by symmetry @V
=
=
0.
)
E = @V
@y
@x
@z ẑ.
z
r
x
1
10
4πϵ
" #
2qz
1
1
2z
”3/2 ẑ =
(a) E = − 4πϵ
2q ✓− 21 ◆“
"
#3/2 ẑ (agrees with Prob. 2.2a).
0
d 2
2
4πϵ0 z12 + " d #22qz
1
1 z +( 2 ) 2z
2
(a) E =
2q
ẑ (agrees with Ex. 2.1).
⌘3/2 ẑ = 4⇡✏
2 3/2
4⇡✏0 +
2 ⇣ 2
0 z2 + d
d 2
,
z
+
2
1 √ 21
1√ 1
λ
√1
√1
2z
−
2z
ẑ
(b) E = − 4πϵ
2
2 2
0
z 2 +L2
(−L+ z 2 +L2 ) 2 z 2 +L2
⇢ (L+ z +L )
1+
1
1
1
1
,
√ 1
p2 −L−√z2 +L
p √ 1 p ẑ (agrees
(b) E = = − λ √ z p −L+ z2 +L
2z 2 ẑ = 2Lλ
2z with
ẑ Ex. 2.1).
2
2
2
2
2
2
2
2
2
2
4⇡✏4πϵ
2
2
+L
( L +4πϵz0 z+ zL2 +
) L2 z + L2
0 0 (L
z 2+
+L2 z + L )(z +Lz )−L
(
)
p
p
z
L + z 2 +-L2 L
z2 +
2L
1
. L2
,
+
p
= σ 1p 1
ẑ =
ẑ (agrees with Ex. 2.2).
σ2
z2
2
2
2
2
(z +1L−)√ L
4⇡✏0 zProb.
z +2.6).
L2
0 √ z 2 +2L
ẑ (agrees with
(c) E = − 2ϵ4⇡✏
2 R +z 2z − 1 ẑ = 2ϵ
0
R2 + z 2
0
⇢

1
1
z
the right-hand
in (a)1 is ẑ−q,
0 , which, naively,
suggests
E = −∇V
p charge 2z
(c) EIf =
= then V
1 =p
ẑ (agrees
with Prob.
2
2
2
2✏0 2 toRProb.
2✏0 is that R
+ z 2.2b. The point
z 2 know V on the z axis, and from this we cannot
we +
only
∂V
hope to compute Ex = − ∂V
∂x or Ey = − ∂y . That was OK in part (a), because we knew from symmetry that
If
the
right-hand
charge
in
(a)
is
q,
then
V = 0 ,sowhich,
naively,
E =is insuﬃcient
rV = 0, in
Ex = Ey = 0. But now E points in the x direction,
knowing
V onsuggests
the z axis
determine E. 2
with the answer to Prob. 2.2. The point is that we only know V on the z axis, and from this we cannot hope
@V
to compute Ex = @V
@x or Ey =
@y . That was OK in part (a), because we knew from symmetry that
Ex = Ey = 0. But now E points in the x direction, so knowing V on the z axis is insufficient to determine E.
Problem 2.26
✻
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
b
h
1
V (a) =
4⇡✏0
Z
0
p
2h
✓
2⇡r
r
(where r =
r
◆
2⇡ 1 p
h
p ( 2h) =
dr =
4⇡✏0 2
2✏0
p
/ 2)
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
h
r̄
r
a
✲
34
1
V (b) =
4⇡✏0
Z
p
0
2⇡ 1
p
4⇡✏0 2
=
=
p
2 2✏0
2h
Z
✓
p
2⇡r
r̄
2h
0
q
h2 +
CHAPTER 2. ELECTROSTATICS
◆
dr
p
h2 +
r
2
r
p
(where
r
2
2h r
p
r̄
2h r
=
q
h2 +
r
p
2
2h r )
dr
q
h
+ p ln(2 h2 +
2
r
p
2
2h r + 2 r
p
p
2h
2h)
0

p
p
p
h
h
= p
h + p ln(2h + 2 2h
2h) h p ln(2h
2h)
2 2✏0
2
2
p !
p
p i
h h
h
2+ 2
h
p ln(2h + 2h) ln(2h
p
= p
2h) =
ln
=
ln
4✏0
4✏0
2 2✏0 2
2
2
p
p i
h
hh
=
ln(1 + 2). ) V (a) V (b) =
1 ln(1 + 2) .
2✏0
2✏0
(2 +
p
2)2
2
!
3
Problem
2.27
Problem
2.27
Cut the Cut
cylinder
into slabs,
shownasinshown
the figure,
the cylinder
intoasslabs,
in theand
figure, and
use result
of result
Prob. of
2.25c,
z !with
x and
⇢ dx:
use
Prob.with
2.25c,
z → x!and
σ → ρ dx:
V =
=
! "# \$
z+L/2
R
pz+L/2
% &√
'
ρ R2 + x2 2x dx2
R + x − x dx
2ϵ0
z L/2
#
z−L/2
p
⇥ p
⇤ z+L/2 )*z+L/2
√
( √
⇢ 1
21 +
2 +2R2 ln(x
ρ
2 R2 + x2 ) 2 x2 2
2 *
2
x
R
x
+
x z−L/2
R + x z) −
2✏0 2 = 2ϵ0 2 x R + x + R ln(x +
L/2
8
9 3
r
2
3
9
8
r
2
q
q
L 2
<
=
z+ L + R2 +(z+
q
q
L +2 )R2 + z+ L 2
=
< 2
2
⇢
z+
2
L
L 2
L 2
L 2
2
(
4 2
5 2zL2 ) . 5
r
2
ρ
z+
R
+
z+
z
R
+
z
+R
ln
2
L
L
L
L
(
)
(
)
(
)
(
)
2
2
4
r2
4✏0 : = 2
2 R +(z+ 2 ) −(z− ) 2 R +(z− ) +R ln
−2zL .
4ϵ0 :(z+ 2 )
2
2
2 z L + R2 + z L
2
;
( L +2 )R2 +(z− L;
2
z−
)
⇢
2✏0
=
z− L
2
V =
2
(Note:
2
2
z + L2− &z++zL '2L2+ &z=− Lz 2'2
(Note:
2
2
L2
4
=zL
−z 2 −
+ z−
zL
2
L2
L zL +2 4 =
+
z
− zL
4
2
L
"#
!
\$
✲" ✛
\$!
x
dx
2
L
+2zL.)
4 = −2zL.)
2
s
(s
◆2
✓
◆2
2
,
,
L 2&
+✓
'
'2
z
+
.
.z2 L2 &
@V
ẑ⇢
L
L2
L 2 2
2
2
z+ 2 R + z 2
z − L2
qL
E = rV = ẑ
= ∂V
Rẑρ+ z +2
+ qL
2
2
/
/
@z= −ẑ 4✏0 = −
2 z+ 2 +
2 z− 2 −
R +
E = −∇V
L
L
'2 − R +
'2
∂z
4ϵ0
2R + z +22 &
2R + z 2 2 &
R + z + L2
R + z − L2
L
L
z+ 2
z 2
"
)
1 +0 q
1+ q
z− L #
z+ L
2
L 2
1
2
2+
1 (+z qL2 )2 2 L 2
1
+
R2 +
z+q
R
(
)
2
2
L 2
2
2
R +(z+ 2 )
R +(z− 2 ) 2L
q
+R
2 q
2
2
−
+R
2+ /
2+ /
' − 2L
z + L2 + RL
z +2L2 & z L '2L2 + RL
z 2L2 &
L 2
z + 2 + R + z{z+ 2
z − 2 + R + }z − 2
|
\$! 1
"
#
1
q
1 q
1
2
/
− / L 2&
'22 +
'2
R2 + z +2L2 &
R
z 22
R + z + L2
R + z − L2
8 s
9
s
✓
◆2
✓
◆2
<
=
⎧
⎫
, L
, Lẑ⇢
.2
⎬
E=
2 ẑρ
R2 ⎨
+ z + - 2L .R22 + z
2L
4✏E0 :
2 z+
2 z − L ; − 2L
=−
2 R2 +
− 2 R2 +
⎭
4ϵ0 ⎩
2
2
⎡
, c 2012
Education,
- Pearson
2
protected
under.all
as they.currently
exist. No portion of this material may be
ρ ⎣
L
L
2 + z + in any +
form R
or2by
without
⎦ ẑ. permission in writing from the publisher.
+anyzmeans,
−
=
L − Rreproduced,
2ϵ0
2
2
Problem 2.28
z
✻
Orient axes so P is on z axis.
P
V =
1
%
ρ
dτ.
=
2
Here√ρ is constant, dτ = r sin θ dr dθ dφ,
z
θ r
✲
1
Hello
35
CHAPTER 2. ELECTROSTATICS
2
⇢ 4
=
L
2✏0
s
✓
L
R2 + z +
2
◆2
+
s
✓
L
2
R2 + z
◆2
3
5 ẑ.
Problem 2.28
Orient axes so P is ⇢
on z axis.
R ⇢
Here ⇢pis constant, d⌧ = r2 sin ✓ dr d✓ d ,
1
V = 4⇡✏0 r d⌧.
r = z 2 + r2 2rz cos ✓.
V =
R⇡
0
p
⇢
4⇡✏0
R
2
pr sin ✓ dr d✓ d
z 2 +r 2 2rz cos ✓
d✓ =
sin ✓
z 2 +r 2 2rz cos ✓
)V =
But ⇢ =
· 2⇡ · 2
q
4
3,
3 ⇡R
(
Rz
0
so V (z) =
Contents
0
p
1
rz
=
⇢
4⇡✏0
R 2⇡
;
1
rz
+
⇣
3q
1
2✏0 4⇡R3
=
1
4⇡✏0
1
2
4⇡✏ r
R 00
R
⇢
r
d⌧ =
⇢(r )[ 4⇡ 3 (r
1
4⇡✏0
0
R
2
)
z2
3
R2
r )] d⌧ =
✲y
x❂
=
⌘
⇢(r0 ) r2
θ r
φ
p
p
⇡
1
2rz cos ✓ 0 = rz
r2 + z 2 + 2rz
r2 + z 2
1
⇢
2/z , if r < z,
|r z|) =
2/r , if r > z.
r r dr
z
Problem 2.29
r2 V =
z
r2 + z 2
RR 1
r
P
d = 2⇡.
(r + z
1 2
z r dr
z
✻
n
⇢
✏0
1 z3
z 3
⇣
+
R2 z 2
2
3
z2
R2
⌘
o
=
⇢
2✏0
⇣
R2
z2
3
q
8⇡✏0 R
⌘
.
✓
3
=
q
8⇡✏0 R
1
d⌧ (since ⇢ is a function of r0 , not r)
r
1
✏0 ⇢(r).
; V (r) =
r2
R2
2rz
◆
. X
X
Problem 2.30.
Problem 2.30.
(a)
(a)Ex.
Ex.2.5:
2.4: EEabove
= 2✏2ϵσ00n̂;
n̂;EEbelow
= −2✏2ϵσ00n̂n̂(n̂
(n̂always
alwayspointing
pointingup);
up);EEabove
Ebelow
=✏0ϵσ0n̂.
n̂.X!
above=
below=
above −E
below=
Ex.
E ==✏0ϵσ0. .X!
Ex.2.6:
2.5: At
Ateach
eachsurface,
surface,EE==00one
oneside
sideand
andEE==✏0ϵσ0 other
otherside,
side,so
so ∆E
22
R
Prob.
E == ✏0ϵσ0r̂.r̂.X!
Prob.2.11:
2.11: EEout
= ✏0ϵσR
= ✏0ϵσ0r̂r̂; ;EEinin==00; ;so
so ∆E
out=
rr2 2r̂r̂=
0
✻
s
(b)
(b)
H!
Outside: E·da
E·da==E(2⇡s)l
E(2πs)l== 1ϵ10QQenc
= ϵ10(2⇡R)l
(2πR)l)
(atsurface).
surface).
enc=
Outside:
✏0
✏0
✏0 s
✏0
✻
R
c
σ
⃝2005
This material is
Inside:
soEE==
∆E
enc
Inside:
QQenc
==0,0,
so
0.0. )∴
E ==✏0ϵof0ŝ.ŝ.
X!material may be
protected
laws
as
they
currently
exist.
No portion
this
" #\$ under
%
reproduced, in any form or by any means, without permission in writing from the publisher.
l
2
(c) Vout = R✏R0 r2 σ = R✏Rσ
(at surface); Vin = R0 ; so Vout = Vin . X
(c) Vout = ϵ0 r = 0ϵ0 (at surface); Vin = ✏Rσ
ϵ0 ; so Vout = Vin . !
@Vout
@rout =
∂V
∂r =
R2
✏0Rr22σ =
− ϵ0 r 2 =
@Vin
in
out
(at surface); @V
= 0 ; so @V
= ✏0σ. X
@rout
@r
∂Vin
∂V
∂Vin
−✏0ϵσ0 (at surface); @r
=
0
;
so
−
∂r
∂r
∂r = − ϵ0 . !
protected under all copyright laws as they currently exist. No portion of this material may be
Problemin2.31
reproduced,
any form or by any means, without permission in writing from the publisher.
1
4πϵ0
&
qi
rij
1
4πϵ0
'
−q
a
√q
2a
+
,
+
2
q
1
.
∴ W4 = qV =
−2 + √
4πϵ0 a
2
(a) V =
=
1
)
−q2
*
+
1
−q
a
(
)
=
q2
q
4πϵ0 a
q2
)
−2 +
*
√1
2
*
.
(1)
(4)
−
+
+
−
36
CHAPTER 2. ELECTROSTATICS
Problem 2.31
(a) V =
P
qi
rij
(b) W1 = 0, W2 =
Wtot =
2
1 q
4⇡✏0 a
=
n
q
a
1
4⇡✏0
1
4⇡✏0
n
⇣
1+
q2
a
p1
2
⌘
+
pq
2a
+
✓
◆
2
q
1
) W4 = qV =
2+ p
.
4⇡✏0 a
2
1
4⇡✏0
; W3 =
q
a
o
⇣
⇣
q
4⇡✏0 a
=
2
pq
2a
q2
a
2+
p1
2
⌘
.
⌘
; W4 = (see (a)).
✓
◆
o
1 2q 2
1
2 + p12 =
2+ p
.
4⇡✏0 a
2
1
1
4⇡✏0
(1) r
(2)
r+
+
r(4)
r
(3)
Problem 2.32
Conservation of energy (kinetic plus potential):
1
1
1 qA qB
2
2
mA vA
+ mB vB
+
= E.
2
2
4⇡✏0 r
At release vA = vB = 0, r = a, so
E=
1 qA qB
.
4⇡✏0 a
When they are very far apart (r ! 1) the potential energy is zero, so
1
1
1 qA qB
2
2
mA vA
+ mB vB
=
.
2
2
4⇡✏0 a
Meanwhile, conservation of momentum says mA vA = mB vB , or vB = (mA /mB )vA . So
1
1
2
mA vA
+ mB
2
2
vA =
s
✓
mA
mB
◆2
1
qA qB
2⇡✏0 (mA + mB )a
2
vA
=
✓
1
2
✓
◆
mA
;
mB
mA
mB
◆
2
(mA + mB )vA
=
vB =
s
1 qA qB
.
4⇡✏0 a
1
qA qB
2⇡✏0 (mA + mB )a
✓
◆
mB
.
mA
Problem 2.33
From Eq. 2.42, the energy of one charge is
W =
1
1
X
1
1
1 ( 1)n q 2
q 2 X ( 1)n
qV = (2)
=
.
2
2
4⇡✏0 na
4⇡✏0 a 1
n
n=1
(The factor of 2 out front counts the charges to the left as well as to the right of q.) The sum is
can get it from the Taylor expansion of ln(1 + x):
ln(1 + x) = x
1 2 1 3
x + x
2
3
ln 2 (you
1 4
x + ···
4
with x = 1. Evidently ↵ = ln 2 .
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
2
37
CHAPTER 2. ELECTROSTATICS
Problem 2.34
R
(a) W = 12 ⇢V d⌧ . From Prob. 2.21 (or Prob. 2.28): V =
 3
q⇢
r
3
4⇡r dr =
3
4✏0 R
3
0
✓ 2◆
2
q⇢ 2
qR
q
1
3q
=
R =
=
.
4
3
5✏0
5✏0 3 ⇡R
4⇡✏0 5 R
1 1 q
W = ⇢
2 4⇡✏0 2R
(b) W =
R
✏0
2
Z
R
✓
r2
R2
◆
2
E 2 d⌧ . Outside (r > R) E =
1 q
4⇡✏0 r 2 r̂
⇢
2✏0
⇣
R2
1 r5
R2 5
R
0
Problem 2.32
!
(a) W = 21 ρV dτ . From Prob. 2.21 (or P
⌘
⇣
⌘
2
q
r2
1
= 4⇡✏
3 Rr 2
&
3
\$ R%
0 2R
1 1 q
r2
W = ρ ◆
3 − 2 4πr2 dr
✓
2 R4πϵ
R
3 0 2R 0
q⇢
=
R3
% 2
4✏0 R
qρ5 2 qR2 q
1
3q
=
=
R =
5ϵ0
5ϵ0 43 πR3
4πϵ0 5 R
; Inside (r < R) E =
(b) W =
q
1
4⇡✏0 R3 rr̂.
ϵ0
2
!
E 2 dτ . Outside (r > R) E =
*\$
∞
1
1 2
ϵ0
2
q
(r 4π dr) +
∴
W
=
(Z
)
2
4
Z R ⇣ ⌘2
1
2
(4πϵ
)
r
0
R
✏0
1
1
r
*%
)W =
q2
(r2 4⇡ dr) +
(4⇡r2 dr)
&)∞
% 5&
4
2 (4⇡✏0 )2
R3
R r
0
1 q2
1 ))
1
r
=
− ) + 6
(✓
)
◆
✓
◆
✓
◆
R
1
r R
R
5
1 q2
1
1
r5
1 q2 1
1
1 3 q 2 4πϵ0 2
=
+ 6
=
+
=
.X
4⇡✏0 2
r R
R
5 0
4⇡✏0 2 R 5R
4⇡✏0 5 R
,!
.
V E·da + V E 2 dτ , whe
(c) W = ϵ20
S
H
R
arbitrary. Let’s use a sphere of radius a > R.
(c) W = ✏20
V E·da + V E 2 d⌧ , where V is large enough to enclose all the charge, but otherwise
S
1 q
*\$ %
arbitrary. Let’s use a sphere of radius a > R. Here V = 4⇡✏0 r .
&%
&
1 q
1 q
ϵ0
r2 sin
W
(Z ✓
)
2
◆✓
◆
◆2= 2
Z R
Z a✓
4πϵ
r
4πϵ
r
0
0
✏0
1 q
1 q
1 q
2r=a
W =
r2 sin ✓ d✓ d +
E 2 d⌧ +
(4⇡r/
dr)
2
2
4⇡✏0 r
4⇡✏0 r2
4⇡✏
r
0
0
R
ϵ0
q2 1
q2
4π
r=a
=
4π
+
+
2
2 5R
⇢
✓
◆a
2
(4πϵ
)
a
(4πϵ
)
(
2
2
0
0
✏0
q
1
q
4⇡
1
1
2
/
0
=
4⇡
+
+
4⇡q
2
1 q
1
1
1
1
2 (4⇡✏0 )2 a
(4⇡✏0 )2 5R (4⇡✏0 )2
r R
=
=
+
− +
⇢
4πϵ
2
a
5R
a
R
4π
0
1 q2 1
1
1
1
1 3 q2
=
+
+
=
.X
4⇡✏0 2 a 5R a R
4⇡✏0 5 R
As a → ∞, the contribution from the surf
⇣
⌘
"
#
2
1 q
1 q2 while
As a ! 1, the contribution from the surface integral 4⇡✏0 2a goes to 4πϵ
zero,
the picks
volume
( 6a − 1)
up integral
the slack.
0 2a 5R
⇣
⌘
2
1 q
6a
1) picks up the slack.
Problem 2.33
4⇡✏0 2a ( 5R
Problem 2.35
dW = dq̄ V = dq̄
q̄ =
✓
4 3
r3
⇡r ⇢ = q 3
3
R
1
4⇡✏0
◆
q̄
,
r
(q̄ = charge on sphere of radius r).
(q = total charge on sphere).
4⇡r2
3q
dq̄ = 4⇡r dr ⇢ = 4 3 q dr = 3 r2 dr.
R
⇡R
✓ 3 3◆ ✓
◆
1
qr
1 3q 2
1 3q 2 4
dW =
r
dr
=
r dr
3
3
4⇡✏0 R
r R
4⇡✏0 R6
✓ 2◆
Z
1 3q 2 R 4
1 3q 2 R5
1
3q
W =
r dr =
=
.X
6
6
4⇡✏0 R 0
4⇡✏0 R 5
4⇡✏0 5 R
2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
✯
r
✮
dq̄
q̄
c
⃝2005
Pearson Ed
protected under al
reproduced, in any
38
Problem 2.36
R
(a) W = ✏20 E 2 d⌧.
⇣
⌘2 R
b
q
W = ✏20 4⇡✏
a
0
E=
1
r2
2
CHAPTER 2. ELECTROSTATICS
1 q
4⇡✏0 r 2
(a < r < b), zero elsewhere.
✓
◆
Rb 1
q2
1 1
2
q2
4⇡r2 dr = 8⇡✏
=
.
2
a r
0
8⇡✏0 a b
2
q
1 q
1 q
1 q
1
(b) W1 = 8⇡✏
, W2 = 8⇡✏
, E1 = 4⇡✏
So
2 r̂ (r > a), E2 = 4⇡✏ r 2 r̂ (r > b).
0 b
0 r
0
⇣
⌘02 a 2
⇣
⌘
2
R
R
q
q2
1
1
2 1 1
2
E1 · E2 = 4⇡✏0
E1 · E2 d⌧ =
q b r4 4⇡r dr = 4⇡✏
.
r 4 , (r > b), and hence
4⇡✏0
0b
R
2
q
1
1
1
Wtot = W1 + W2 + ✏0 E1 · E2 d⌧ = 8⇡✏
q 2 a1 + 1b 2b = 8⇡✏
a
b .X
0
0
Problem 2.37
z
r
b
r
q2
a q
x
E1 =
1 q1
r̂;
4⇡✏0 r2
y
q1
E2 =
where (from the figure)
r
=
Therefore
1 q2
4⇡✏0 r 2
p
r2 + a2
Wi =
r̂
;
Wi = ✏0
2ra cos ✓,
q1 q2
2⇡
(4⇡)2 ✏0
Z
q1 q2
(4⇡✏0 )2
cos
(r
a cos ✓)
r
3
r
It’s simplest to do the r integral first, changing variables to
2 r d r = (2r
As r : 0 ! 1,
The
r
r
: a ! 1, so
2a cos ✓) dr
Wi =
q1 q2
8⇡✏0
integral is 1/a, so
Wi =
Z
⇡
0
q1 q2
8⇡✏0 a
1
a
Z
0
⇡
1
r
2
1
r2
(r
r
2
cos r2 sin ✓ dr d✓ d ,
a cos ✓)
r
.
sin ✓ dr d✓.
:
(r
)
✓Z
=
Z
dr
sin ✓ d✓ =
a cos ✓) dr =
◆
r dr
.
sin ✓ d✓.
q1 q2
.
4⇡✏0 a
Of course, this is precisely the interaction energy of two point charges.
Problem 2.38
(a)
R
=
q
;
4⇡R2
a
=
q
;
4⇡a2
b
=
q
.
4⇡b2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.34
!
1 q
(a) W = ϵ20 E 2 dτ.
E = 4πϵ
2 (a < r < b), zero elsewhere.
0 r
'
&
"
#2 ! \$ %
!b 1
q2
1 1
b 1 2
q
q2
ϵ0
2
W = 2 4πϵ0
4πr dr = 8πϵ0 a r2 =
.
−
a r2
8πϵ0 a b
2
2
1 q
1 q
1 q
1 −q
, W2 = 8πϵ
, E1 = 4πϵ
(b) W1 = 8πϵ
So
2 r̂ (r > a), E2 = 4πϵ
2 r̂ (r > b).
0 b
0 r
0 r
"
"
#02 a 2
#
CHAPTER 2. ELECTROSTATICS
39
2
!
!
∞
q2
−q
1
1
E1 · E2 = 4πϵ
E1 · E2 dτ = − 4πϵ
.
q 2 b r14 4πr2 dr = − 4πϵ
r 4 , (r > b), and hence
0
0
0b
\$
\$
%
%
!
2
q
1
1
2
1
1
WRtot
= W1 + W
− 1b .!
= R8πϵ
R 2b + ϵ10 qE1 · E2 Rdτa = 8πϵ0 qR2 R a +1 b −
1 ⇣q
q
q⌘
0
0 0 a
q b
(b) V (0) =
E·dl
=
dr
(0)dr
dr
(0)dr
=
+
.
2
2
1
b
a 4⇡✏0 r
R
4⇡✏0 b R a
Problem
2.35 1 4⇡✏0 r
q
−q
q
(a) σR =
; σa =
; σ R=a
. RR
! 0 (the charge
“drains
(0)2 = b 1 (0)dr
4πR2 o↵”); V4πa
4πb2
a
q⌘
.
a
!a
! R\$ 1 q %
!0
!0
!b \$ 1 q %
q
q#
1 "q
Problem 2.39
+
−
dr
−
dr
−
(b) V (0) = − ∞ E·dl = − ∞ 4πϵ
(0)dr
−
(0)dr
=
2
2
b
a 4πϵ0 r
R
0 r
4πϵ0 b R a
qa
qb
qa + qb
(a) a =
;
;
.
b =
R =
!0
!a
! R\$ 1 q %
1 "q
q
4⇡a2
4⇡b2
4⇡R2
(c) σb → 0 (the charge “drains oﬀ”); V (0) = − ∞ (0)dr − a 4πϵ
−
2 dr − R (0)dr =
0 r
4πϵ0 R a
1 qa + qb
(b) Eout = Problem2 2.36
r̂, where r = vector from center of large sphere.
4⇡✏0 r
qa
qb
qa + qb
σa = −
; qσb = −
; σR =
.
2
2
2
1 q(a)
1
a
b
4πa
4πb r (r ) is 4πR
(c) Ea =
r̂
,
E
=
r̂
,
where
the
vector
from center of cavity a (b).
a
b
b
a
b
4⇡✏0 ra2
4⇡✏0 rb2
1 qa + qb
r̂, where r = vector from center of large sphere.
(b) Eout =
4πϵ
r2
0
(d) Zero.
(c)
b
(e)
R
1 q
4⇡✏0 r 2
R0
(0)dr =
R
dr
1 ⇣q
4⇡✏0 R
1 qb(but not Ea or Eb ); force on qa and qb still zero.
changes (but not a 1or qba); Eoutside changes
(c) Ea =
r̂a , Eb =
r̂b , where ra (rb ) is the vector from center of cavity a (b).
2
4πϵ0 ra
4πϵ0 rb2
Problem 2.40
(a) No. For example, if it is very close to the wall, it will induce charge of the opposite sign on the wall,
(d) Zero.
and it will be attracted.
(b) No. Typically
itRwill
be attractive,
seeσbfootnote
12changes
for an extraordinary
counterexample.
(e) σ
changes
(but not but
σa or
); Eoutside
(but not Ea or
Eb ); force on qa and qb still zero.
Problem 2.41 Problem 2.37
Between
plates, the
E =plates
0; outside
= So
σ/ϵ0 = Q/ϵ0 A. So
Between the plates,
E = the
0; outside
E = the
/✏0plates
= Q/✏E0 A.
2
2
✏0 2
✏0 PQ=2 ϵ0 E 2 =
Q2ϵ0 Q = Q .
P = E =
=
.2 2
2ϵ0 A2
2
2 ✏20 A2 2 2✏0 A22 ϵ0 A
Problem 2.42 Problem 2.38
1 Q
Inside, E = 0; outside,
E=
= 0;4⇡✏
Inside, E
outside,
2 r̂; soE =
0 r
Eave =
1 1 Q
1 1 EQ
f
ave = 2z4πϵ
2 r̂; )
=0 R(E
ave zz;
2 4⇡✏0 R2 r̂; f
1 Q
4πϵ0 r 2 r̂;
Q )z ; σ =
==
σ(Eave
2.
4⇡R
z✻
so
✒E
θ
Q
4πR2 .
!
!\$ Q % 1 \$ 1 Q %
R
RF Q
= f1z da1= Q 4πR
cos θ R2 sin θ dθ dφ
2
Fz = fz da = z4⇡R
cos2 ✓ 2R24πϵ
sin0 ✓Rd✓
d
2 2 4⇡✏ R2
0
\$
%
\$ Q %2 \$ 1
%(π/2
\$ %
!
Q2
2
1 ⇡/2
1 2 Q 2
(2 = Q
R 1 Q 2 2π π/2 sin
2θ dθ
sin
θ
θ
cos
=
=
.
Q 2 = ⇡/2
Q
Q
2
1
1
1
1
πϵ0 4R = 2
2πϵ0 4R.
0=
= 2✏0 4⇡R 2⇡ 0 2ϵ0sin4πR
✓ cos ✓ d✓0= ⇡✏0 4R
32πR2 ϵ0
2 sin ✓ 0
2⇡✏0 4R
32⇡R2 ✏0
Problem 2.43 ⃝2005
c
protected
under
as they
No portion
this material
may be law:
Say
the
charge
on
the
inner
cylinderlaws
is Q,
for acurrently
length exist.
L. The
field isofgiven
by Gauss’s
R
reproduced, in1any form or1 by any means, without
Q 1 permission in writing from the publisher.
E·da = E · 2⇡s · L = ✏0 Qenc = ✏0 Q ) E = 2⇡✏0 L s ŝ. Potential di↵erence between the cylinders is
V (b)
V (a) =
Z
a
b
E·dl =
Q
2⇡✏0 L
As set up here, a is at the higher potential, so V = V (a)
Z
a
b
1
ds =
s
V (b) =
Q
ln
2⇡✏0 L
Q
2⇡✏0 L
ln
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
b
a
.
✓ ◆
b
.
a
cylinder is Q, for a length L. The field is given by Gauss’s law:
!Say the charge on the inner
E·da = E · 2πs · L = ϵ10 Qenc = ϵ10 Q ⇒ E = 2πϵQ0 L 1s ŝ. Potential diﬀerence between the cylinders is
V (b) − V (a) = −
"
b
a
Q
E·dl = −
2πϵ0 L
"
b
a
Q
1
ds = −
ln
s
2πϵ0 L
As set up here, a is at the higher potential, so V = V (a) − V (b) =
C=
Q
V
=
2πϵ0 L
b ,
ln( a
)
Q
2πϵ0 L
2πϵ0
% &.
ln ab
40
so capacitance
per unit length is
ln
# \$
b
.
a
%b&
a .
CHAPTER 2. ELECTROSTATICS
Problem
2.400 L
2⇡✏0
C = VQ = 2⇡✏
.
b , so capacitance per unit length is
ln( a
)
ln ab
ϵ0 2
E Aϵ.
(a) W = (force)×(distance) = (pressure)×(area)×(distance)
=
2
Problem 2.44
(
'
2
(b) W = (energy per unit volume)×(decrease in volume) = ✏ϵ00 E22 (Aϵ). Same as (a), confirming that the
(a) W = (force)⇥(distance) = (pressure)⇥(area)⇥(distance) =
E A✏.
2
energy lost is equal to the work done.
⇣
⌘
2
Problem 2.41
(b) W = (energy per unit volume)⇥(decrease in volume) = ✏0 E2 (A✏). Same as (a), confirming that the
energyFrom
lost Prob.
is equal
to the
the field
workatdone.
2.4,
height z above the center of a square loop (side a) is
Problem 2.45
1
4λaz
)
E=
ẑ.
%
2&
4πϵ
a2 (side a) is
a
0
2
From Prob. 2.4, the field at height z above the center
loop
z of
+ a4square
z2 +
2
✲ ✛ da
2
✲ ✛ da
2
1
4 az
da
over a from q
0 to ā: ẑ.
Here λ → σ 2 (see figure), and we integrate
E=
2
2
4⇡✏0 z 2 + a
✛
✲
z2 + a
4
a
2
" ā
1
a2
a da
Here !
(see
figure),
and we&integrate
over
a from
)
E =da
. Let
u = 0 to
2σz
, soā:a da = 2 du.
2
%
2
2
4πϵ0
4
0
z 2 + a4
z 2 + a2
Z "
* 2
+√
,-ā2 /4
ā 2
1
a da du
a2
ā /4
σz
2u
+ z2
q√
E = = 12 4σz
z
. Let
u
=
,
so
a
da
=
2
du.
−1
tan
4⇡✏
a2
a2 2 =
0
2
0
2 z2 +
4πϵ
πϵ0 z4
z
2u +
0
0 z +(u4+ z )
2 z
0
)
"
!#ā2 /4
. Z ā2 /4
+ ā2
,
/
p
2
12σ
du
z 2
2u + z 2
2 +z
p − tan−1
==
4 ztan−1
= (1) ; tan 1
4⇡✏
⇡✏0 z
z
(u +zz 2 ) 2u + z 2
πϵ0 0
0
0
q
(
! *
)
ā2
0
2
2
2
2 + z 2σ
1
=
tan 1
tan −1
(1) 1;+ a − π ẑ.
⇡✏0
zE = πϵ0 tan
2z 2
4
✛
a+da
✲
1
2
%
&
2σ
!
r tan−1 (∞) −# π = 2σ π − π = σ . !
a → ∞ (infinite plane):"E = πϵ
4
πϵ0 2
4
2ϵ0
0
2
2
a
⇡
a2
1
1
p
E=
tan
1 + −12 √
ẑ =
tan
ẑ.
2 + (a2 series:
z ≫ a (point charge):
and
as azTaylor
1 4+ x − π4 , ⇡✏
⇡✏0 Let f (x) = tan2z
/2)
0 expand4z
1 2 ′′
⇥ f (x)1= f (0) +
⇤ ′ (0)
f (0) + · · ·
2 +⇡ x ⇡
tan (1) ⇡4 xf
= ⇡✏
2
2
4 = 2✏0 . X
0
p
1
a (point charge): Let f (x)
+ x ⇡4 Inc.,
, andUpper
expand
asRiver,
a Taylor
series:
c = tan
⃝2005
Pearson 1Education,
NJ. All
rights reserved.
a ! 1 (infinite plane): E =
z
2
⇡✏0
This material is
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any
without permission in writing from the publisher.
1 means,
0
2 00
f (x) = f (0) + xf (0) + x f (0) + · · ·
2
Here f (0) = tan
1
(1)
⇡
4
=
⇡
4
⇡
4
= 0; f 0 (x) =
f (x) =
Thus (since
a2
2z 2
= x ⌧ 1), E ⇡
2
⇡✏0
⇣
1 a2
4 2z 2
⌘
=
1
1p1
1+(1+x) 2 1+x
=
1p
,
2(2+x) 1+x
so f 0 (0) = 14 , so
1
x + ( )x2 + ( )x3 + · · ·
4
1
a2
4⇡✏0 z 2
=
q
1
4⇡✏0 z 2 .
X
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
5
Here f (0)
tan−1 (1) − π4 = π4 −
CHAPTER
2. =
ELECTROSTATICS
π
4
= 0; f ′ (x) =
1√1
1
1+(1+x) 2 1+x
=
1√
411+x ,
2(2+x)
so f ′ (0) = 14 , so
1
x + ( )x2 + ( )x3 + · · ·
4
⇢
✓
◆ !
◆
✓
◆
"@ ✓
1 @
1
@ k sin ✓ cos
2 sin ✓ cos ✓ sin
2 3k 2σ
q
a2
1 1
a2
1 σa2k
1
⇢ = ✏(since
r·E
=
✏
r
+
sin
✓
+
=
Thus
=
x
≪
1),
E
≈
=
.
!
0
2z 2 0
2z 2✓ @✓ 4πϵ0 z 2
4πϵ0r z 2
r2 @r
r πϵ0 r4sin
r sin ✓ @
r

3
2
Problem12.42
1 2k sin (2 sin ✓ cos ✓ sin ✓)
1 ( k sin ✓ sin )
= ✏0 2 3k +
\$ + r sin ✓ %& r
r
r sin ✓# 1 ∂ \$ A %r
1
∂ B sin θ cos φ
ρ =⇥ϵ0 ∇·E = ϵ0
r2
+
⇤θ ∂φk✏0 ⇥
⇤
k✏
22∂r
0
2
r
r
r
sin
= 2 3+
2 sin (2 cos ✓ sin ✓) sin( = 2 3 + rsin (4 cos2 ✓ 2 + 2 cos2 ✓ 1)
'
r
ϵ0r
1
1 B sin θ
= ϵ⇥0 2 A +
(− sin
φ) = 2 (A − B sin φ).
⇤
3k✏
3k✏
0
0
sin 2θ ✓ r1) =
r cos 2✓).
= 2 1 +r sin (2r cos
(1 + sin
r
r2
Problem 2.46
f (x) =
Problem 2.43
1 Q
From
Prob. 2.12, the field inside a uniformly charged sphere is: E = 4πϵ
Problem
2.47
3 r. So the force per unit volume
0 R
)
*
) Q *) Q *
Q
2
1
Q
3
From
the
field inside
a uniformly
E =z 4⇡✏
force per unit volume
3 r. So
r,charged
and thesphere
force is:
in the
direction
on the
dτ is:
is f =Prob.
ρE =2.12,
4
3 r =
3
3
0 R
is f = ⇢E =
Q
4πϵ0 R
Q
3
4⇡✏0 R3 r = ✏0
3 πR
ϵ0 4πR
2
Q
r,
4⇡R3
and the force in the z direction on d⌧ is:
\$
%2
3
Q
dFz = fz dτ =
r cos θ(r2 sin θ dr dθ dφ).
✓
◆2 3
3 ϵ0Q 4πR
dFz = fz d⌧ =
r cos ✓(r2 sin ✓ dr d✓ d ).
3
✏0 4⇡R
The total force on the “northern” hemisphere
is:
4
3
3 ⇡R
%2 + R is: + π/2
\$
+ 2π
+ the “northern”
The total force on
hemisphere
Q
3
3
r
dr
cos
θ
sin
θ
dθ
dφ
Fz = fz dτ =
✓ ϵ0 4πR
◆2 Z3 R 0 Z ⇡/2 0
Z
Z 2⇡ 0
3
Q\$
-π/2 .
%2 \$3 4 % , cos2 ✓ sin
Fz = fz d⌧ =
3 3 Q 0 r dr
R 0 sin θ -- ✓ d✓ 0 d 3Q2
✏0 = 4⇡R
.
(2π) =
3
✓ ϵ0 4πR
◆2 ✓
◆4
2 !-0
64πϵ0 R2
⇡/2
3
Q
R4
sin2 ✓
3Q2
=
(2⇡) =
.
✏0 4⇡R3
4
2 0
64⇡✏0 R2
Problem 2.44
+
+
1
σ
1 σ
σR
1 σ
Problem 2.48
2
Vcenter = Z
da =
Z R da = 4πϵ R (2πR ) = 2ϵ
4πϵ
η
4πϵ
0
0
0
1
1
1
R 0
Vcenter =
da =
da =
(2⇡R2 ) =
r
4⇡✏0
4⇡✏0 R
2✏0
(4⇡✏/0 R
Z
+
2
da =
sin2 ✓sin
d✓,θ dθ,
da2⇡R
= 2πR
1
1
σ
Vpole V=pole =
da ,da
with
, with 2 2 2 2 2 2
2 2
2 2
4⇡✏0 4πϵr
η
r η= =R R+ +R R −2R2R
coscos
✓=
2R2R
(1(1 −
coscos
✓).θ).
θ=
0
Z
+
2
⇡/2 -π/2
1
) 2⇡/2
sin ✓ sin
d✓ θ dθ
R σRp √
σ(2πR
) π/2
1(2⇡R
p √
p √
= =
= p= √(2 1(2 cos
1 −✓)cos θ)4⇡✏0 4πϵ
0
0
cos
1 −✓ cos θ2 2✏20 2ϵ0
R0 2R 20
0 1
p
√
R σR
R σR
R σR
p= √
= p= √
(1 (1
0) −
= 0)
.
Vcenter
= = ( 2 ( 1).
. ) Vpole
∴ Vpole
− Vcenter
2 − 1).
2✏0 2ϵ0
2✏0 2ϵ0
2✏0 2ϵ0
r
R
R
θ
Problem
Problem
2.492.45
determine
the electric
and outside
the sphere,
Gauss’s
FirstFirst
let’s let’s
determine
the electric
field field
insideinside
and outside
the sphere,
usingusing
Gauss’s
law: law:
/
0
+ Z
+
4
I
Z
Z r + r (
4 πkr
2
2
3
⇡kr
(r < (r
R),< R),
2
ϵ0 E·da
= ϵ20E
4πr
=Q
=d⌧ =
ρ dτ (kr̄)r̄
= (kr̄)r̄
θ dr̄
= 4πkr̄3 dr̄ r̄=dr̄ =
✏0 E·da
= ✏0 4⇡r
=E
Qenc
=enc ⇢
sin ✓ sin
dr̄ d✓
d dθ=dφ
4⇡k
4
πkR
(r
4
0
⇡kR (r > R).> R).
0
c
⃝2005
Pearson Education, Inc., Upper
No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
So Eprotected
= 4✏k0 r2under
r̂ (r all
< R);
E laws
= 4✏kR
r̂ (r
> R). exist.
asr 2they
currently
0
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
42
CHAPTER 2. ELECTROSTATICS
Method I :
◆2
◆2
Z ✓
Z ✓
✏0 R kr2
✏0 1 kR4
E 2 d⌧ (Eq. 2.45) =
4⇡r2 dr +
4⇡r2 dr
2 0
4✏0
2 R
4✏0 r2
)
✓
◆2 (Z R
⇢
✓
◆1
✓
◆
Z 1
✏0
k
1
⇡k 2 R7
1
⇡k 2 R7
6
8
8
7
= 4⇡
r dr + R
dr =
+R
=
+R
2
2 4✏0
8✏0
7
r R
8✏0
7
0
R r
W =
=
✏0
2
Z
⇡k 2 R7
.
7✏0
Method II :
W =
For r < R, V (r) =
=
)W =
=
1
2
Z
⇢V d⌧
(Eq. 2.43).
(
◆
✓
◆
Z r ✓ 2◆
kR4
kr
k
1
4
E·dl =
dr
dr =
R
4✏0 r2
4✏0
4✏0
r
1
1
R
✓
◆
✓
◆
3
3
3
k
r
R
k
r
R3 +
=
R3
.
4✏0
3
3
3✏0
4

✓
◆
◆
Z
Z ✓
1 R
k
r3
2⇡k 2 R
1 6
(kr)
R3
4⇡r2 dr =
R3 r 3
r dr
2 0
3✏0
4
3✏0 0
4
⇢
✓ ◆
2
4
7
2 7
2 7
2⇡k
R
1R
⇡k R
6
⇡k R
R3
=
=
.X
3✏0
4
4 7
2 · 3✏0 7
7✏0
Z
Z
r
R
✓
R
r3
+
3
1
r
R
)
Problem 2.50
E=
rV =
@
A
@r
✓
r
e
r
◆
r̂ =
A
⇢
)e
r(
r
e
r
r̂ = Ae
r2
r
(1 + r)
r̂
.
r2
⇢ = ✏0 r·E = ✏0 A e r (1 + r)r· rr̂2 + rr̂2 ·r e r (1 + r) . But r· rr̂2 = 4⇡ 3 (r) (Eq. 1.99), and
e r (1 + r) 3 (r) = 3 (r) (Eq. 1.88). Meanwhile,
@
r e r (1 + r) = r̂ @r
e r (1 + r) = r̂
e r (1 + r) + e r
= r̂( 2 re r ).

2
2
So rr̂2 ·r e r (1 + r) = r e r , and ⇢ = ✏0 A 4⇡ 3 (r)
e r .
r
⇢ Z
✓
◆
Z
Z
Z 1
e r
3
2
2
2
r
Q = ⇢ d⌧ = ✏0 A 4⇡
(r) d⌧
4⇡r dr = ✏0 A 4⇡
4⇡
re
dr .
r
0
⇣
⌘
R1
2
But 0 re r dr = 12 , so Q = 4⇡✏0 A 1
= zero.
2
Problem 2.51
Let u ⌘ s/R. Then
1
V =
4⇡✏0
Z
r
da =
2 R
V =
4⇡✏0
Z
0
4⇡✏0
1
Z
0
Z
R
0
Z
0
2⇡
p
R2 + s2
⇡
p
u
1 + u2
2u cos
1
2Rs cos
d
!
s ds d .
du.
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
43
CHAPTER 2. ELECTROSTATICS
The (double) integral is a pure number; Mathematica says it is 2. So
V =
R
.
⇡✏0
Problem 2.52
(a) Potential of + is V+ = 2⇡✏0 ln sa+ , where s+ is distance from + (Prob. 2.22). 7
z
Potential of
is V = + 2⇡✏0 ln sa , where s is distance from
.
✻
(x, y, z)
! "
s−
✓ λ◆
s
−
∴ Total V = s ln
.
) Total V =
ln 2πϵ0 . s+
7
s+
2⇡✏0 # s+
#
z
a
a
p s+ = (y − a)2 + z 2 , and
p s− = (y + a)2 + z 2 , so
✲
Now
✻
(x, y, z)
2 , and s =
y
Now s+ = (y a)2!+ z"
(y + a)2 + z 2 , so
−λ
λ
\$
%
!
"
√
λ
s
−
s
2
2
◆
 λ
−
(y+a)2 +z 2
∴ Total V =
ln✓ pλ(y+a). 2 +z
(y + ln
a)2 (y
+ z+2 a) + z .
(x,2πϵ
y, z)
=
s+0 ln √2(y−a)
0ln= 2πϵ
p
2 +z 2
2
2
.
V (x, y, z)V =
=
ln
s+
4πϵ
2⇡✏0
(y a)2 +z 2
(y 0 a)2 (y
+ z−2 a) + z
#
#4⇡✏0
2
2
2
2
a
a
✲
Now s+ = (y − a) + z , and s− = (y +2 a)2 + z , so
y
(y+a) +z
(4πϵ0 V0 /λ)
−λ
λ
% = k = constant. That is:
! √ are given
" by (y−a)2 +z\$2 = e 2
(b) Equipotentials
2
2
2 λ
2
2
(yV +
(y+a)2 +z(y+a)2 +z
(4⇡✏
/ )a) + z
2 constant.
2That is:
2
√
(b)V (x,
Equipotentials
are+given
= e+
k=
y, z) =
ln
.
ln
=
2 +z
y 2 2πϵ
+λ2ay
a2(y−a)
+ zby
=(y2k(y
− 22ay
a20+0 z 2 ) =
⇒
a)
2 +z
2 y (k
2 − 1) + z (k − 1) + a (k − 1) − 2ay(k + 1) = 0, or
0
&
'
4πϵ
(y
−
0
2
2 2 2 2
2
2
2
2 a) + z 2
2
k+1
2
y + 2ayy+ +
a z+ +
z a⇣= −
k(y
+ a + z ) ) y (k 1) + z (k 1) + a (k 1) 2ay(k + 1) = 0, or
2ay
⌘ 2ay
k−1 = 0. The equation for a circle, with center at (y0 , 0) and radius R, is
k+1
2
2
y 2 + z 2 +(ya−
2ay
=
0.
The
a2circle,
with
center at (y0 , 0) and radius R, is
2
2 2equation
(4πϵ
0 /λ)
+ zk2 1=byR2(y+a)
, or 2y+z
+
z2 +
(y002Vfor
−
R
)−
0 ) given
0 =&0. 'That is:
(b) Equipotentialsyare
=
k 2yy
= constant.
2 = e
2
2
2
2 (y−a)
2 +z 2
2
(y y ) +2z =2R the
, orequipotentials
R ) 2yy
k+1
0 =⇣0.
2y + z + (y
y 2 + 2ay0 +Evidently
a + z &= k(y'
− 2ay + a2 0+are
z 2 )circles,
⇒ y 2 (kwith
− 1)y+
z⌘2a(k −
1) +and
a2 (k − 1) − 2ay(k + 1) = 0, or
0 =
k−1
k+1
Evidently
the
equipotentials
are
circles,
with
y
=
a
and
&
'
0
2
k+1
k 1
a circle,
at (y20+2k−1)
y2 + z 2 + a
a22 −
(k2 +2k+1−k
k+1
4kR, is
2
k−1
⇣a2 =⌘a22 for
= 2ay
y02 − R
⇒=
R20.=The
y02 −equation
−(ka22with
= a2center
a2 (k−1)
2 , or
2
k−1
(k−1)
+2k+1
k
+2k
1) 2 2 4k
k+1
2
2
2
2
2
2
2
2
2
2
2 )√R
2 = y2
2 =a2
2
a
=
y
R
a
a
=
a
=
a
,
or
2
2
(y − y0 )0 + z =
Rk , or 0 + z + (y0 −k R1 ) − 2yy0 =&0. ' (k 1)
(k 1)
2a
pR =
|k−1| ; or, in terms of V0 :
2a kthe equipotentials
and
Evidently
are circles, with y0 = a k+1
R = |k
k−1
1| ; or, in terms of V0 :
&
'2
2
2
+2k−1)
2
2 k+1
! = a2 (k−1)
"4k 2 , or
a2 = y02 − R2 ⇒ R2 4πϵ
= y002V0−
− a2 =
a2 0(kV0+2k+1−k
2
2πϵ0 V0 /λ
−2πϵ
/λ (k−1)
/λa = a
k−1
✓
◆2πϵ0 V0
+0 /e
+2⇡✏
1 0 V0 / e
2⇡✏0 V
√ e4⇡✏0 V0 / e+ 1
e
=a +
= 2⇡✏
y0 = a 4πϵ V /λ e
a coth
.
0 V0
/λ
0 a
0 /λ − e−2πϵ
= ka ; 4⇡✏
= 0aV0coth
. λ
R =y02a
or,
in terms
of
V−
e 0=
e2πϵ0 V2⇡✏
0: 1
|k−1| e
0 V0 /
0 V0 /
1
e2⇡✏0 V0 /
e
!
"
2πϵ0 V0 /λ
✓
◆2πϵ0 V0
2
a
2⇡✏0 V0 / e
e = 2a
2 V /λ
! a
"( 2πϵ0 V0 ) = 2⇡✏
R
a csch
.
= a −2πϵ
=
0 V0
0
0 V0 /λ
0 /λ
0 V00 /λ
R = 2a
=0 V0 /λ )2πϵ
a csch
. λ
−0 V10 /λ +(ee2πϵ
− /e−2πϵ
ea2πϵ
V0 =
e4πϵ4⇡✏
+e14πϵ0 V=
2⇡✏0sinh
0 V0
λ
e 0 V0 /
1 a (e2⇡✏0 V0 /
e 2⇡✏0 V0=
) a coth
sinh
=
y0 = a 4πϵ
.
λ
e 0 V0 /λ − 1
e2πϵ0 V0 /λ − e−2πϵ0 V0 /λ
z
!
"
✻
e2πϵ0 V0 /λ
2πϵ0 V0
2
a
(
) = a csch
R = 2a 4πϵ V /λ
.
= a 2πϵ V /λ
=
λ
e 0 0 −1
(e 0 0 − e−2πϵ0 V0 /λ )
sinh 2πϵλ0 V0
z
✒
R
✻
✒
R
−λ
✲
y
λ y0
−λ
λ y0
✲
y
Problem
2.48
c 2012 Pearson
Education,
protected under all copyright laws as they currently
No 1
portion of this material may be
d2 exist.
V
ρ
reproduced,
in any
any
means,
without
permission
inρ.
writing from the publisher.
(Eq.
2.24),
so
=
−
(a)
∇2 Vform
= or
− ϵby
0
dx2
ϵ0
*
Problem 2.48
2qV
1
2
2
.1
(b) qV
= 2 mv → v d= V
ρ
2
=m− ρ.
(a) ∇ V = − ϵ0 (Eq. 2.24), so
dx2
ϵ0
dx
(c) dq = Aρ dx*; dq
dt = aρ dt = Aρv = I (constant). (Note: ρ, hence also I, is negative.)
2qV
.
(b) qV = 12 mv 2 → v =
c
⃝2005
Pearson Education,
44
CHAPTER 2. ELECTROSTATICS
Problem 2.53
(a) r2 V =
(b) qV =
⇢
✏0
1
2
2 mv
(Eq. 2.24), so
! v=
r
d2 V
=
dx2
1
⇢.
✏0
2qV 8
.
m
!
!
dq
2
d2 V (Note:
−1/2
V A⇢ dx 1;
I
m
1 dx
I = A⇢v
I = I m(constant).
(c)(d)
dq d=
=−a⇢
⇢, hence
also
I, −
is negative.)
,
where
β
=
=
βV
=
−
ρ
=
=
−
⇒
dt
dt
2
dx
ϵ0
ϵ0 Av
ϵ0 A
2qV
ϵ0 A
2q .
dx2
q
q
(Note: I is negative,
so β isI positive;
q isd2positive.)
2
V
1/2
I
m
I
m
(d) ddxV2 = ✏10 ⇢ = ✏10 Av
= ✏0 A 2qV
)
=
V
,
where
=
✏
A
2q .
0
dx2
:
(e) Multiply by V ′ = dV
dx
(Note: I is negative, so is positive; q is positive.)
"
"
′
dV
1 ′
0 dV dV
(e) Multiply by V
V ′ = dx=:βV −1/2
⇒ V ′ dV ′ = β V −1/2 dV ⇒ V 2 = 2βV 1/2 + constant.
dx
dx
2
Z
Z
0
0
dV
dV
1
V 1/2 is )
V 0 dV 0 =
V 1/2
dVat)cathode
V 2 =is2zero),
V 1/2 so
+ constant.
But V (0) = VV ′0(0) ==0 (cathode
at potential
zero, and
field
the constant is zero, and
dx
dx
2
#
#
′
dV
2 V 0 (0) =
1/4
But V (0)
(cathode
zero,dV
and
V =
= 4βV 1/20⇒
β Vpotential
⇒ V −1/4
= field
2 βat
dx;cathode is zero), so the constant is zero, and
= 2is at
dx"
"
#
#
p 4 3/4
p
0
dV
−1/4 1/2
x+
V )=
=
V 2V
= 4 V dV
)2 β = 2dx ⇒V 1/4
V 21/4βdV
=constant.
2
dx;
3
dxZ
Z
p
p
4 3/4
1/4
dVso=this
2 constant
dx )is also
V zero.
=2
x + constant.
But V V
(0) = 0,
3
\$
%4/3
\$ %2/3
%1/3
\$
3 #constant is also zero.
3#
9
81I 2 m
But V (0) = 0,
3/4so this
4/3
4/3
V
=
β x, so V (x) =
β
x , or V (x) =
β
x
=
x4/3 .
2 A2 q
2
2
4
32ϵ
0
✓
◆4/3
✓ ◆2/3
✓
◆1/3
3p
3p
9
81I 2 m
4/3
4/3
V 3/4 =
x, so V (x) =
x
,
or
V
(x)
=
x
=
x4/3 .
2
2 & x '4/3
4
32✏20 A2 q
(see graph).
Interms of V0 (instead of I): V (x) = V0
d
( )
⇣ x ⌘4/3
V✻
Without space-charge, V would increase linearly: V (x) = V0 xd .
Interms of V0 (instead of I): V (x) = V0
(see graph).
d
V0
4ϵ0 V
d2 V
4 1 −2/3
Without space-charge,
V would1 increase
linearly: V (x)
=0 V0 xd .
ρ = −ϵ0 2 = −ϵ0 V0 4/3 · x
= −
.
2
2/3
dx
3 3
d
9(d x)
without
1 4 1 2/3
4✏0 V0
*d2 V
'
&
⇢ = ✏0 2q 2√ = ✏0#
V0 4/3 · xx 2/3 =
.
with
3 3
d 0 /m
9(d2 x)2/3
.
v = dx V =
2qV
d
r m
✲
⇣ x ⌘2/3
2q p & p '
x
d
v=
V =81I 2 m
2qV1/3
/m
.
2
2
0
32ϵ0 A q 3
81md4 2
2
(f) V (d) = m
V0 = 32ϵ2 A2 q
d4/3d⇒ V03 = 32ϵ
I
;
I
=
V
;
2 A2 q
81md4 0
0
0
*
⇣
⌘
√
1/3
√
2
4
2q 32✏20 A2 q 3
4 2 ϵ A q 81I3/2
0A
3/2
81md4ϵ
(f) V I(d)
V03 = K32✏
.
== 9V√0m0=d2 32✏
=20 A2 q I22 ; I 2 =
V020 Am2 q= KVd04/3 ,)
where
4 V0 ;
9d r m 81md
p
p
4✏0 A 2q
4 2 ✏ A q 3/2
3/2
Problem
I = 9pm02.49
V0 = KV0 , where K =
.
d2
9d2
m
"
&
'
1
η −η/λ
ρη̂η̂
(a) E = 2.54
1+
e
dτ.
Problem
4πϵ0 η 2
λ
✓
◆
Z
r e r / d⌧.
1
⇢ r̂
(a)(b)E Yes.
= The field2 of 1a+point charge
at the origin is radial and symmetric, so ∇×E = 0, and hence this is also
r
4⇡✏0
true (by superposition) for any collection of charges.
" r
"
r ' −r/λ
1 c 2012rPearson
1 & Education,
(c) V = −
1+
e Inc.,
E·dl = − protected
q
2
under
all
copyright laws as they currently exist. No portion of this material may be
4πϵ
r
λ
0
∞
∞
reproduced,
in any form or+"
by any means, without permission
in writing
,from the publisher.
" ∞
"
∞
1
1 &
r ' −r/λ
q
1 ∞ 1 −r/λ
1 −r/λ
=
q
1+
e
dr =
e
dr +
e
dr .
4πϵ0 r r2
λ
4πϵ0
r2
λ r r
r
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 2. ELECTROSTATICS
45
(b) Yes. The field of a point charge at the origin is radial and symmetric, so r⇥E = 0, and hence this is
also true (by superposition) for any collection of charges.
Z r
Z r
1
1 ⇣
r ⌘ r/
9
(c) V =
E·dl =
q
1
+
e
dr
4⇡✏0 1 r2
9
1
⇢Z
Z 1
Z
1
1
⇣
⌘
1
1
r
q
1
1
1
!ee−r/λ
!
r/
−r/λ
−r/λ q
dr! =
e r/ dr +
e r/Therefore
dr .
− λ1−r/λ
←−
exactly right
term.
Now r1=
= −re2 r 1 +
2 e 4⇡✏
!0dr
r 1dr
4⇡✏
r2 to kill the last
r
0
e−r/λ
1 r−r/λ
e
r
r
dr
←−
exactly
right
to
kill
the
last
term.
Therefore
Now r2 e
dr = − r − λ
R
R e r/ "r
r/
\$
1
#
∞
Now r12 e r/ dr = e r
dr
exactly
right
to
kill
the
last
term.
Therefore
−r/λ
rq
" ##
e−r/λ
#∞q\$ e
−r/λ
V (r) =
−(q
=
. −r/λ
#
e
)
#
r r/
#4πϵ0= r qr/ e
V4πϵ
(r)0 q=
−
r 1
q e4πϵ0 r .
4πϵ0e
r #=r
V (r) =
.
4⇡✏0
r& r ' 4⇡✏0 r
&
'
%
R −R/λ'
q
1
1
R −R/λ'
(d)
IE·da%=
!R
!2 =
1✓+ q & e◆
.
q 2 1 1✓+1 & e◆
R 4π
(d)
−R/λ
2
1
1
R
q
4πϵ
!
R
!
λ
ϵ
λ R1 + RR/ e−R/λ .
0 =
0R
R/ e
24π
1
+
!
!
=
q
(d) S E·da E·da
=
q! 2 R
1 2+
eλ
4⇡ R =
1+
eλ
.
( S
( R
S
4⇡✏( 4πϵ
R0 !
✏0 ϵ0 ) −r/λ
+,R
* r
q 0 R e−r/λ
q
q e
)
,R
2
−r/λ
(
(
(

R
R
*
ZV dτ =
Z qR r/
Z re
−r/λ
−r/λ
r e4π
!
dr =
−
1
dr
=
−
Rq
q 2r/ e ⇣λ r
r⌘ R +
2 ϵ0 q
−r/λ
q
e
q
e
4πϵ
!
r
ϵ
(1/λ)
0
0
2
r/
V
=0
!
dr0 = re redr = dr =
−10 − 1
V d⌧ V= dτ r r4⇡'rdr4π
=.
2
2
! 0 0&
ϵ0 )(1/λ)
λ
V
4⇡✏0 4πϵ
r
✏0' 0 ϵ0. 0
✏0 (1/
0
V
0
R&
0
2 q
−R/λ
⇢ q
◆1 R .
=λ
+
−e
1✓+
ϵ q 2
λ R
−e−R/λ
= 02 = λ ϵ e R/
1 + 1 ++λ1 +
. 1 .
0
✏0
'
&
'
.
-&
%
(
q
1
q
⇢✓ R
◆
✓ R& ◆
I
Z
-&
−R/λ'
−R/λ'
%
(
1+
= ..
∴
E·da + 2 1 V dτ
=
q 1+
Re RR/− −R/λ
Re RR/+ 1−R/λ
q qed
1
q
λ+ V+ V d⌧ V
ϵ=0 dτ = 1 +
λ 1 +e
λ 1 +e
ϵ=01 .= qqed
)S ∴
E·da E·da
1−
+
e
e+ 1 +
. qed
2
2
✏0 ϵ0
✏0 ϵ0
λ
λ
S
Vλ
S
V
(e)(e)
Does
thethe
result
in (d)
hold
forfor
a nonspherical
surface?
Suppose
we we
R
Does
result
in (d)
hold
a nonspherical
surface?
Suppose
2 2
S R
(e)
Does
the
result
in
(d)
hold
for
a
nonspherical
surface?
Suppose
we
make
a
“dent”
in
the
sphere—pushing
a
patch
(area
R
sin
θ
dθ
dφ)
make a “dent” in the sphere—pushing a patch (area R sin ✓2 d✓ d )
S
q
2 2
make
a
“dent”
in
the
sphere—pushing
a
patch
(area
R
sin
θ
dθ
dφ)
from
R
out
to
S
(area
S
sin
θ
dθ
dφ).
from radius R out to radius S (area S sin ✓2 d✓ d ).
q
from radius R out to radius S (area S sin θ dθ dφ).
&
'
&
'
.
%
1
1⇢ q
.
✓+ S & e◆
✓+ R & e◆
−S/λ ' 2
−R/λ' 2
%
I
1
(S
sin
θ
dθ
dφ)
−
1
(R
sin
θ
dθ
dφ)
∆ E·da
=
SS/ −S/λ
1λ R RR/ −R/λ
1λ S
q S 2q1
2 1
2
2
2
2
4πϵ
Rdφ)
1
+
e
(S
sin
θ
dθ
−
1
+
e
(R
sin
θ
dθ
dφ)
∆ E·da
=
E·da
= 0 )&
1
+
e
(S
sin
✓
d✓
d
)
1
+
e
(R
sin
✓
d✓
d
)
'
,
λ&
λ
S02S 'S 2
R2 R 2
q 4⇡✏0 4πϵ
R& ◆
,dφ.
)&
✓
◆
✓
−S/λ '
−R/λ'
=
e
−
1
+
e
sin
θ
dθ
1
+
q λ S
S
q
4πϵ
λ R1 +eRR/ e−R/λ
sindθ dθ
= 0=
1 + 1 +e S/ e−S/λ1−
+
sin ✓ d✓
. dφ.
λ
λ
4⇡✏0 4πϵ0
( SZ
(
( Z−r/λ
S (
1 1 Z
1 1q q
( 1 1q q e e r/
(r2 sin
Sdr
−r/λ
θ
dr
dθ
dφ
=
sin
θ
dθ
dφ
re−r/λ
V
dτ
=
2
1
1
q
e
1
q
2
2
2
V
d⌧
=
r
sin
✓
dr
d✓
d
=
sin
✓
d✓
d
re r/
dr
2
λ 2∆
λ dτ
4πϵ
r r
λ dφ
4πϵ
2=
2=
0
0
r sin θ dr dθ
sin θRdθRdφ
re−r/λ dr
V
4⇡✏
4⇡✏
0
0
2
2
2
*
*
++#
λ
λ
4πϵ
r
λ
4πϵ
0
0
S
R
q q
⇣
⇣ r ⌘⌘
−r/λ
* + *r ## Sr ++#S
=−
θ qdθ✓ dφ
#
=4πϵ0 sinsin
d✓ d e e r/ 1 −r/λ
1 λ+ R
=)&
−
sin θ dθ dφ &
e
1+ R #
4⇡✏
0
✓ 0S ' ◆
✓ R ' ◆λ ,R
q q 4πϵ
)&
'
' θ dθ,dφ.
−S/λ
=−
1 q+1 + Se e SS/− 1 +1 +&Re−R/λ
=4πϵ0
eRR/sin
sin ✓ d✓ d .
−S/λ λ
λ
=−
1+
e
− 1+
e−R/λ sin θ dθ dφ.
4⇡✏
0
4πϵ0
λ
λ
!
/ H
1 1 R
1 1
exactly
compensates
forfor
thethe
change
in inE·da,
we we
getget
thethe
total
using
So So
thethe
change
in in
change
V 1d⌧
compensates
change
E·da,
q for
total
using
! exactly
/andand
λ2 2V dτ
ϵ0 q✏0for
Vdid
dτdid
exactly
compensates
for Any
theAny
change
in surface
E·da,
and
we built
getupϵ1up
q
for
the
total
using
So the
change
inasλ2as
thethe
dented
sphere,
justjust
we we
with
thethe
perfect
sphere.
closed
surface
cancan
be be
built
by
successive
dented
sphere,
with
perfect
sphere.
closed
by
successive
0
the of
dented
sphere,
as
we did
with
theall
perfect
sphere.
Any closed
surface
be charges
built
upinside,
byinside,
successive
distortions
the
sphere,
sojust
the
result
holds
forfor
all
shapes.
By
superposition,
if there
arecan
many
distortions
of the
sphere,
so the
result
holds
shapes.
By
superposition,
if there
are
many
charges
1
distortions
of .the
sphere,outside
so the do
result
for all(in
shapes.argument
By superposition,
if therethat
are many charges inside,
total
Qenc
Charges
not holds
contribute
✏0 enc
for this
thethe
total
is isϵ10 Q
do not
contribute
(in the
the argument above
above we
wefound
found that
H. !1Charges1 outside
R
/the total is
Q
.
Charges
outside
do
not
contribute
(in
the
argument
above
we
found
that
1 E·da
for
this
volume
+
V
d⌧
=
0—and,
again,
the
sum
is
not
changed
by
distortions
of
the
surface,
as this
enc
2
volume E·da +/ λ2 ϵ0V dτ =! 0—and,
again, the sum is not changed by distortions of the surface, as long as q for
1
long
as
q
remains
outside).
So
the
new
“Gauss’s
Law”
holds
for
any
charge
configuration.
volume
E·da
+
V
dτ
=
0—and,
again,
the
sum
is
not
changed
by
distortions
of
the
surface,
as
long
as q
remains outside. So the new
λ2 “Gauss’s Law” holds for any charge configuration.
remains outside. So the new “Gauss’s Law” holds for any charge configuration.
1 This material is
1 reserved.
c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights
V =
(f) protected
In diﬀerential
“Gauss’s
law”currently
∇·ENo+portion
or,1putting
under allform,
laws as they
of this
may be it all in terms of E:
1 ρ,material
2
λ
ϵ
0 Vthe
= publisher.
ρ, or, putting it all in terms of E:
(f) Inindiﬀerential
form,
“Gauss’s
law”
+ from
reproduced,
any means,
without
permission
writing
(any form or by
2
λ
ϵ0
1
( 1 ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇2 V + 1 V = 1 ρ.
=
∇·E − 2 E·dl
1
1
λ
λ22 V + ϵ10 V = 1 ρ.
0
= ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇
∇·E − 2 ϵE·dl
λ
ϵ0
λ2
ϵ0
∆
c
⃝2005
protected ⃝2005
under
all
laws as Inc.,
they Upper
currently
exist.River,
No portion
this material
c
Pearson
Education,
NJ. Allofrights
reserved.may
Thisbematerial is
reproduced,
in any form
any means,
without
in writing
from the
protected
underorall
laws
as theypermission
currently exist.
No portion
of publisher.
this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
=−
q
4πϵ0
1+
S
λ
e−S/λ − 1 +
R
λ
e−R/λ sin θ dθ dφ.
)
*
So the change in λ12 V dτ exactly compensates for the change in E·da, and we get ϵ10 q for the total using
the dented sphere, just as we did with the perfect sphere. Any closed surface can be built up by successive
distortions of the sphere, so the result holds for all shapes. By superposition, if there are many charges inside,
the total is ϵ10 Qenc . Charges outside do not contribute (in the argument above we found that
for this
)
*
volume E·da + λ12 V dτ = 0—and, again, the sum is not changed by distortions of the surface, as long as q
46new “Gauss’s Law” holds for any charge CHAPTER
2. ELECTROSTATICS
remains outside. So the
configuration.
1
1
(f) In diﬀerential form, “Gauss’s law” reads: ∇·E + 2 V = ρ, or, putting it all in terms of E:
1
1λ
ϵ0
(f) In di↵erential form,!“Gauss’s law” reads: r·E + 2 V = ⇢, or, putting
it all in terms of E:
✏0
1
1
1
1
Z −
∇·E
E·dl
=
ρ.
Since
E
=
−∇V
,
this
also
yields
“Poisson’s
equation”: −∇2 V + 2 V = ρ.
1
1
1λ
λ2 1
ϵ0
ϵ0
2
r·E
E·dl = ⇢. Since E = rV , this also yields “Poisson’s equation”: r V + 2 V = ⇢.
2
✏0
✏0
ρ
dτ
λ
r/
ϵ
0
)V
=−
ρ
R
ρ
r e−
ρ
−
1
λ2
λ
−
(∇ 2
R
0
E=
×
V =−
✲
E·dl
dτ
∇
/
rλ
ρ ;
ϵ0
E=−∇V
✛
r )e
=
dl
✢
V
+
r̂ (1
r2
R E·
4π 1
ϵ
0
R
1
2
λ
=
1 0
ϵ
4π
−
·E
V
E=
∇
✣❪
❫
E
(g) ReferProblem
to ”Gauss’s
law” in di↵erential form (f). Since E is zero, inside a conductor (otherwise charge would
2.50
move, and in such a direction as to cancel the field), V is constant (inside), and hence ⇢ is uniform, throughout
∂
= ϵ0“extra”
∇·E = ϵcharge
everywhere).
= ϵreside
0 ∂x (ax)
0 a (constant
the volume. ρAny
must
on the surface.
(The fraction at the surface depends on , and
The
same
charge
density
would
be
compatible
(as far as Gauss’s law is concerned) with E = ayŷ, for
on the shape of the conductor.)
a
instance,
Problem
2.55 or E = ( 3 )r, etc. The point is that Gauss’s law (and ∇×E = 0) by themselves do not determine
the field —like any diﬀerential equations, they must be supplemented by appropriate boundary conditions.
@
⇢ = ✏Ordinarily,
(ax)are
= so
✏0 a“obvious”
(constantthat
everywhere).
these
we impose them almost subconsciously (“E must go to zero far from
0 r·E = ✏0 @x
the
source
charges”)—or
we
appeal
to
symmetry
ambiguity
(“the with
field E
must
be theforsame—in
The same charge density would be compatible (as far to
asresolve
Gauss’sthe
law
is concerned)
= ayŷ,
a
magnitude—on
both
sides
of
an
infinite
plane
of
surface
charge”).
But
in
this
case
there
are no natural
instance, or E = ( 3 )r, etc. The point is that Gauss’s law (and r⇥E = 0) by themselves do not determine
boundary
conditions,
and
no
persuasive
symmetry
conditions,
to
fix
the
The
question
“What is the
the field —like any di↵erential equations, they must be supplemented by appropriate boundary conditions.
electric
field
produced
by
a
uniform
charge
density
filling
all
of
space?”
is
simply
ill-posed
:
it
does
Ordinarily, these are so “obvious” that we impose them almost subconsciously (“E must go to zero far from not give
the source charges”)—or we appeal to symmetry to resolve the ambiguity (“the field must be the same—in
magnitude—on both sides of an infinite
of surface
charge”).
inRiver,
this NJ.
case
are no This
natural
c plane
⃝2005
Pearson
Education,
Inc., Upper But
All there
rights reserved.
material is
underconditions,
as they
currently
exist.
No
portion of
this material
boundary conditions, and no persuasiveprotected
symmetry
to
fix
the
The
question
“What
is
themay be
reproduced, in any form or by any means, without permission in writing from the publisher.
electric field produced by a uniform charge density filling all of space?” is simply ill-posed : it does not give
us
information
to determine the answer. (Incidentally, it won’t help to appeal to Coulomb’s law
⇣ sufficient
⌘
R r̂
1
E = 4⇡✏0 ⇢ r 2 d⌧ —the integral is hopelessly indefinite, in this case.)
Problem 2.56
Compare Newton’s law of universal gravitation to Coulomb’s law:
F=
Evidently
1
4⇡✏0
G
m1 m2
r̂;
r2
F=
1 q1 q2
r̂.
4⇡✏0 r2
! G and q ! m. The gravitational energy of a sphere (translating Prob. 2.34) is therefore
Wgrav =
3 M2
G
.
5 R
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
47
CHAPTER 2. ELECTROSTATICS
Now, G = 6.67 ⇥ 10 11 N m2 /kg2 , and for the sun M = 1.99 ⇥ 1030 kg, R = 6.96 ⇥ 108 m, so the sun’s
gravitational energy is W = 2.28 ⇥ 1041 J. At the current rate this energy would be dissipated in a time
t=
W
2.28 ⇥ 1041
=
= 5.90 ⇥ 1014 s = 1.87 ⇥ 107 years.
P
3.86 ⇥ 1026
Problem 2.57
First eliminate z, using the formula for the ellipsoid:
(x, y) =
Q
1
p
2
2
4
2
2
4
4⇡ab c (x /a ) + c (y /b ) + 1
(x2 /a2 )
(y 2 /b2 )
.
Now (for parts (a) and (b)) set c ! 0, “squashing” the ellipsoid down to an ellipse in the x y plane:
(x, y) =
1
Q
p
2⇡ab 1
(x/a)2
(y/b)2
.
(I multiplied by 2 to count both surfaces.)
(a) For the circular disk, set a = b = R and let r ⌘
p
x2 + y 2 .
(r) =
(b) For the ribbon, let Q/2b ⌘ ⇤, and then take the limit b ! 1:
(c) Let b = c, r ⌘
Q
1
p
.
2⇡R R2 r2
(x) =
⇤
1
p
.
2⇡ a2 x2
p
y 2 + z 2 , making an ellipsoid of revolution:
x2
r2
+
= 1,
a2
c2
with
=
Q
1
p
.
4⇡ac2 x2 /a4 + r2 /c4
The charge on a ring of width dx is
dq = 2⇡r ds,
2x dx 2r dr
dr
Now
+ 2 =0)
=
2
a
c
dx
(x) =
where ds =
p
p
dx2 + dr2 = dx 1 + (dr/dx)2 .
r
c2 x
c4 x2
c2 p 2 4
, so ds = dx 1 + 4 2 = dx
x /a + r2 /c4 . Thus
2
a r
a r
r
dq
Q
1
c2 p 2 4
Q
p
= 2⇡r
x /a + r2 /c4 =
. (Constant!)
dx
4⇡ac2 x2 /a4 + r2 /c4 r
2a
protected under all copyright laws as they currently exist. No portion of this material may be
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48
CHAPTER 2. ELECTROSTATICS
Problem 2.58
y
( _a2, ÷3 _a2 )
b
(0,-a)
q
r
x
( _a2,- ÷3 _a2)
(a) One such point is on the x axis (see diagram) at x = r. Here the field is

q
1
Ex =
4⇡✏0 (a + r)2
2
cos ✓
= 0,
b2
Now,
(a/2)
cos ✓ =
b
Therefore
r
;
b =
2
⇣a
r
2
⌘2
+
or
2 cos ✓
1
=
.
b2
(a + r)2
!2
3
a
= (a2
2
p
2[(a/2) r]
1
. To simplify, let
=
(a + r)2
(a2 ar + r2 )3/2
(1 2u)
1
=
,
2
3/2
(1 + u)2
(1 u + u )
or
(1
ar + r2 ).
r
⌘u:
a
2u)2 (1 + u)4 = (1
u + u2 )3 .
protected under all copyright laws as they currently exist. No portion of this material may be
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49
CHAPTER 2. ELECTROSTATICS
Multiplying out each side:
1
6u2
4u3 + 9u4 + 12u5 + 4u6 = 1
3u + 6u2
7u3 + 6u4
3u5 + u6 ,
or
3u
12u2 + 3u3 + 3u4 + 15u5 + 3u6 = 0.
u = 0 is a solution (of course—the center of the triangle); factoring out 3u we are left with a quintic equation:
1
4u + u2 + u3 + 5u4 + u5 = 0.
According to Mathematica, this has two complex roots, and one negative root. The two remaining solutions are
u = 0.284718 and u = 0.626691. The latter is outside the triangle, and clearly spurious. So r = 0.284718 a.
(The other two places where E = 0 are at the symmetrically located points, of course.)
y
a a
_,_
( ÷2
÷2 )
b+
b_
q_
q+
r
x
(b) For the square:
q
Ex =
4⇡✏0
where
cos ✓± =
Thus
✓
cos ✓+
2 2
b+
p
(a/ 2) ± r
;
b±
◆
cos ✓
2 2
b
b2± =
✓
a
p
2
◆2
=0
+
✓
)
a
p ±r
2
cos ✓+
cos ✓
=
,
b2+
b2
◆2
= a2 ±
p
2 ar + r2 .
p
p
(a/ 2) + r
(a/ 2) r
p
p
=
.
(a2 + 2 ar + r2 )3/2
(a2
2 ar + r2 )3/2
To simplify, let w ⌘
p
2 r/a; then
1+w
=
(2 + 2w + w2 )3/2
(2
1 w
,
2w + w2 )3/2
or
(1 + w)2 (2
2w + w2 )3 = (1
w)2 (2 + 2w + w2 )3 .
Multiplying out the left side:
8
8w
4w2 + 16w3
10w4
2w5 + 7w6
4w7 + w8 = (same thing with w !
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w).
50
CHAPTER 2. ELECTROSTATICS
The even powers cancel, leaving
8w
16w3 + 2w5 + 4w7 = 0,
or
4
8v + v 2 + 2v 3 = 0,
where v ⌘ w2 . According to Mathematica, this cubic equation has one negative root, one root that is spurious
(the point lies outside the square), and v = 0.598279, which yields
r
v
r=
a = 0.546936 a .
2
y
(-a cos(2p/5), a sin(2p/5))
(a cos(p/5), a sin(p/5))
b
a
For the pentagon:
Ex =
where
cos ✓ =
q
4⇡✏0
c
q
r
✓
x
1
cos ✓
+2 2
(a + r)2
b
a cos(2⇡/5) + r
,
b
2
f
cos
2
=
cos
c2
◆
= 0,
a cos(⇡/5)
c
r
;
2
b2 = [a cos(2⇡/5) + r] + [a sin(2⇡/5)] = a2 + r2 + 2ar cos(2⇡/5),
c2 = [a cos(⇡/5)
2
2
r] + [a sin(⇡/5)] = a2 + r2
2ar cos(⇡/5).
1
r + a cos(2⇡/5)
r
+2
+2
3/2
2
2
2
(a + r)2
[a + r + 2ar cos(2⇡/5)]
[a + r2
a cos(⇡/5)
3/2
2ar cos(⇡/5)]
= 0.
Mathematica gives the solution r = 0.688917 a.
For an n-sided regular polygon there are evidently n such points, lying on the radial spokes that bisect
the sides; their distance from the center appears to grow monotonically with n: r(3) = 0.285, r(4) = 0.547,
r(5) = 0.689, . . . . As n ! 1 they fill out a circle that (in the limit) coincides with the ring of charge itself.
Problem 2.59 The theorem is false. For example, suppose the conductor is a neutral sphere and the external
field is due to a nearby positive point charge q. A negative charge will be induced on the near side of the sphere
(and a positive charge on the far side), so the force will be attractive (toward q). If we now reverse the sign of
q, the induced charges will also reverse, but the force will still be attractive.
If the external field is uniform, then the net force on the induced charges is zero, and the total force on the
conductor is QEe , which does switch signs if Ee is reversed. So the “theorem” is valid in this very special case.
protected under all copyright laws as they currently exist. No portion of this material may be
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51
CHAPTER 2. ELECTROSTATICS
Problem 2.60 The initial configuration consists of a point charge q at the center, q induced on the inner
surface, and +q on the outer surface. What is the energy of this configuration? Imagine assembling it piece-bypiece. First bring in q and place it at the origin—this takes no work. Now bring in q and spread it over the
surface at a—using the method in Prob. 2.35, this takes work q 2 /(8⇡✏0 a). Finally, bring in +q and spread it
over the surface at b—this costs q 2 /(8⇡✏0 b). Thus the energy of the initial configuration is
Wi =
q2
8⇡✏0
✓
1
a
1
b
◆
.
The final configuration is a neutral shell and a distant point charge—the energy is zero. Thus the work
necessary to go from the initial to the final state is
W = Wf
Wi =
q2
8⇡✏0
✓
1
a
1
b
◆
.
Problem 2.61
y
R
(2pj/n)
rj
x
R
Suppose the n point charges are evenly spaced around the circle, with the jth particle at angle j(2⇡/n).
According to Eq. 2.42, the energy of the configuration is
1
Wn = n qV,
2
where V is the potential due to the (n
1) other charges, at charge # n (on the x axis).
n
X1 1
1
V =
q
,
4⇡✏0 j=1 r j
r
j
= 2R sin
✓
j⇡
n
◆
(see the figure). So
Wn =
n 1
q2 n X
1
q2
=
⌦n .
4⇡✏0 R 4 j=1 sin(j⇡/n)
4⇡✏0 R
protected under all copyright laws as they currently exist. No portion of this material may be
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52
CHAPTER 2. ELECTROSTATICS
Mathematica says
9
⌦10 =
10 X
1
= 38.6245
4 j=1 sin(j⇡/10)
10
⌦11 =
11 X
1
= 48.5757
4 j=1 sin(j⇡/11)
11
⌦12 =
If (n
12 X
1
= 59.8074
4 j=1 sin(j⇡/12)
1) charges are on the circle (energy ⌦n
Wn = [⌦n
1q
1
2
/4⇡✏0 R), and the nth is at the center, the total energy is
+ (n
1)]
q2
.
4⇡✏0 R
For
n = 11 :
⌦10 + 10 = 38.6245 + 10 = 48.6245 > ⌦11
n = 12 :
⌦11 + 11 = 48.5757 + 11 = 59.5757 < ⌦12
Thus a lower energy is achieved for 11 charges if they are all at the rim, but for 12 it is better to put one at
the center.
protected under all copyright laws as they currently exist. No portion of this material may be
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53
CHAPTER 3. POTENTIAL
Chapter 3
Potential
Problem 3.1
p
The argument is exactly the same as in Sect. 3.1.4, except that since z < R, z 2 + R2 2zR = (R z),
q
1
1 q
instead of (z R). Hence Vave =
[(z + R) (R z)] =
. If there is more than one charge
4⇡✏0 2zR
4⇡✏0 R
1 Qenc
inside the sphere, the average potential due to interior charges is
, and the average due to exterior
4⇡✏0 R
Qenc
charges is Vcenter , so Vave = Vcenter + 4⇡✏
. X
0R
Problem 3.2
A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV .
But we know that Laplace’s equation allows no local minima for V . What looks like a minimum, in the figure,
must in fact be a saddle point, and the box “leaks” through the center of each face.
Problem 3.3
Laplace’s equation in spherical coordinates, for V dependent only on r, reads:
✓
◆
1 d
dV
dV
dV
c
c
r2 V = 2
r2
= 0 ) r2
= c (constant) )
= 2 ) V =
+ k.
r dr
dr
dr
dr
r
r
Example: potential of a uniformly charged ✓
sphere.◆
1
d
dV
dV
dV
c
In cylindrical coordinates: r2 V =
s
=0)s
=c)
=
)
V = c ln s + k.
s ds
ds
ds
ds
s
Example: potential of a long wire.
Problem 3.4
Refer to Fig. 3.3, letting ↵ be the angle between r and the z axis. Obviously, Eave points in the
direction, so
I
Z
1
1
q
1
Eave =
E
da
=
ẑ
2
2
r 2 cos ↵ da.
4⇡R
4⇡R 4⇡✏0
By the law of cosines,
R2 = z 2 +
r
2
r
2
= R2 + z 2
2 r z cos ↵
2Rz cos ✓
)
)
r 2 R2 ,
2r z
cos ↵
z 2 + r 2 R2
z
=
=
2
3
2
r
2z r
(R + z 2
cos ↵ =
z2 +
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reproduced, in any form or by any means, without permission in writing from the publisher.
R cos ✓
.
2Rz cos ✓)3/2
ẑ
54
Eave =
=
Z
q
z R cos ✓
R2 sin ✓ d✓ d
2
2
2
16⇡ R ✏0
(R + z 2 2Rz cos ✓)3/2
Z ⇡
Z
qẑ
z R cos ✓
qẑ
sin
✓
d✓
=
8⇡✏0 0 (R2 + z 2 2Rz cos ✓)3/2
8⇡✏0
ẑ
(where u ⌘ cos ✓). The integral is
1
I = p
2
R R + z2
✓
1
1
=
R |z R|
So
CHAPTER 3. POTENTIAL
1
2Rzu 1
◆
1
z+R
1
2Rz 2
1
1
z Ru
du
(R2 + z 2 2Rzu)3/2
✓p
R2 + z 2

1
|z
2Rz 2
◆ 1
R2 + z 2
2Rzu + p
R2 + z 2 2Rzu
1
✓
◆
1
1
(z + R) + (R2 + z 2 )
.
|z R| z + R
R|
(a) If z > R,
✓
◆

✓
1
1
1
1
1
2
2
I =
(z
R)
(z
+
R)
+
(R
+
z
)
R z R z+R
2Rz 2
z R
✓
◆

1
2R
1
2R
2
=
2R + (R2 + z 2 ) 2
= 2.
R z 2 R2
2Rz 2
z
R2
z
1
z+R
◆
1 q
ẑ,
4⇡✏0 z 2
the same as the field at the center. By superposition the same holds for any collection of charges outside the
sphere.
(b) If z < R,
✓
◆

✓
◆
1
1
1
1
1
1
2
2
I =
(R z) (z + R) + (R + z )
R R z
z+R
2Rz 2
R z
z+R
✓
◆

1
2z
1
2z
= 0.
=
2z + (R2 + z 2 ) 2
R R2 z 2
2Rz 2
R
z2
Eave =
So
Eave = 0.
By superposition the same holds for any collection of charges inside the sphere.
Problem 3.5
H
Same as proof of second uniqueness theorem, up to the equation S V3 E3 · da =
each
surface, either V3 = 0 (if V is specified on the surface), or else E3? = 0 (if @V
@n =
R
(E3 )2 = 0, and hence E2 = E1 . qed
V
Problem 3.6
identity:
Z Putting U = T = V3 into Green’s
I
⇥
⇤
2
V3 r V3 + rV3 · rV3 d⌧ =
V3 rV3 · da. But r2 V3 = r2 V1
V Z
S
I
So
E32 d⌧ =
V3 E3 · da, and the rest is the same as before.
V
r2 V 2 =
R
(E3 )2 d⌧ . But on
E? is specified). So
V
⇢
⇢
+
= 0, and rV3 =
✏0 ✏0
S
Problem 3.7
Place image charges +2q at z = d and q at z = 3d. Total force on +q is

✓
◆
✓
◆
q
2q
2q
q
q2
1 1
1
1
29q 2
F=
+
+
ẑ
=
+
ẑ
=
ẑ.
4⇡✏0 (2d)2
(4d)2
(6d)2
4⇡✏0 d2
2 8 36
4⇡✏0 72d2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
E3 .
55
CHAPTER 3. POTENTIAL
Problem 3.8
(a) From Fig. 3.13:
q0
r
0
=
=
Therefore:
r
=
p
r2 + a2
2ra cos ✓;
r
0
=
p
r 2 + b2
2rb cos ✓.
Therefore:
R
q
R2
p
(Eq. 3.15), while b =
(Eq. 3.16).
a r2 + b2 2rb cos ✓
a
q
q
q
= q
.
4
2
2
a
R
R
ar
2+
2
r
2r
cos
✓
+
R
2ra
cos
✓
R
a2
a
R
1
V (r, ✓) =
4⇡✏0
✓
q
r
+
q0
r
0
◆
q
=
4⇡✏0
(
p
1
r2 + a2
1
2ra cos ✓
p
R2 + (ra/R)2
Clearly, when r = R, V ! 0.
@V
(b) = ✏0 @V
(Eq. 2.49). In this case, @V
@n
@n = @r at the point r = R. Therefore,
✓
◆⇢
q
1 2
(✓) = ✏0
(r + a2 2ra cos ✓) 3/2 (2r 2a cos ✓)
4⇡✏0
2
✓ 2
◆
1
a
3/2
2
2
+
R + (ra/R)
2ra cos ✓
2r 2a cos ✓
2
R2
r=R
⇢
q
=
(R2 + a2 2Ra cos ✓) 3/2 (R a cos ✓) + R2 + a2 2Ra cos ✓
4⇡

q
a2
=
(R2 + a2 2Ra cos ✓) 3/2 R a cos ✓
+ a cos ✓
4⇡
R
q
=
(R2 a2 )(R2 + a2 2Ra cos ✓) 3/2 .
4⇡R
Z
Z
q
qinduced =
da =
(R2 a2 ) (R2 + a2 2Ra cos ✓) 3/2 R2 sin ✓ d✓ d
4⇡R

⇡
q
1
2
2
2
=
(R
a )2⇡R
(R2 + a2 2Ra cos ✓) 1/2
4⇡R
Ra
0

q 2
1
1
p
=
(a
R2 ) p
.
2a
R2 + a2 + 2Ra
R2 + a2 2Ra
p
But a > R (else q would be inside), so R2 + a2 2Ra = a R.

q 2
1
1
q
q
=
(a
R2 )
=
[(a R) (a + R)] =
( 2R)
2a
(a + R) (a R)
2a
2a
=
2ra cos ✓
3/2
✓
a2
R
qR
= q0 .
a
(c) The force on q, due to the sphere, is the same as the force of the image charge q 0 , to wit:
✓
◆
1
qq 0
1
R 2
1
1
q 2 Ra
F =
=
q
=
.
4⇡✏0 (a b)2
4⇡✏0
a
(a R2 /a)2
4⇡✏0 (a2 R2 )2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
)
.
a cos ✓
◆
4
4
CHAPTER 3. SPECIAL TECHNIQUES
CHAPTER 3. SPECIAL TECHNIQUES
(c) The force on q, due to the sphere, is the same as the force of the image charge q ′ , to wit:
!
"
′
(c) The force on q, due
charge
CHAPTER
POTENTIAL
qq ′ sphere,1is the same
R 2 as the 1force of the image
q 2 Ra q , to3.wit:
1 to56the
1
−
F =
=
q
=
−
.
!
"
2
2
2
2
2
2
′
2
4πϵ01 (a −qq
b)
4πϵ01
a R (a − R 1/a)
4πϵ01 (a −q RRa)
− q2
F =
=
=−
.
2
2
2
− b)we
4πϵ
a
(a − R /a)
4πϵ0 (a2 − R2 )2
0 (a
0
To bring
bring qq in
in from
from infinity
infinity 4πϵ
to a,
a,
then,
we do
do
work
To
to
then,
work
To bring q in from infinity
%&a
\$
#aa to a, then, we do work
2
2
2
&a
q
R  11
R
2RZ
q R #a aa
qq2 R
11
qq2 R
&%&a= − 11
W
=
da
=
−
.
\$
2
2
&
2
2
2
2R2 ).
W = 4πϵ
da
=
=
2
2
2
&
4πϵ
2
4πϵ
2(a
−
(a
−
R
)
(a
−
R
)
2
2
q
R
a
q
R
1
1
1
q
R2 )
2
0
0
0
2
2
2
∞&
4⇡✏
4⇡✏
2
4⇡✏
2(a
R
(a
R
)
(a
R
)
0
0
0
1
W =
da =
= −
−
.
∞
4πϵ10
4πϵ0
2 (a2 − R2 ) &∞
4πϵ0 2(a2 − R2 )
(a2 − R2 )2
∞
Problem 3.8
Problem
3.9
2
CHAPTER 3. SPECIAL TECHNIQUES
Place a 3.8
second image charge,
q ′′ , at the center of the sphere;
Problem
00
′′at the
this
will
alter
thecharge,
fact
that
is an
Place
a second
image
q , qthe
center
of equipotential,
thethe
sphere;
a−b
Place
a not
second
image
charge,
, atsphere
the
center
of
sphere;
*'
(
)
1 q ′′
this
will
not
alter
the
fact
that
the
sphere
is
an
equipotential,
this
will
not
alter
the
fact
that
the
sphere
is
an
equipotential,
a−b
but merely increase that potential from zero to V0 =
00 ;′′
′′
′
*'
)
1
q
q
q
q(
4πϵ01 Rq
but′′but
merely
increase
thatthat
potential
from
zerozero
to to
V0 V=
; ;
' ′′
()
*
merely
increase
potential
from
0 =
′
4⇡✏4πϵ
q
qa
q
q = 4πϵ0 V0 R at center of sphere.
0 R0 R
'
()
*
′′ 4⇡✏0 V0 R at center of sphere.
q 00 =
q
=
4πϵ
V
R
at
center
of
sphere.
a
0
For a neutral0 sphere,
q ′ + q ′′ = 0.
! ′′
!
"
0 ′ ′ 00 ′′"
′
ForFor
a neutral
q +
0. 0. qq
a neutral
sphere,
q qq+ q= =
1 sphere,
1
q
1
F =
q ✓ !2 ′′+
=
−
"
! 2+
"
◆
✓
2
′
′
0 b)
0qq
4πϵ
a00 q (a −
4πϵ
a1 1 (a −
b)
q
1 2◆
01
0
1
q
q
qq
1
q +
− +
+
+ 2
F F= =qq4πϵ
= = 4πϵ
′ q
2
2b)2
2 a2−(a
−
−2b)2
−2 b)(a (a b)
q(−Rq/a)
/a)(2a
R(a/a)
0b(2a2 a
4⇡✏(R
b)
0 0 a
= 4⇡✏0 ′ 2 a
=
2
2
2
2
2
2
2a (a
4πϵ
(a
− b)
4πϵ0
− R /a)
b(2a
− b) q( q(−Rq/a)
(R
/a)(2a
−2R
0 a
qq 0qq
b(2a
b)
R
/a)/a)
= Rq/a) (R /a)(2a
= = 4πϵ 2 a2!(a −
=
2
2 (a −2R2 /a)
2
"
2
2
2
b)
4πϵ
a
3
4⇡✏0 qa02 (a Rb) (2a2 −
4⇡✏
R02 ) 0 a (a R /a)
.
=−
"
!
3 2 2 2 22
2✓ ◆3 (a
2− R
4πϵ
(2a
−2)R
q 2 q0 Ra R (2a
R
) ).
=
−
=
2
2. 2
a (a2(a R
−2R
4⇡✏4πϵ
)2 )
0 0 abecause
(Drop the minus
sign,
the problem
asks for the force of attraction.)
(Drop the
minus sign, because the problem asks for the force of attraction.)
Problem
3.9
(Drop
the minus
sign, because the problem asks for the force of attraction.)
Problem3.9
3.9
Problem
(a) Image 3.10
problem: λ above, −λ below. Potential was found in Prob. 2.47:
Problem
z
(a)(a)
Image
problem:
below.
was
found
in in
Prob.
2.52:
Image
λ above, −λ
below.
Potential
was
found
Prob.
2.47:
z
2λ above,
λ Potential
✻problem:
V (y, z) z=
ln(s− /s+ ) =
ln(s2− /s2+ )
✻
zy
4πϵ
4πϵ
z s+ (y, z)
2λ
λ
0
0
✻
2
2
+
2
V (y,✻
z)✣ = y
ln(s
/s
)
=
ln(s
/s
)
✻
−
+
λ
2
2
+
− +/s ) =
y4πϵ
V (z
(y,+
z)d)
=2 ,
ln(s
ln(s /s+ )
4πϵ
+
ss+− (y, z)
0
d
λ 0✲ y 2 +
✣
✣
4⇡✏
4⇡✏
✲y
+
0
=
lnλλx+2 2
,0
⇢
2
2
d
2
2
d
4πϵ0λ ✲
y
+
(z
−
d)
y
+
(z
+
d)
s
− ✲
y + (z + d)
✲xx
−
y
=
ln
d
4πϵ0
y 2 + (z − d)2 = 4⇡✏0 ln y 2 + (z d)2
∂V
∂V
∂V
−
(b) σ = −ϵ0
. Here
=
, evaluated at z = 0.
∂n∂V
∂n∂V ∂z∂V
+
,&
(b) σ = −ϵ@V
.
Here
=
, evaluated at z = 0.
0
&
∂n λ @V∂n @V1∂z
&,&
(b) =
at −
z = 0. 1
σ(y)✏[email protected]−ϵ. 0Here @n+2= @z , evaluated
2(z
+
d)
2(z
−
d)
& &
2
2
2
4πϵ0λ y + (z +
y + (z −
1 d)
1 d)
z=0
&
σ(y) = −ϵ0+ ⇢
2(z
+
d)
−
2(z
−
d)
&
4πϵ0d y 2 +1 (z−d
+ d)2,
y 2 +1 (z − d)2
λd
2λ
z=0
(y) =
2(z,
+ d)
2(z d)
−
=
.
= −✏0 4⇡✏ +2 y 2 2+−(z +
2
2
2
2
2
2
2
d)
y
+
(z
d)
4π2λ 0y + d
y +−d
d
π(y +λdd )
z=0
= −
− 2
.
= − ⇢
2
2
2
2
2
4π
y
+
d
y
+
d
π(y
+
d
)
dl parallel to the y axis:
Check: =
Total2 charge dinduced onda strip
= of width
.
2 + d2
2 + d2
2 + d2 )
4⇡
y
y
⇡(y
∞
\$ of width
%&
#
Check: Total charge
induced on a strip
l
parallel
to the y/ axis:- . &∞
1
lλd 1
lλd π
π .0
lλd
−1 y
&
∞
dy
=
−
=
− y axis:
q
=
−
tan
− −
\$
%&
# induced
ind Total charge
∞to the
.
/
&
2
2
Check:
on
a
strip
of
width
l
parallel
&
πlλd y + 1d
πlλd d 1
πlλd 2 π
2 π .0
−1 d y −∞
&
dy
=
−
=
−
−
qind = − −∞
tan
−
&
πZ1 y 2 + d2
π d
2
⇣ ⌘d 1 −∞ l dπh ⇡ 2 ⇣ ⇡ ⌘i
= −λl.
λind = −λ,
l d Therefore
l d as1 it should
−∞ 1
1 y be.
qind =
dy
=
tan
=
⇡
y 2 + d2 λind = ⇡−λ, das it should
d be. 1
⇡ 2
2
= −λl.
Therefore
Chapter 1
Special Techniques
=
l.
1
Therefore
c
⃝2005
protected
under
laws as they currently exist. No portion of this material may be
, as all
should be.
indc =
⃝2005 Pearson
NJ. All
reserved.
This
material is
reproduced,
in anyEducation,
form or by Inc.,
any means,
withoutRiver,
permission
inrights
writing
from the
publisher.
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
CHAPTER 3. SPECIAL TECHNIQUES
CHAPTER 3. POTENTIAL
5
57
Problem3.10
3.11
Problem
Theimage
imageconfiguration
configurationisisasasshown.
shown.
The
y
✻
q
!(
q
1
1
q
1
1
p
(x,y)y)==
"p
VV(x,
++"
✲x
4⇡✏0 0
+zz2 2
4πϵ
(x(x− a)a)2 2++(y(y− b)b)2 2++zz2 2
(x(x++a)a)2 2++(y(y++b)b)2 2+
#)
11
11
q
−q
p
− "p
−"
..
2 + (y
2+z
2
2 + (y + b)
2
2
2
2
(x
+
a)
b)
(x
a)
(x + a) + (y − b) + z
(x − a) + (y + b)2 2++zz2 2
CHAPTER
55
CHAPTER3.3.SPECIAL
SPECIALTECHNIQUES
TECHNIQUES
For this to work, θ must be and integer divisor of 180◦. Thus 180◦, 90◦ , 60◦ , 45◦ , etc., are OK, but no
⇢
2
others. It works for 45◦ , say, qwith
the charges
as1shown.
1
1
45◦ line
F=
x̂
ŷ + p
[cos ✓ x̂ + sin ✓ ŷ] (4)
, (2)
2
2
2
2 )20),
+ −
4⇡✏0 the(2a)
(Note the strategy: to make
x axis an(2b)
equipotential
(2 a (V
+ b=
Problem
3.10p
Problem
p reflection point. To make the
you
place the3.10
image
charge (1) in the
yy
where
cos
✓
=
a/
a2 + b2 , is
sin
✓asshown.
=shown.
b/ a2 + b2 .
The
image
configuration
as
◦
−
The
image
configuration
is
+ ✻✻
45 line an equipotential, you place charge (2) at the image point.
✲x
⇢

−q
qq
+
− (1)
−q
2
But that screws up !the
x
axis,
so
you
must
now
insert
image
(3)
to
q
a
1
b
1
!
◦
F
=
x̂
+
ŷ
.
q
1
1
balance (2). Moreover,
to make the 1452 line2 V
= 0 ayou
also need
(4),
2
b2
(a + b )3/2 ++
(a2 +1 b2 )3/2
"
VV
(x,(x,
y)y)== q "" 16⇡✏0
− +
✲✲x
"0 you 2need
2+
2 2V =
2+
2 2
to balance
(1).4πϵ
But
restore
the
x 2axis
to
(5)
x
(5) (3)
2
2
2
2
0 0now,
(xto
−
a)
(y
−
b)
+
z
(x
+
a)
+
(y
+
b)
z
4πϵ
(x − a) + (y − b) + z
(x + a) + (y + b) #+
z
#

to balance (4), and so on.  2
◦
2
2
2
why1it works
1 q
1q1
1 1
q
q
1
1q for θ = 45
−q−q
q.
−
. .
p
W−"
="
+ 2 +2 −p
= 1
−""
2
2
2
2
2
2
2
2
◦
2
4⇡✏
(2a)
(2b)
16⇡✏
a
b ⇐No good
2+
2+
2
2+
2+
2+ b line
(x
+
a)
(y
−
b)
+
z
(x
−
a)
+
(y
+
b)
z
(2
a
+
b
)
a
0a)+
0b)+
135
(x
+
(y
−
b)
z
(x
−
a)
(y
+
z
■
The reason this doesn’t work for arbitrary angles is that you are even+
(3) (0)
◦region
◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦
tually
forced
to
place
an
image
charge
within
the
original
of
◦
+ are
θ θmust
divisor
ofof180
. .Thus
, etc.,
ForFor
this
to to
work,
✓ must
bebe
an
integer
divisor
of 180
. 180
Thus
180180
,180
90, 90
,, 60
,, 60
45, 45
etc.,
are
OK,
butbut
no
Forthis
this
towork,
work,
must
beand
andinteger
integer
divisor
Thus
90, 60
,, 45
, etc.,
areOK,
OK,
butnono
✲x
interest,
and
that’s
not
images
must
go
outside
the
re◦ allowed—all
◦say,
others.
It
works
for
45
,
with
the
charges
as
shown.
others.
It
works
for
45
,
say,
with
the
charges
as
shown.
◦
others. It works for 45 , say, with the charges as shown.
◦
4545line
−
gion, or you’re no longer dealing with the same problem at all.)
line
(4)(4)
(2)(2)
(1)
−
(Note
thethe
strategy:
to to
make
thethe
xtheaxis
an an
equipotential
(V (V=
0),=0),0),
(Note
x xaxis
(Note
thestrategy:
strategy:
tomake
make
axis
anequipotential
equipotential
(V=
(2) + +− −
youyou
place
thethe
image
charge
(1)(1)
in
the
reflection
point.
ToTo
make
thethe
charge
point.
youplace
place
theimage
image
charge
(1)in
inthe
thereflection
reflection
point.
Tomake
make
the why it doesn’t work for θ = 135◦
◦ ◦ an equipotential, you place charge (2) at the image point.
−−
++✲
45 45line
line
an
equipotential,
you
place
charge
(2)
at
the
image
point.
453.11
line an equipotential, you place charge (2) at the image point.
✲x x
Problem
++
− (1)
ButBut
that
screws
up
the
x
axis,
so
you
must
now
insert
image
(3)
to
− (1)
that
screws
up
the
x
axis,
so
you
must
now
insert
image
(3)
to
%
\$
But that screws up the x axis, so you
must
now
2insert
2 image (3) to
√
λ ◦ line
(x
+you
a) also
+ y need
balance
(2).(2).
Moreover,
make
the
=
(4),(4),
balance
Moreover,
to
make
V V0=
From
Prob.
2.47
(with
y0to→
d):
V the
=45
where
a2 = y0 2 − R2 ⇒ −a += d2 − R2 ,
, need
balance
(2).
Moreover,
to
make
the45line
45◦lnV
line
=0 0you
youalso
also
need
(4),
2
2
(x to
a)
y need
− +
0axis
to to
balance
(1).(1).
ButBut
now,
to to
restore
the4πϵ
xthe
to
V−to
=
(5)(5)
now,
the
x xaxis
V V0=+you
(5)(5)
(3)(3)
tobalance
balance
(1).
But
now,
torestore
restore
axis
=0 0you
youneed
need
(5)
and to to
balance
(4),
and
so
on.
balance
(4),
and
so
on.
◦
to&balance (4), and so on. '
◦
why
it
works
forfor
θ=
4545
)
(
why it works
θ=
d
2πϵ0 V0
2πϵ0 V0
a coth(2πϵ0 V0 /λ) = d
No good
⇒ (dividing)
λ=
, or 135
= cosh
◦
◦
−1 + ⇐ . No good
■■
areason
csch(2πϵ
Vdoesn’t
=work
R
135line
line
The
reason
angles
is isthat
evenRis that
λeven0doesn’t
0 /λ)
cosh
(d/R)
The
this
work
forarbitrary
arbitrary
angles
that
youare
are
even+ ⇐
The
reason
thisthis
doesn’t
work
forfor
arbitrary
angles
youyou
are
(3)
(3) (0)(0)
tually
forced
charge
within
region
++
tually
forced
toplace
place
animage
image
charge
within
theoriginal
original
region
tually
forced
to to
place
an an
image
charge
within
thethe
original
region
of ofof
✲✲x
interest,
and
that’s
not
allowed—all
images
must
go
outside
the
reinterest,
and
that’s
not
allowed—all
images
must
go
outside
the
reinterest,
and
that’s
not
allowed—all
images
must
go
outside
the
rex
Problem 3.12
−
gion,
dealing
with
problem
−
gion,
oryou’re
you’re
nolonger
longer
dealing
with
thesame
same
problem
atall.)
all.)
gion,
or or
you’re
no no
longer
dealing
with
thethe
same
problem
at at
all.)
(1)(1)
−−
(2)(2)
+a
∞
*
2
◦
−nπx/a
why
it
doesn’t
for3.34).
θ=
135
why it doesn’t
work
for
θ=
135◦
V0 (y) sin(nπy/a)
dy work
(Eq.
V (x, y) =
Cn e
sin(nπy/a) (Eq. 3.30), where Cn =
a
Problem 3.12
n=1

Problem
3.11
0
Problem
3.11
2
p
(x\$ +
+2 y22 2 %2 %
\$ a)
2
2 √ R2 ,
(x
+
a)
+
y
From Prob. 2.52&(with y0 ! d): V = ' λlnλ
,
where
a2 =2 y20 2 2 R
)
a
=
d√
(x
+
a)
+
y
2
2,2
2
2
2 R ⇒
2
2−
From
(with
V V=
a a==y0y0−
a a== d2d−
RR
ln(x
4⇡✏
+2 y 2 , ,where
FromProb.
Prob.2.47
2.47
d):a/2
=4πϵ
where
,
ln (x a)
−
R
⇒
0
+V
, fory00y→
yd):
<
0<→
0(with
2
2
−−
a)a)++
yy
0 0
4πϵ
(x
In this case V0 (y) =
. Therefore,
and
−V
,
for
a/2
<
y
<
a
0
✓
◆
and
and ⇢ a&coth(2⇡✏ V / ) = d '
2⇡✏
2⇡✏0 V0
((0 V0 ))
0 0
&⎧
'(dividing) d ⎫=
)
cosh
, 0or
=
d
2πϵ
2πϵ
V
a
coth(2πϵ
V
/λ)
=
d
0 V00V.03
0
0
0
0
!
#
d
2πϵ
V
1 2πϵ
a/2
a
coth(2πϵ
V
/λ)
=
d
a
0
0
0
⎪+
3a/2λ λ=
a csch(2⇡✏
+ ⇒⇒(dividing)
R⎪
0 V0 / ) = R
, ,oror
cosh
cosh
(d/R)
⎬ ==
(dividing)
=
cosh
−1
3
3a . .
cos(nπy/a)
2V
2 a⎨csch(2πϵ
cos(nπy/a)
−1
V
/λ)
=
R
RR 0
λλ
0 00V0 /λ) = R
cosh
(d/R)
a
csch(2πϵ
3
3
cosh (d/R)
sin(nπy/a) dy =
−
Cn = V0
sin(nπy/a) dy −
+
⎪
⎪
a ⎩
a
(nπ/a) 30
(nπ/a) 3a/2
⎭
Problem 3.13
0
a/2
Problem
3.12
Problem
3.12
4
5 nπ 6
5 nπ 67 2V 4 Za
5 nπ 67
1
2VX
0
0 2
n
n⇡x/a + cos(0) + cos(nπ) − cos
−
cos
=
1
+
(−1)
−
2
cos
.
V (x, =
y) =
C
e
sin(n⇡y/a)
(Eq.
3.30),
where
C
=
V
(y)
sin(n⇡y/a)
dy (Eq.
3.34).
n
n
∞∞
nπ *
2
2
nπ a 2 +a0+a
2
n=1 *
2
0
V0V(y)
sin(nπy/a) dy (Eq. 3.34).
VV
(x,(x,
y)y)== CnCen−nπx/a
sin(nπy/a)
e−nπx/a
sin(nπy/a) (Eq.
(Eq.3.30),
3.30), where
where CnCn==
0 (y) sin(nπy/a) dy (Eq. 3.34).
c
⃝2005
Pearson Education,
n=1
n=1
−q
0 0
protected
under all
laws Upper
currently
exist.
of this material
may beis
c 2012 Pearson
Education,
River,
NJ. No
Allportion
rights reserved.
This material
reproduced,
in anyall
form
or by &
any
means,
without
permission
in
writing
from
the
publisher.
protected under
laws
as
they
currently
exist.
No
portion
of
this
material
may
be
''
& means, without permission
reproduced, in any form or by any
writing from the publisher.
+V
0 , 0for
+V
, for0 0<<y y<<a/2
a/2 . in
InInthis
case
V
(y)
=
Therefore,
this case 0V0 (y) = −V0 , for a/2 < y < a
. Therefore,
−V0 , for a/2 < y < a
⎧⎧
⎫⎫
!!
a/2
a a
⎪
⎪
3a/2
3a3 ##
a/2
+
+
⎪
⎨
⎬⎪
+
+
⎬ 2V2V
3 33a/2 cos(nπy/a)
3 3a
cos(nπy/a)
22 ⎨
0 0
cos(nπy/a)
cos(nπy/a)
3
3 3
3 ++
dydy ==
−−
CnCn== V0V0
sin(nπy/a)
dydy
−− sin(nπy/a)
sin(nπy/a)
sin(nπy/a)
3
3 3
3
⎪⎪
⎪
aa ⎩
aa
(nπ/a)
(nπ/a)
⎪
(nπ/a)
(nπ/a)
⎭
0 0
a/2
⎩
⎭
a/2
0 0
a/2
a/2
4
5
6
5
67
4
5
67
5 nπ 6
5 nπ 67 2V2V
4
5 nπ 67
2V0 4
nπ
nπ
nπ
0
nn
== 2V0 −−
cos
++
cos(0)
++
cos(nπ)
−−
cos
== 0 1 +
(−1)
−−
2 cos
. .
cos
cos(0)
cos(nπ)
cos
1
+
(−1)
2
cos
nπnπ
22
22
nπnπ
22
58
⇢
In this case V0 (y) =
CHAPTER 3. POTENTIAL
+V0 , for 0 < y < a/2
V0 , for a/2 < y < a
8
> Za/2
2 <
Cn = V0
sin(n⇡y/a) dy
a >
:
Za
0
=
a/2
. Therefore,
2V0
sin(n⇡y/a) dy =
>
a
;
⇣ n⇡ ⌘
2V0 n
cos
+ cos(0) + cos(n⇡)
n⇡
2
The term in curly
8
>
>n = 1
<
n=2
n=3
>
>
:
n=4
9
>
=
cos
brackets is:
:
:
:
:
⇢
Cn =
So
8V0 /n⇡,
0,
8V0
⇡
V (x, y) =
2
=
a/2
cos(n⇡y/a)
(n⇡/a)
0
2V0 n
1 + ( 1)n
n⇡
cos(n⇡y/a)
+
(n⇡/a)
2 cos
⇣ n⇡ ⌘o
2
a
a/2
)
.
9
2 cos(⇡/2) = 0, >
>
=
2 cos(⇡) = 4,
etc. (Zero if n is odd or divisible by 4, otherwise 4.)
2 cos(3⇡/2) = 0, >
>
;
2 cos(2⇡) = 0,
1 1
1+1
1 1
1+1
Therefore
⇣ n⇡ ⌘o
(
X
n = 2, 6, 10, 14, etc. (in general, 4j + 2, for j = 0, 1, 2, ...),
otherwise.
e
n⇡x/a
n=2,6,10,...
1
sin(n⇡y/a)
8V0 X e
=
n
⇡ j=0
sin[(4j + 2)⇡y/a]
.
(4j + 2)
(4j+2)⇡x/a
Problem 3.14
V (x, y) =
So
(y) =
✏0
@
@x
4✏0 V0
=
a
4V0
⇡
⇢
X
1
e
n
n=1,3,5,...
n⇡x/a
4V0 X 1 n⇡x/a
e
sin(n⇡y/a)
⇡
n
X
sin(n⇡y/a).
(Eq. 3.36);
=
✏0
x=0
=
✏0
@V
@n
4V0 X 1 n⇡
(
)e
⇡
n
a
(Eq. 2.49).
n⇡x/a
sin(n⇡y/a)
x=0
n=1,3,5,...
Or, using the closed form 3.37:
✓
◆
2V0
sin(⇡y/a)
1
V (x, y) =
tan
)
⇡
sinh(⇡x/a)
=
sin(n⇡y/a)
=
2✏0 V0 sin(⇡y/a) cosh(⇡x/a)
a sin2 (⇡y/a) + sinh2 (⇡x/a)
2V0
✏0
⇡ 1+
=
x=0
1
sin2 (⇡y/a)
sinh2 (⇡x/a)
✓
sin(⇡y/a)
sinh2 (⇡x/a)
◆
⇡
cosh(⇡x/a)
a
x=0
2✏0 V0
1
.
a sin(⇡y/a)
[Comment: Technically, the series solution for is defective, since term-by-term di↵erentiation has produced
a (naively) non-convergent sum. More sophisticated definitions of convergence permit one to work with series
of this form, but it is better to sum the series first and then di↵erentiate (the second method.)]
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
59
CHAPTER 3. POTENTIAL
Summation of series Eq. 3.36
V (x, y) =
Now sin w = Im eiw , so
where Z ⌘ e
⇡(x iy)/a
. Now
X 1
4V0
I, where I ⌘
e
⇡
n
n=1,3,5,...
I = Im
X1
e
n
n⇡x/a in⇡y/a
e
n⇡x/a
= Im
sin(n⇡y/a).
X1
Z n,
n
8
9
Z Z <X
1
1
=
X 1
X
1
Zn =
Z (2j+1) =
u2j du
;
n
(2j + 1)
0 : j=0
1,3,5,...
j=0
✓
◆
1
1
1+Z
1
1
=
du = ln
= ln Rei ✓ = (ln R + i ✓)7 ,
CHAPTER 3. SPECIAL
1 TECHNIQUES
u2
2
1 Z
2
2
ZZ
0
1+Z
i ✓i θ =1+Z
whereReRe
. Therefore
where
= 1 1−Z
Z . Therefore
soso
#
\$#
\$
⇢!
"
iy)/a
−π(x−iy)/a
−π(x+iy)/a
iy)/a
−π(x−iy)/a
1+
e e⇡(x
11−
e e⇡(x+iy)/a
1+
11
11
1+
ZZ 1 +
e e⇡(x
1+
1+
\$#
\$
I I==Im
RR
++
i ✓)
==
== #
i θ) == ✓.θ. But
But
Im (ln(ln
iy)/a
−π(x−iy)/a
iy)/a
−π(x−iy)/a
−π(x+iy)/a
22
22
11−
ZZ 1 1 −
e e⇡(x
11−
e e⇡(x
11−
e e⇡(x+iy)/a
#
\$
−πx/a i⇡y/a
−iπy/a
−2πx/a
−πx/a
−2πx/a
1+
e e⇡x/a
e eiπy/a −
e ei⇡y/a
e e2⇡x/a
1+
−
1+
2ie
sin(⇡y/a)
e e2⇡x/a
1+
2ie⇡x/a
sin(πy/a) −
==
==
, ,
%
%
%2
%2
2
2
iy)/a %
iy)/a %
−π(x−iy)/a
−π(x−iy)/a
1%1 −
e e⇡(x
1%1 −
e e⇡(x
Therefore
Therefore
−πx/a
2e2e⇡x/a
sin(⇡y/a)
2 sin(⇡y/a)
sin(⇡y/a)
sin(πy/a)
2 sin(πy/a)
sin(πy/a)
tan
✓ θ==
== ⇡x/a
==
. .
tan
−2πx/a
−πx/a sinh(⇡x/a)
11−
e e2⇡x/a
e eπx/a −
e e⇡x/a
sinh(πx/a)
✓&
◆'
✓&
◆'
1
0 0
sin(πy/a) , and V (x, y) = 2V2V
sin(πy/a) .
1−1 sin(⇡y/a)
1−1 sin(⇡y/a)
I I== 1tan
tan
, and V (x, y) = ⇡
.
tan sinh(⇡x/a)
2 2 tan sinh(⇡x/a)
sinh(πx/a)
π
sinh(πx/a)
Problem
Problem3.15
3.14
2
@2V
@ 2V2
∂
V
with
boundary
conditions
(a)(a) 2 + + ∂2 V= 0,
= 0,
with
boundary
conditions
@x∂x2 @y∂y 2
8⎧
9⎫
(i)(i) V V
(x,
0)0)==0,0, >
>
(x,
>⎪
>
⎪
⎪
<
=⎪
⎨
⎬
(ii)
(x,
a)a)==0,0,
(ii)V V
(x,
(iii)
(0,(0,
y)y)==0,0, >
>
(iii)V V
⎪
>⎪
>
⎪
:
;⎪
⎩
⎭
(iv)
(b,(b,
y)y)==V0V(y).
(iv)V V
0 (y).
y
✻
a
V0 (y)
V =0
b
z✙
✲x
AsAsininEx.
Ex.3.4,
3.4,separation
separationofofvariables
variablesyields
yields
# kxkx
kx \$
VV
(x,
y)y)==AeAe
++
Be
kyky++
DD
cos
ky)
. .
(x,
Be−kx(C(Csinsin
cos
ky)
Here
A, (ii))
Here(i))
(i)⇒DD==0,0,(iii))
(iii)⇒BB== −A,
(ii)⇒kakais isananinteger
integermultiple
multipleofof⇡:π:
⇣/
⌘
n⇡x/a
n⇡x/a 0
nπx/a e −nπx/a sin(n⇡y/a) = (2AC) sinh(n⇡x/a) sin(n⇡y/a).
VV
(x,
y)
=
AC
e
(x, y) = AC e
−e
sin(nπy/a) = (2AC) sinh(nπx/a) sin(nπy/a).
c But
2012 Pearson
Inc., Upper
River,general
NJ. All rights
This material
is
(2AC) Education,
is a constant,
the most
linearreserved.
combination
of separable
protected under all copyright laws as they currently exist. No portion of this material may be
(ii), (iii) inisany form or by any means, without permission in writing from the publisher.
reproduced,
V (x, y) =
∞
1
solutions consistent with (i),
Cn sinh(nπx/a) sin(nπy/a).
n=1
It remains to determine the coeﬃcients Cn so as to fit boundary condition (iv):
1
2
Cn sinh(nπb/a) sin(nπy/a) = V0 (y). Fourier’s trick ⇒ Cn sinh(nπb/a) =
a
2a
V0 (y) sin(nπy/a) dy.
60
CHAPTER 3. POTENTIAL
But (2AC) is a constant, and the most general linear combination of separable solutions consistent with (i),
(ii), (iii) is
1
X
V (x, y) =
Cn sinh(n⇡x/a) sin(n⇡y/a).
n=1
It remains to determine the coefficients Cn so as to fit boundary condition (iv):
X
Cn sinh(n⇡b/a) sin(n⇡y/a) = V0 (y). Fourier’s trick ) Cn sinh(n⇡b/a) =
2
a
Za
V0 (y) sin(n⇡y/a) dy.
0
Therefore
2
Cn =
a sinh(n⇡b/a)
Za
V0 (y) sin(n⇡y/a) dy.
0
2
(b) Cn =
V0
a sinh(n⇡b/a)
Za
0
2V0
sin(n⇡y/a) dy =
⇥
a sinh(n⇡b/a)
V (x, y) =
4V0
⇡
⇢
0, if n is even,
if n is odd.
2a
n⇡ ,
X
sinh(n⇡x/a) sin(n⇡y/a)
.
n sinh(n⇡b/a)
n=1,3,5,...
Problem 3.16
Same format as Ex. 3.5, only the boundary conditions are:
8
9
(i) V = 0 when x = 0, >
>
>
>
>
>
>
(ii) V = 0 when x = a, >
>
>
>
>
<
=
(iii) V = 0 when y = 0,
> (iv) V = 0 when y = a, >
>
>
>
>
>
(v) V = 0 when z = 0, >
>
>
>
>
:
;
(vi) V = V0 when z = a.
This time we want sinusoidal fuctions in x and y, exponential in z:
X(x) = A sin(kx) + B cos(kx),
Y (y) = C sin(ly) + D cos(ly),
p
Z(z) = Ee
k2 +l2 z
+ Ge
p
k2 +l2 z
.
(i)) B = 0; (ii)) k = n⇡/a; (iii)) D = 0; (iv)) l = m⇡/a; (v)) E + G = 0. Therefore
p
Z(z) = 2E sinh(⇡ n2 + m2 z/a).
Putting this all together, and combining the constants, we have:
V (x, y, z) =
1 X
1
X
n=1 m=1
Cn,m sin(n⇡x/a) sin(m⇡y/a) sinh(⇡
p
n2 + m2 z/a).
It remains to evaluate the constants Cn,m , by imposing boundary condition (vi):
i
p
XXh
V0 =
Cn,m sinh(⇡ n2 + m2 ) sin(n⇡x/a) sin(m⇡y/a).
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
61
CHAPTER 3. POTENTIAL
According to Eqs. 3.50 and 3.51:
⇣ p
⌘ ✓ 2 ◆2 Za Za
2
2
Cn,m sinh ⇡ n + m =
V0
sin(n⇡x/a) sin(m⇡y/a) dx dy =
a
0
0
(
0,
if n or m is even,
16V0
, if both are odd.
⇡ 2 nm
)
Therefore
16V0
V (x, y, z) = 2
⇡
p
sinh ⇡ n2 + m2 z/a
1
p
sin(n⇡x/a) sin(m⇡y/a)
.
nm
sinh ⇡ n2 + m2
n=1,3,5,... m=1,3,5,...
X
X
Consider the superposition of six such cubes, one with V0 on each of the six faces. The result is a cube
with V0 on its entire surface, so the potential at the center is V0 . Evidently the potential at the center of the
original cube (with V0 on just one face) is one sixth of this: V0 /6. To check it, put in x = y = z = a/2:
16V0
V (a/2, a/2, a/2) = 2
⇡
p
sinh ⇡ n2 + m2 /2
1
p
sin(n⇡/2) sin(m⇡/2)
.
nm
sinh ⇡ n2 + m2
n=1,3,5,... m=1,3,5,...
X
X
Let n ⌘ 2i + 1, m ⌘ 2j + 1, and note that sinh(2u) = 2 sinh(u) cosh(u). The double sum is then
S=
1 1
i
h p
1 XX
( 1)i+j
sech ⇡ (2i + 1)2 + (2j + 1)2 /2 .
2 i=0 j=0 (2i + 1)(2j + 1)
Setting the upper limits at i = 3, j = 3 (or above) Mathematica returns S = 0.102808, which (to 6 digits) is
equal to ⇡ 2 /96, confirming (at least, numerically) that V (a/2, a/2, a/2) = V0 /6.
Problem 3.17
1 d3
1 d2
1 d2
3
2
2
2
2
x
1
=
3
x
1
2x
=
x x2 1
3
2
2
8 · 6 dx
48 dx
8 dx
i 1 d ⇥
⇤
1 d h 2
2
2
2
=
x
1 + 2x x
1 2x =
x
1 x2 1 + 4x2
8 dx
8 dx
⇤ 1⇥
⇤
1 d ⇥ 2
2
=
x
1 5x
1 =
2x 5x2 1 + x2 1 10x
8 dx
8
1
1
5
3
=
5x3 x + 5x3 5x =
10x3 6x = x3
x.
4
4
2
2
P3 (x) =
We need to show that P3 (cos ✓) satisfies
1 d
sin ✓ d✓
where P3 (cos ✓) =
1
2
cos ✓ 5 cos2 ✓
✓
sin ✓
dP
d✓
◆
=
l(l + 1)P, with l = 3,
3 .
⇤
dP3
1⇥
=
sin ✓ 5 cos2 ✓ 3 + cos ✓(10 cos ✓( sin ✓) =
d✓
2
3
=
sin ✓ 5 cos2 ✓ 1 .
2
1
sin ✓ 5 cos2 ✓
2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
3 + 10 cos2 ✓
62
@
@✓
✓
sin ✓
1 d
sin ✓ d✓
dP3
d✓
✓
◆
3 d ⇥ 2
sin ✓ 5 cos2 ✓
2 d✓
⇥
3 sin ✓ cos ✓ 5 cos2 ✓
=
=
dP
sin ✓
d✓
◆
1
1
⇥
3 cos ✓ 5 cos2 1
=
=
Z1
CHAPTER 3. POTENTIAL
3·4·
P1 (x)P3 (x) dx =
1
Z1
⇤
3⇥
2 sin ✓ cos ✓ 5 cos2 ✓
2
⇤
5 sin2 ✓ .
=
5 1
1
cos ✓ 5 cos2 ✓
2
(x)
1
5x3
2
cos2 ✓
3 =
3x dx =
⇤
=
⇤
1 + sin2 ✓ ( 10 cos ✓ sin ✓)
3 cos ✓ 10 cos2 ✓
6
l(l + 1)P3 . qed
1 5
x
2
x3
1
1
=
1
(1
2
1+1
1) = 0. X
1
Problem 3.18
(a) Inside: V (r, ✓) =
1
X
Al rl Pl (cos ✓)
(Eq. 3.66) where
l=0
(2l + 1)
Al =
2Rl
Z⇡
V0 (✓)Pl (cos ✓) sin ✓ d✓
(Eq. 3.69).
0
In this case V0 (✓) = V0 comes outside the integral, so
(2l + 1)V0
Al =
2Rl
Z⇡
Pl (cos ✓) sin ✓ d✓.
0
But P0 (cos ✓) = 1, so the integral can be written
Z⇡
P0 (cos ✓)Pl (cos ✓) sin ✓ d✓ =
0
Therefore
Al =
Plugging this into the general form:
⇢
⇢
0, if l 6= 0
2, if l = 0
0, if l 6= 0
V0 , if l = 0
(Eq. 3.68).
.
V (r, ✓) = A0 r0 P0 (cos ✓) = V0 .
The potential is constant throughout the sphere.
1
X
Bl
Outside: V (r, ✓) =
Pl (cos ✓) (Eq. 3.72), where
rl+1
l=0
(2l + 1) l+1
Bl =
R
2
Z⇡
V0 (✓)Pl (cos ✓) sin ✓ d✓
(Eq. 3.73).
0
=
(2l + 1) l+1
R V0
2
Z⇡
0
Pl (cos ✓) sin ✓ d✓ =
⇢
0,
if l 6= 0
RV0 , if l = 0
.
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
63
CHAPTER 3. POTENTIAL
Therefore V (r, ✓) = V0
(b)
R
1
(i.e. equals V0 at r = R, then falls o↵ like ).
r
r
V (r, ✓) =
8 1
9
X
>
>
>
l
>
>
Al r Pl (cos ✓), for r  R (Eq. 3.78) >
>
>
<
=
l=0
1
X
>
Bl
>
>
Pl (cos ✓), for r
>
:
rl+1
>
>
R (Eq. 3.79) >
>
;
l=0
where
Bl = R2l+1 Al
,
(Eq. 3.81)
and
1
Al =
2✏0 Rl
1
=
2✏0 Rl
1
Z⇡
0 (✓)Pl (cos ✓) sin ✓ d✓
(Eq. 3.84)
0
1
0
Z⇡
Pl (cos ✓) sin ✓ d✓ =
0
⇢
0,
R
Therefore
V (r, ✓) =
Note: in terms of the total charge Q = 4⇡R2
if l 6= 0
if l = 0
8
R 0
>
,
>
>
< ✏0
.
9
for r  R >
>
>
=
>
2
>
>
: R 0 1 , for r
✏0 r
0,
V (r, ✓) =
Problem 3.19
0 /✏0 ,
>
>
>
R;
8
9
1 Q
>
>
>
>
,
for
r

R
>
>
< 4⇡✏0 R
=
>
>
>
:
1 Q
, for r
4⇡✏0 r
⇥
V0 (✓) = k cos(3✓) = k 4 cos3 ✓
>
>
;
R>
.
.
⇤
3 cos ✓ = k [↵P3 (cos ✓) + P1 (cos ✓)] .
(I know that any 3rd order polynomial can be expressed as a linear combination of the first four Legendre
polynomials; in this case, since the polynomial is odd, I only need P1 and P3 .)

✓
◆
5↵
3
1
3
3
3
4 cos ✓ 3 cos ✓ = ↵
5 cos ✓ 3 cos ✓ + cos ✓ =
cos ✓ +
↵ cos ✓,
2
2
2
so
4=
5↵
8
)↵= ;
2
5
3=
Therefore
V0 (✓) =
3
↵=
2
3 8
· =
2 5
k
[8P3 (cos ✓)
5
12
)
5
3P1 (cos ✓)] .
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
=
12
5
3=
3
.
5
64
CHAPTER 3. POTENTIAL
Now
V (r, ✓) =
8 1
9
X
>
>
>
l
>
>
Al r Pl (cos ✓), for r  R (Eq. 3.66) >
>
>
<
=
l=0
1
X
>
Bl
>
>
Pl (cos ✓), for r
>
:
rl+1
l=0
where
(2l + 1)
Al =
2Rl
Z⇡
V0 (✓)Pl (cos ✓) sin ✓ d✓
>
>
R (Eq. 3.72) >
>
;
,
(Eq. 3.69)
0
8 ⇡
9
Z
Z⇡
=
(2l + 1) k <
=
8
P
(cos
✓)P
(cos
✓)
sin
✓
d✓
3
P
(cos
✓)P
(cos
✓)
sin
✓
d✓
3
l
1
l
;
2Rl 5 :
0
0
⇢
k (2l + 1)
2
2
k 1
=
8
3
=
[8 l3 3 l1 ]
l3
l1
5 2Rl
(2l + 1)
(2l + 1)
5 Rl
⇢
8k/5R3 , if l = 3
=
(zero otherwise).
3k/5R, if l = 1
Therefore
V (r, ✓) =
 ⇣ ⌘
3k
8k 3
k
r 3
rP1 (cos ✓) +
r
P
(cos
✓)
=
8
P3 (cos ✓)
3
3
5R
5R
5
R
3
⇣r⌘
R
P1 (cos ✓) ,
or
k
5
⇢ ⇣ ⌘
r 31⇥
8
5 cos3 ✓
R 2
3 cos ✓
⇤
3
⇣r⌘
R
cos ✓
⇢ ⇣ ⌘
k r
r 2⇥
) V (r, ✓) =
cos ✓ 4
5 cos2 ✓
5R
R
⇤
3
3
(for r  R). Meanwhile, Bl = Al R2l+1 (Eq. 3.81—this follows from the continuity of V at R). Therefore
Bl =
⇢
8kR4 /5, if l = 3
3kR2 /5, if l = 1
(zero otherwise).
So
V (r, ✓) =
" ✓ ◆
4
3kR2 1
8kR4 1
k
R
P
(cos
✓)
+
P
(cos
✓)
=
8
P3 (cos ✓)
1
3
5
r2
5 r4
5
r
3
✓
R
r
◆2
#
P1 (cos ✓) ,
or
k
V (r, ✓) =
5
✓
R
r
◆2
( ✓ ◆
2
R ⇥
cos ✓ 4
5 cos2 ✓
r
3
⇤
3
)
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65
CHAPTER 3. POTENTIAL
(for r
R). Finally, using Eq. 3.83:
(✓) = ✏0
1
X
(2l + 1)Al Rl
1
l=0
⇥
⇤
Pl (cos ✓) = ✏0 3A1 P1 + 7A3 R2 P3
 ✓
◆
✓
◆
3k
8k
✏0 k
= ✏0 3
P1 + 7
R2 P3 =
[ 9P1 (cos ✓) + 56P3 (cos ✓)]
5R
5R3
5R

✏0 k
56
✏0 k
=
9 cos ✓ +
5 cos3 ✓ 3 cos ✓ =
cos ✓[ 9 + 28 · 5 cos2 ✓ 28 · 3]
5R
2
5R
=
⇥
✏0 k
cos ✓ 140 cos2 ✓
5R
Problem 3.20
Use Eq. 3.83:
(✓) = ✏0
1
X
⇤
93 .
(2l+1)Al R
l 1
l=0
Putting them together:
2l + 1
Pl (cos ✓). But Eq. 3.69 says: Al =
2Rl
Z⇡
V0 (✓)Pl (cos ✓) sin ✓ d✓.
0
1
✏0 X
(✓) =
(2l + 1)2 Cl Pl (cos ✓),
2R
with Cl =
l=0
Z⇡
V0 (✓)Pl (cos ✓) sin ✓ d✓. qed
0
Problem 3.21
Set V = 0 on the equatorial plane, far from the sphere. Then the potential is the same as Ex. 3.8 plus the
potential of a uniformly charged spherical shell:
✓
E0 r
V (r, ✓) =
R3
r2
◆
cos ✓ +
1 Q
.
4⇡✏0 r
Problem 3.22
1
1
1
hp
X
X
X
Bl
Bl
Bl
(a) V (r, ✓) =
P
(cos
✓)
(r
>
R),
so
V
(r,
0)
=
P
(1)
=
=
r 2 + R2
l
l
rl+1
rl+1
rl+1
2✏0
l=0
l=0
l=0
p
p
1
1
2
2
2
2
Since r > R in this region, r + R = r 1 + (R/r) = r 1 + (R/r)
(R/r)4 + . . . , so
2
8

1
X
Bl
1 R2
=
r
1
+
rl+1
2✏0
2 r2
1 R4
+ ...
8 r4
l=0
Comparing like powers of r, I see that B0 =

R2 1
V (r, ✓) =
4✏0 r
"
R2
=
1
4✏0 r
1 =
R2
, B1 = 0, B2 =
4✏0
2✏0
✓
R2
2r
◆
R4
+ ... .
8r3
R4
, . . . . Therefore
16✏0
R2
P2 (cos ✓) + . . . ,
4r3
1
8
✓
R
r
◆2
3 cos ✓
2
#
1 + ... ,
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(for r > R).
i
r .
66
(b) V (r, ✓) =
1
X
l=0
CHAPTER 3. POTENTIAL
Al rl Pl (cos ✓) (r < R). In the northern hemispere, 0  ✓  ⇡/2,
V (r, 0) =
1
X
Al rl =
l=0
Since r < R in this region,
p
r 2 + R2 = R
1
X
l=0
Comparing like powers: A0 =
V (r, ✓) =
2✏0

R

R
=
1
2✏0
2✏0
2✏0

1
1 + (r/R)2 = R 1 + (r/R)2
2
p

1 r2
Al r =
R+
2✏0
2R
1 r4
+ ...
8 R3
l
R, A1 =
rP1 (cos ✓) +
i
r .
hp
r 2 + R2
2✏0
, A2 =
4✏0 R
1 ⇣ r ⌘2
cos ✓ +
3 cos2 ✓
R
4 R
r .
, . . . , so
1 2
r P2 (cos ✓) + . . . ,
2R
⇣r⌘
1
(r/R)4 + . . . . Therefore
8
(for r < R, northern hemisphere).
1 + ... ,
In the southern hemisphere we’ll have to go for ✓ = ⇡, using Pl ( 1) = ( 1)l .
V (r, ⇡) =
1
X
( 1)l Al rl =
l=0
2✏0
hp
r 2 + R2
i
r .
(I put an overbar on Al to distinguish it from the northern Al ). The only di↵erence is the sign of A1 :
A1 = +( /2✏0 ), A0 = A0 , A2 = A2 . So:
V (r, ✓) =

1 2
R + rP1 (cos ✓) +
r P2 (cos ✓) + . . . ,
2✏0
2R

⇣r⌘
R
1 ⇣ r ⌘2
=
1+
cos ✓ +
3 cos2 ✓
2✏0
R
4 R
Problem 3.23
V (r, ✓) =
(for r < R, southern hemisphere).
1 + ... ,
8 1
9
X
>
>
>
>
>
Al rl Pl (cos ✓), (r  R) (Eq. 3.78), >
>
>
<
=
l=0
1
X
>
Bl
>
>
Pl (cos ✓), (r
>
:
rl+1
l=0
>
>
R) (Eq. 3.79), >
>
;
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67
CHAPTER 3. POTENTIAL
where Bl = Al R2l+1 (Eq. 3.81) and
1
Al =
2✏0 Rl
=
=
1
2✏0 Rl
1
0 (✓)Pl (cos ✓) sin ✓ d✓
(Eq. 3.84)
0
1
0
2✏0 Rl
Z⇡
1
8
9
>
>
Z⇡
< Z⇡/2
=
Pl (cos ✓) sin ✓ d✓
Pl (cos ✓) sin ✓ d✓
0
>
>
:
;
0
⇡/2
8 1
9
Z0
<Z
=
Pl (x) dx
Pl (x) dx .
:
;
0
(let x = cos ✓)
1
Now Pl ( x) = ( 1)l Pl (x), since Pl (x) is even, for even l, and odd, for odd l. Therefore
Z0
Pl (x) dx =
1
Z0
Pl ( x) d( x) = ( 1)
l
1
Al =
2✏0
⇥
1
0
Rl 1
( 1)l
⇤
Pl (x) dx =
0
8
0,
>
>
<
>
>
: ✏0
0
Rl 1
Z1
0
So A0 = A2 = A4 = A6 = 0, and all we need are A1 , A3 , and A5 .
Z1
P1 (x) dx =
Z1
1
P3 (x) dx =
2
Z1
1
P5 (x) dx =
8
0
Z1
x2
x dx =
2
0
0
Z1
1
=
0
5x
0
0
✓
x4
5
4
x2
3
2
◆
1
0
1
=
2
✓
1
x6
63x
70x + 15x dx =
63
8
6
0
✓
◆
1 21 35 15
1
1
=
+
=
(36 35) =
.
8 2
2
2
16
16
Z1
5
Therefore
A1 =
3
Pl (x) dx, if l is odd >
>
;
✓
5
4
3
2
◆
=
x4
x2
70 + 15
4
2
◆
1
.
8
1
0
✓ ◆
✓
◆
✓ ◆
1
1
1
0
0
; A3 =
;
A
=
; etc.
5
2
4
✏0 2
✏0 R
8
✏0 R
16
0
and
B1 =
9
if l is even >
>
=
1
.
2
1
3x dx =
2
3
Pl (x) dx,
0
and hence
Z1
Z1
0
✏0
R3
✓ ◆
✓
◆
✓ ◆
1
1
1
0 5
0 7
; B3 =
R
; B5 =
R
; etc.
2
✏0
8
✏0
16
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.
68
Thus
V (r, ✓) =
Problem 3.24
8
>
>
>
<
>
>
>
:
CHAPTER 3. POTENTIAL
9

1 ⇣ r ⌘2
1 ⇣ r ⌘4
>
P1 (cos ✓)
P3 (cos ✓) +
P5 (cos ✓) + ... ,
(r  R), >
>
=
2✏0 "
4 R
8 R
#
✓
◆
✓
◆
2
4
3
1 R
1 R
>
0R
P1 (cos ✓)
P3 (cos ✓) +
P5 (cos ✓) + ... , (r R). >
>
;
2✏0 r2
4 r
8 r
0r
1 @
s @s
✓
s
@V
@s
◆
+
◆
+
1 @2V
= 0.
s2 @ 2
Look for solutions of the form V (s, ) = S(s) ( ):
✓
◆
1 d
dS
1 d2
s
+ 2 S 2 = 0.
s ds
ds
s d
Multiply by s2 and divide by V = S :
s d
S ds
✓
s
dS
ds
1 d2
= 0.
d 2
Since the first term involves s only, and the second only, each is a constant:
✓
◆
s d
dS
1 d2
s
= C1 ,
= C2 , with C1 + C2 = 0.
S ds
ds
d 2
Now C2 must be negative (else we get exponentials for
geometrically they must— when is increased by 2⇡).
C2 =
k 2 . Then
d2
=
d 2
k2
)
= A cos k + B sin k .
Moreover, since ( + 2⇡) = ( ), k must be an integer: k = 0, 1, 2, 3, . . . (negative integers are just repeats,
but k =✓0 must
◆ be included, since = A (a constant) is OK).
d
dS
s
s
= k 2 S can be solved by S = sn , provided n is chosen right:
ds
ds
s
d
snsn
ds
1
= ns
d n
(s ) = n2 ssn
ds
1
= n2 sn = k 2 S ) n = ±k.
Evidently the general solution is S(s) = Csk + Ds k , unless k = 0, in which case we have only one solution
to a second-order equation—namely, S = constant. So we must treat k = 0 separately. One solution is a
constant—but what’s the other? Go back to the diferential equation for S, and put in k = 0:
✓
◆
d
dS
dS
dS
C
ds
s
=0)s
s
= constant = C )
=
) dS = C
) S = C ln s + D (another constant).
ds
ds
ds
ds
s
s
So the second solution in this case is ln s. [How about ? That too reduces to a single solution,
case k = 0. What’s the second solution here? Well, putting k = 0 into the equation:
d2
d
=0)
d 2
d
= constant = B )
= A, in the
= B + A.
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1
69
CHAPTER 3. POTENTIAL
Contents
But a term of the form B is unacceptable, since it does not return to its initial value when
by 2⇡.] Conclusion: The general solution with cylindrical symmetry is
1
X
⇥ k
V (s, ) = a0 + b0 ln s +
s (ak cos k + bk sin k ) + s
k=1
1 Special Techniques
k
is augmented
⇤
(ck cos k + dk sin k ) .
Yes: the potential of a line charge goes like ln s, which is included.
Problem
Problem3.25
3.24
Picking
V V= =
0 on
thethe
yz yz
plane,
with
E0 E
in0the
direction,
we have (Eq. 3.74):
Picking
0 on
plane,
with
in x
the
x direction,
⇢!
"
(i)(i)V V==
0,0,
when
whens =
s =R,R,
(ii)
== −E
E0 s0cos
R.R.
0 x0 x
(ii)V V!→ E
−E
s cos,φ,forfors s ≫
y
−
−
−
−
−
2
✻
+
+
φ+
+
+
s
✲x
Evidently a0 = b0 = bk = dk = 0, and ak = ck = 0 except for k = 1:
\$
#
c1
cos φ.
V (s, φ) = a1 s +
s
=⇒E0
Evidently a0 = b0 = bk = dk = 0, and ak = ck = 0 except for k = 1:
✢
⇣
(i)⇒ c1 = −a1 R2 ; (ii)→ a1 = −E0 . Therefore
c1 ⌘
z
V (s, ) = a1 s +
cos .
s
(
'% &
&
%
2
2
E
R
R
0
− 1 cos φ.
(s, φ)a=
(i)) c1 = a1 R2 ; V(ii)!
E0 s. +
Therefore cos φ, or V (s, φ) = −E0 s
1 = −E
s
s
"✓ ◆
#
✓
◆
2
R
) E0 R 2
)
%
&
) s
V (s, ) =
E0∂V
s +)
cos , or 2 V (s, ) = E
1 cos .
) s = −ϵ0 E0 − R − 1 cos φ) 0 = s2ϵ0 E0 cos φ.
σ = −ϵ0
)
∂s )s=R
s2
s=R
✓
◆
@V
R2
Problem 3.25
= ✏*
= ✏0 E0
1 cos
= 2✏0 E0 cos .
0∞
@sk s=R
s2
s=R
Inside: V (s, φ) = a0 +
s (ak cos kφ + bk sin kφ) . (In this region ln s and s−k are no good—they blow
k=1
Problem 3.26
1
up at s = 0.)
X
∞k
k
Inside: V (s, ) = a0 + *
s (a
1 k cos k + bk sin k ) . (In this region ln s kand s are no good—they blow
+
Outside: V (s, φ) = a0 k=1
(c
cos
kφ
+
d
sin
kφ).
(Here
ln
s
and
s
are
no
good at s → ∞).
k
k
sk
k=1
up at s = 0.)
1
X
&)
%
1
) ln s and sk are no good at s ! 1).
Outside: V (s, ) = a0 +
(ck cos k +∂V
dkout
sin k ∂V
).in(Here
)
−
(Eq. 2.36).
sk σ = −ϵ0
k=1
∂s
∂s )s=R
◆
✓
@Vout
@Vin
Thus
=
✏
(Eq. 2.36).
"
0
∞ !
*
@s
@s
k
s=R
a sin 5φ = −ϵ0
− k+1 (ck cos kφ + dk sin kφ) − kRk−1 (ak cos kφ + bk sin kφ) .
R
k=1
Thus
⇢
1
X
k
1
a sin
5 = Inc.,
(ck cos
dk sin
k ) This
kRkmaterial
(ak is
cos k + bk sin k ) .
c
⃝2005
Pearson
Education,
River,
NJ.kAll +
rights
reserved.
k+1
Rcurrently exist. No portion of this material may be
as
they
k=1
reproduced, in any form or by any means, without permission in writing from the publisher.
✓
◆
1
4
Evidently ak = ck = 0; bk = dk = 0 except k = 5; a = 5✏0
d5 + R b5 . Also, V is continuous at s = R:
R6
1
a0 +R5 b5 sin 5 = a0 + 5 d5 sin 5 . So a0 = a0 (might as well choose both zero); R5 b5 = R 5 d5 , or d5 = R10 b5 .
R
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70
CHAPTER 3. POTENTIAL
a
aR6
; d5 =
. Therefore
4
10✏0 R
10✏0
Combining these results: a = 5✏0 R4 b5 + R4 b5 = 10✏0 R4 b5 ; b5 =
⇢
a sin 5
V (s, ) =
10✏0
s5 /R4 , for s < R,
R6 /s5 , for s > R.
Problem 3.27 Since r is on the z axis, the angle ↵ is just the polar angle ✓ (I’ll drop the primes, for simplicity).
Monopole term:
Z
Z 
1
⇢ d⌧ = kR
(R 2r) sin ✓ r2 sin ✓ dr d✓ d .
r2
But the r integral is
ZR
(R
2r) dr = Rr
R
0
r2
= R2
R2 = 0.
0
So the monopole term is zero.
Dipole term:

Z
Z
1
r cos ✓⇢ d⌧ = kR (r cos ✓) 2 (R
r
But the ✓ integral is
Z⇡
sin3 ✓
3
sin2 ✓ cos ✓ d✓ =
0
2r) sin ✓ r2 sin ✓ dr d✓ d .
⇡
=
1
(0
3
1

1
(R
r2
=
R4
3
0
So the dipole contribution is likewise zero.
✓
◆
Z
Z
3
1
1
r2
cos2 ✓
⇢ d⌧ = kR r2 3 cos2 ✓
2
2
2
r integral:
Z
R
r2 (R
2r) dr =
0
✓ integral:
Z⇡
3 cos ✓
{z
2
|
1
✓
r3
R
3
0 3(1 sin2 ✓) 1=2 3 sin2 ✓
integral:
= 2
⇣⇡⌘
2
◆
sin ✓ d✓ = 2
2
}
r4
2
3
R
0
Z⇡
0) = 0.
2r) sin ✓ r2 sin ✓ dr d✓ d .
R4
=
2
sin ✓ d✓
2
0
✓
3⇡
8
Z2⇡
◆
✓
=⇡ 1
3
Z⇡
R4
.
6
sin4 ✓ d✓
0
9
8
◆
=
⇡
.
8
d = 2⇡.
0
The whole integral is:
1
kR
2
✓
R4
6
◆⇣
⇡⌘
k⇡ 2 R5
(2⇡) =
.
8
48
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71
CHAPTER 3. POTENTIAL
For point P on the z axis (r ! z in Eq. 3.95) the approximate potential is
1 k⇡ 2 R5
.
4⇡✏0 48z 3
V (z) ⇠
=
Problem 3.28
z
r
R
f'
r'
y
x
For a line charge, ⇢(r0 ) d⌧ 0 ! (r0 ) dl0 , which in this case becomes R d 0 .
r
r0
r · r0
cos ↵
n=0:
=
=
=
=
r sin ✓ cos x̂ + r sin ✓ sin ŷ + r cos ✓ ẑ,
R cos 0 + R sin 0 , so
rR sin ✓ cos cos 0 + rR sin ✓ sin sin 0 = rR cos ↵,
sin ✓(cos cos 0 + sin sin 0 ).
Z
⇢(r ) d⌧ ! R
0
0
Z
2⇡
d
0
= 2⇡R ;
V0 =
0
n=1:
Z
Z
r0 cos ↵ ⇢(r0 ) d⌧ 0 ! R cos ↵ R d
0
= R2 sin ✓
Z
2⇡
1 2⇡R
R
=
.
4⇡✏0 r
2✏0 r
(cos cos
0
+ sin sin
0
)d
0
= 0; V1 = 0.
0
n=2:
✓
◆
Z
Z
Z h
i
3
1
R3
2
0 2
0
0
2
2
Rd 0 =
(r ) P2 (cos ↵) ⇢(r )d⌧ ! R
cos ↵
3 sin2 ✓ (cos cos 0 + sin sin 0 )
1 d
2
2
2

✓
◆ Z 2⇡
Z 2⇡
Z 2⇡
Z 2⇡
R3
=
3 sin2 ✓ cos2
cos2 0 d 0 + sin2
sin2 0 d 0 + 2 sin cos
sin 0 cos 0 d 0
d 0
2
0
0
0
0
✓
◆
⇤ ⇡ R3
R3 ⇥
3
1
=
3 sin2 ✓ ⇡ cos2 + ⇡ sin2 + 0
2⇡ =
3 sin2 ✓ 2 = ⇡ R3
cos2 ✓
.
2
2
2
2
So
V2 =
R3
3 cos2 ✓
8✏0 r3
1 =
R3
P2 (cos ✓).
4✏0 r3
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0
72
Problem 3.29
p = (3qa qa) ẑ + ( 2qa
CHAPTER 3. POTENTIAL
2q( a)) ŷ = 2qa ẑ. Therefore
V ⇠
=
1 p · r̂
,
4⇡✏0 r2
and p · r̂ = 2qa ẑ · r̂ = 2qa cos ✓, so
1 2qa cos ✓
.
4⇡✏0
r2
V ⇠
=
(Dipole.)
Problem 3.30
R
R
(a) By symmetry, p is clearly in the z direction: p = p ẑ; p = z⇢ d⌧ ) z da.
p =
=
Z
(R cos ✓)(k cos ✓)R sin ✓ d✓ d = 2⇡R k
2
3
Z⇡
cos ✓ sin ✓ d✓ = 2⇡R k
2
3
0
2 3
⇡R k[1
3
4⇡R3 k
;
3
( 1)] =
p=
✓
cos3 ✓
3
◆
⇡
0
4⇡R3 k
ẑ.
3
(b)
V ⇠
=
1 4⇡R3 k cos ✓
kR3 cos ✓
=
.
2
4⇡✏0 3
r
3✏0 r2
(Dipole.)
This is also the exact potential. Conclusion: all multiple moments of this distribution (except the dipole) are
exactly zero.
Problem 3.31
Using Eq. 3.94 with r0 = d/2 and ↵ = ✓ (Fig. 3.26):
1
r
for
r
, we let ✓ ! 180 + ✓, so cos ✓ !
cos ✓:
1
r
+
◆n
1 ✓
1X d
=
Pn (cos ✓);
r n=0 2r
=
◆n
1 ✓
1X d
Pn ( cos ✓).
r n=0 2r
But Pn ( x) = ( 1)n Pn (x), so
V =
1
q
4⇡✏0
✓
1
r
+
1
r
◆
=
◆n
1 ✓
1 1X d
q
[Pn (cos ✓)
4⇡✏0 r n=0 2r
Therefore
Vdip =
Voct =
2q
4⇡✏0 r
✓
d
2r
◆3
Pn ( cos ✓)] =
2q 1 d
qd cos ✓
P1 (cos ✓) =
,
4⇡✏0 r 2r
4⇡✏0 r2
P3 (cos ✓) =
2q d3 1
5 cos3 ✓
4⇡✏0 8r4 2
while
3 cos ✓ =
2q
4⇡✏0 r
X
n=1,3,5,...
✓
d
2r
◆n
Pn (cos ✓).
qd3 1
5 cos3 ✓
4⇡✏0 8r4
3 cos ✓ .
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73
CHAPTER 3. POTENTIAL
Problem 3.32
(a) (i) Q = 2q,
(ii) p = 3qa ẑ,
(b) (i) Q = 2q,
(ii) p =
(c) (i) Q = 2q,
(iii) V ⇠
=
1
4⇡✏0
Q
r
+
p·r̂
r2
i

1
2q 3qa cos ✓
=
+
.
4⇡✏0 r
r2

1
2q qa cos ✓
+
.
4⇡✏0 r
r2

1
2q 3qa sin ✓ sin
(iii) V ⇠
+
=
4⇡✏0 r
r2
(iii) V ⇠
=
qa ẑ,
(ii) p = 3qa ŷ,
Problem 3.33
(a) This point is at r = a, ✓ =
h
= 0, so E =
⇡
2,
(from Eq. 1.64, ŷ·r̂ = sin ✓ sin ).
p
p
✓ˆ =
( ẑ); F = qE =
3
4⇡✏0 a
4⇡✏0 a3
pq
ẑ.
4⇡✏0 a3
p
2p
2pq
(2r̂) =
ẑ. F =
ẑ.
4⇡✏0 a3
4⇡✏0 a3
4⇡✏0 a3
⇣ ⇡ ⌘i
qp h
pq
V (a, 0, 0)] =
cos(0)
cos
=
.
4⇡✏0 a2
2
4⇡✏0 a2
(b) Here r = a, ✓ = 0, so E =
(c) W = q [V (0, 0, a)
Problem 3.34
Q=
1
q
;
4⇡✏0 r
q, so Vmono =
q
4⇡✏0
V (r, ✓) ⇠
=
✓
p = qa ẑ,
1 a cos ✓
+
r
r2
◆
.
so Vdip =
E(r, ✓) ⇠
=
1 qa cos ✓
. Therefore
4⇡✏0 r2
q
4⇡✏0

⌘
1
a ⇣
ˆ .
r̂
+
2
cos
✓
r̂
+
sin
✓
✓
r2
r3
Problem 3.35 The total charge is zero, so the dominant term is the dipole. We need the dipole moment
of this configuration. It obviously points in the z direction, and for the southern hemisphere (✓ : ⇡2 ! ⇡) ⇢
switches sign but so does z, so
p =
Z
z⇢ d⌧ = 2⇢0
R4 sin2 ✓
4
2
= 4⇡⇢0
Z
⇡/2
✓=0
⇡/2
=
0
r cos ✓ r sin ✓ dr d✓ d = 2⇢0 (2⇡)
2
Z
R
3
r dr
0
⇡⇢0 R4
.
2
Z
⇡/2
cos ✓ sin ✓ d✓
0
Therefore (Eq. 3.103)
E⇡
⌘
⇡⇢0 R4 ⇣
ˆ .
2
cos
✓
r̂
+
sin
✓
✓
8⇡✏0 r3
Problem 3.36
ˆ ✓ˆ = p cos ✓ r̂ p sin ✓ ✓ˆ (Fig. 3.36). So 3(p · r̂) r̂
p = (p · r̂) r̂ + (p · ✓)
ˆ
2p cos ✓ r̂ + p sin ✓ ✓. So Eq. 3.104 ⌘ Eq. 3.103. X
Problem 3.37
Vave (R) =
1
4⇡R2
Z
p = 3p cos ✓ r̂
p cos ✓ r̂ + p sin ✓ ✓ˆ =
V (r) da, where the integral is over the surface of a sphere of radius R. Now da =
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74
R2 sin ✓ d✓ d , so Vave (R) =
1
4⇡
Z
CHAPTER 3. POTENTIAL
V (R, ✓, ) sin ✓ d✓ d .
Z
Z
Z
dVave
1
@V
1
1
=
sin ✓ d✓ d =
(rV · r̂) sin ✓ d✓ d =
(rV ) · (R2 sin ✓ d✓ d r̂)
dR
4⇡
@R
4⇡
4⇡R2
Z
Z
1
1
=
(rV ) · da =
(r2 V ) d⌧ = 0.
4⇡R2
4⇡R2
5
(The final integral, from the divergence theorem, is over the volume of the sphere,
where by assumption the
Laplacian of V is zero.) So Vave is independent of R—the same for all spheres, regardless of their radius—and
2
hence
(takingcan
thealso
limit
R ! 0), Vdirectly.
V (0).
ave (R) =Let
This
integral
beas
integrated
x = uqed
; dx = 2u du.
Problem 3.38 At a point (x, y) on the plane the field of q is
"
\$
%&'0
!0 √
!0
'
u2
u#
u
x
d
−1
2
1
q
' = −d sin−1 (1) = −d π .
√
√ E =du = 2 − r̂ , d and
dx = 2
− u +r =
sinx x̂ + √
'√
y
ŷ
d
ẑ,
2
q
2
2
2
d−x
d − u 4⇡✏0 r 3
d
d
√
d
d
q
d
. Meanwhile, the field of (just below the surface) is
,
2
2
3/2
4⇡✏0 (x(+ y + d2 ))
2✏0
)
(Eq. 2.17). (Of course, this is for a duniform
surface
but as long
is
πd
d
π 2 d2 charge,
2π 3as
d3 ϵwe
0 mare infinitesimally far away
16πϵ
.
t =field inside=the conductor
0 m =so
e↵ectively uniform.) The total
is
zero,
2
2
2A 2
4 2q
q
so its z component is
Therefore
Problem 3.35
Problem 3.39
x✛
q
d
4⇡✏0 (x2 + y 2 + d2 )3/2
+
−
+
−
2✏0
+
=0
+
−
qd
. X
2⇡(x2 + y 2 + d2 )3/2
(x, y) =
)
+
−
q
+
−
−
The image configuration is shown in the figure; the positive image charge forces cancel in pairs. The net
force of the negative image charges is:
*(
1
2
11
1
11
11
+
+
2+
2+
2 + ...
FF == 4⇡✏0qq2
2
2
2 + ...
[2(a
x)]
[2a
+
2(a
x)]
[4a
+
2(a
x)]
4πϵ0
[2(a − x)]
[2a + 2(a − x)]
[4a + 2(a − x)]
&
1
1
1
1
1
1
− (2x)22− (2a + 2x)22− (4a + 2x)22− . .. .. .
(2x)
(2a + 2x)
(4a + 2x)
⇢
2"+
, + 1
,&
1
11
11
11
11
11 qq2
=
+
+
+
.
.
. − 1 2++ 1 2++
+. .. .. . . .
2
2
2
2
+
+
+
.
.
.
+
= 4⇡✏0 4
(a− x)
x)2
(2a− x)
x)2
(3a− x)
x)2
(a++x)
x)2
(2a++x)
x)2
4πϵ0 4
(a
(2a
(3a
xx2
(a
(2a
1
1
q2
2
When a ! 1 (i.e. a
x) only the 1 2 term survives: F =
X (same as for only one plane—
1
q
When a → ∞ (i.e. a ≫ x) only the x2 term survives: F = − 4⇡✏0 (2x)22 ! (same as for only one plane—
x
4πϵ0 (2x)
Eq. 3.12). When x = a/2,
Eq. 3.12). When x = a/2,
⇢

1 q 2"+
1
1
1
1
1
1
,
+
F = 1 q2
+ 1 2 + 1 2 + ...
+ 1 2 + 1 2 + . . ,&
.
= 0. X
1
1
2
2
(a/2) + (3a/2) + (5a/2) + . . . − (a/2) + (3a/2) + (5a/2) + . . .
F = 4⇡✏0 4
= 0. !
2
2
2
2
2
2
4πϵ0 4
(a/2)
(3a/2)
(5a/2)
(a/2)
(3a/2)
(5a/2)
Problem 3.40
Problem
3.36Prob. 2.52, we place image line charges
Following
at y = b and + at y = b (here y is the horizontal
axis,
z vertical).
Following
Prob. 2.47, we place image line charges −λ at y = b and +λ at y = −b (here y is the horizontal
axis, z vertical).
c 2012 Pearson Education, Inc.,
laws as they currently exist. No portion of this material may be
P
reproduced, in any form or by any means,
s1 without permission in writing from the publisher.
s2
−λ
−b
s3
b
+λ
R
☛ -
- ./ 0 - ./ 0
λ
a−b
a−b
2
2
./
0
y0
./
0
a
−λ
c
⃝2005
s4
✲y
When a → ∞ (i.e. a ≫ x) only the
1
1
q2
term survives: F = −
! (same as for only o
2
x
4πϵ0 (2x)2
Eq. 3.12). When x = a/2,
F =
1 q2
4πϵ0 4
"+
, +
,&
1
1
1
1
1
1
+
+
+
.
.
.
−
+
+
+
.
.
.
=
(a/2)2
(3a/2)2
(5a/2)2
(a/2)2
(3a/2)2 (5a/2)2
Problem 3.36
CHAPTER 3. POTENTIAL
75
Following Prob. 2.47, we place image line charges −λ at y = b and +λ at y = −b (here y is the
axis, z vertical).
z
✻
P
s1
s2
−b
−λ
b
+λ
R
- ./ 0 - ./ 0
λ
a−b
a−b
2
2
./
0
y0
./
0
a
−λ
☛ -
In the solution to Prob. 2.52 substitute:
s3
s4
✲y
c
⃝2005
Pearson Education, Inc., Upper
material is
✓
◆2 River,
✓ NJ. All
◆2 rights reserved. This
protected
may be
a under
b
a + b laws asa theyb currentlyaexist.
+ b No portion
R2
2 of this material
areproduced,
!
,inyany
= permission in writing
R ) from
b = the publisher.
.
form or byso
any means, without
0 !
2
2
2
2
a
 ✓ 2◆
✓ 2◆
✓ 2 2◆
s
s
s s
ln 32 + ln 12
=
ln 12 32
4⇡✏0
s4
s2
4⇡✏0
s4 s2
⇢
2
2
2
2
[(y + a) + z ][(y b) + z ]
=
ln
, or, using y = s cos , z = s sin ,
4⇡✏0
[(y a)2 + z 2 ][(y + b)2 + z 2 ]
⇢ 2
(s + a2 + 2as cos )[(as/R)2 + R2 2as cos ]
ln
.
=
4⇡✏0
(a2 + a2 2as cos )[(as/R)2 + R2 + 2as cos ]
V =
Problem 3.41 Same as Problem 3.9, only this time we want q 0 + q 00 = q, so q 00 = q q 0 :
✓ 00
◆
✓
◆
q
q
q0
q2
qq 0
1
1
F =
+
=
+
+
.
4⇡✏0 a2
(a b)2
4⇡✏0 a2
4⇡✏0
a2
(a b)2
The second term is identical to Problem 3.9, and I’ll just quote the answer from there:

2
q2
R2 )
3 (2a
F =
a
R
.
4⇡✏0 a3
(a2 R2 )2
(a) F = 0 ) a(a2
R2 )2 = R3 (2a2
R2 ), or (letting x ⌘ a/R), x(x2 1)2 2x2 + 1 = 0. We want a real
p
root greater than 1; Mathematica delivers x = (1 + 5)/2 = 1.61803, so a = 1.61803R = 5.66311 Å.
(b) Let a0 = x0 R be the minimum value of a. The work necessary is

Z a0
Z 1
Z 1
2
q2
1
R2 )
q2
1
(2x2 1)
3 (2a
W =
F da =
a
R
da
=
dx
3
2
2
2
2
4⇡✏0 a0 a
(a
R )
4⇡✏0 R x0 x
x3 (x2 1)2
1

q2
1 + 2x0 2x30
=
.
4⇡✏0 R 2x20 (1 x20 )
p
Putting in x0 = (1 + 5)/2, Mathematica says the term in square brackets is 1/2 (this is not an accident; see
q2
footnote 6 on page 127), so W =
. Numerically,
8⇡✏0 R
W =
(1.60 ⇥ 10 19 )2
8⇡(8.85 ⇥ 10 12 )(5.66 ⇥ 10
10 )
J = 2.03 ⇥ 10
19
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
J = 1.27 eV.
76
CHAPTER 3. POTENTIAL
Problem 3.42
y
y
V1
V1
a
a
V0
-b
y
0
0
b
V0
=
V0
a
x
+
V0
0
-b
x
b
0
-b
0
0
b
x
The first configuration on the right is precisely Example 3.4, but unfortunately the second configuration is not
the same as Problem 3.15:
y
0
a
0
V0
0
x
b
We could reconstruct Problem 3.15 with the modified boundaries, but let’s see if we can’t twist it around by
an astute change of variables. Suppose we let x ! y, y ! u, a ! c, b ! a, and V0 ! V1 :
y
V1
a
0
0
0
u
c
This is closer; making the changes in the solution to Problem 3.15 we have (for this configuration)
V (u, y) =
4V1
⇡
X
sinh(n⇡y/c) sin(n⇡u/c)
.
n sinh(n⇡a/c)
n=1,3,5...
Now let c ! 2b and u ! x + b, and the configuration is just what we want:
y
V1
a
0
0
0
-b
b
x
The potential for this configuration is
V (x, y) =
4V1
⇡
X
sinh(n⇡y/2b) sin(n⇡(x + b)/2b)
.
n sinh(n⇡a/2b)
n=1,3,5...
protected under all copyright laws as they currently exist. No portion of this material may be
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77
CHAPTER 3. POTENTIAL
(If you like, write sin(n⇡(x + b)/2b) as ( 1)(n
V (x, y) =
4
⇡
1)/2
cos(n⇡x/2b).) Combining this with Eq. 3.42,

1
cosh(n⇡x/a) sin(n⇡y/a)
sinh(n⇡y/2b) sin(n⇡(x + b)/2b)
V0
+ V1
.
n
cosh(n⇡b/a)
sinh(n⇡a/2b)
n=1,3,5...
X
Here’s a plot of this function, for the case a = b = 1, V0 = 1/2, V1 = 1:
1.0
1.0
0.5
0.0
-1.0
0.5
-0.5
0.0
0.5
1.0
0.0
Problem 3.43
X✓
◆
Bl
Since the configuration is azimuthally symmetric, V (r, ✓) =
Al r + l+1 Pl (cos ✓).
r
X Bl
(a) r > b: Al = 0 for all l, since V ! 0 at 1. Therefore V (r, ✓) =
Pl (cos ✓).
rl+1
◆
X✓
Dl
a < r < b : V (r, ✓) =
Cl rl + l+1 Pl (cos ✓). r < a : V (r, ✓) = V0 .
r
We need to determine Bl , Cl , Dl , and V0 . To
✓ do ◆this, invoke boundary conditions as follows: (i) V is
@V
1
continuous at a, (ii) V is continuous at b, (iii) 4
=
(✓) at b.
@r
✏
0
◆
X Bl
X✓
Dl
Bl
Dl
(ii) )
Pl (cos ✓) =
Cl bl + l+1 Pl (cos ✓);
= Cl bl + l+1 ) Bl = b2l+1 Cl + Dl . (1)
bl+1
b8
bl+1
b
9
Dl
>
>
✓
◆
l
<
X
Cl a + l+1 = 0, if l 6= 0, = D = a2l+1 C , l 6= 0,
Dl
l
l
a
(i) )
Cl al + l+1 Pl (cos ✓) = V0 ;
(2)
D0
D
=
aV
aC
>
>
a
0
0
0
0.
: C0 a +
;
=
V
,
if
l
=
0;
0
a1
Putting (2) into (1) gives Bl = b2l+1 Cl a2l+1 Cl , l 6= 0, B0 = bC0 + aV0 aC0 . Therefore
l
Bl = b2l+1 a2l+1 Cl , l 6= 0,
(10 )
B0 = (b a)C0 + aV0 .
(iii) )
X
Bl [ (l + 1)]
1
b
P (cos ✓)
l+2 l
X✓
(l + 1)
Bl
bl+2
or
(l + 1)Bl
Cl lbl
✓
Cl lb
1
l 1
+ Dl
(l + 1)
bl+2
+ Dl
(l + 1)
bl+2
lCl b2l+1 + (l + 1)Dl = 0;
◆
◆
Pl (cos ✓) =
k
✏0
P1 (cos ✓). So
= 0, if l 6= 1;
(l + 1)(Bl
Dl ) =
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reproduced, in any form or by any means, without permission in writing from the publisher.
lb2l+1 Cl .
78
B1 (+2)
Therefore
CHAPTER 3. POTENTIAL
✓
◆
1
2
k
+
C
+
D
= , for l = 1;
1
1 2
b2
b
✏0
C1 +
2
(B1
b3
D1 ) = k.
(l + 1)(Bl Dl ) + lb2l+1 Cl = 0, for l 6= 1,
2
k
(3)
C1 + 3 (B1 D1 ) = .
b
✏0
Plug (2) and (10 ) into (3):
For l 6=
⇥ 0 or 1:
⇤
(l+1) b2l+1 a2l+1 Cl + a2l+1 Cl +lb2l+1 Cl = 0; (l+1)b2l+1 Cl +lb2l+1 Cl = 0; (2l+1)Cl = 0 ) Cl = 0.
Therefore (10 ) and (2) ) Bl = Cl = Dl = 0 for l > 1.
⇤
2 ⇥
For l = 1: C1 + 3 b3 a3 C1 + a3 C1 = k; C1 + 2C1 = k ) C1 = k/3✏0 ; D1 = a3 C1 )
b
D1 = a3 k/3✏0 ; B1 = b3 a3 C1 ) B1 = b3 a3 k/3✏0 .
For l = 0: B0 D0 = 0 ) B0 = D0 ) (b a)C0 +aV0 = aV0 aC0 , so bC0 = 0 ) C0 = 0; D0 = aV0 = B0 .
✓
◆
b3 a3 k
aV0
aV0
k
a3
Conclusion: V (r, ✓) =
+
cos
✓,
r
b.
V
(r,
✓)
=
+
r
cos ✓, a  r  b.
r
3r2 ✏0
r
3✏0
r2

✓
◆
✓
◆
@V
aV0
k
a3
V0
k
✏0
(b) i (✓) = ✏0
= ✏0
+
1
+
2
cos
✓
=
✏
+
cos
✓
= k cos ✓ + V0 .
0
2
3
@r
a
3✏
a
a
✏
a
0
0
a
Z
V0 ✏0
aV
1
Q
1
4⇡a✏
V
aV
?
0
0
0
0
(c)qi =
4⇡a2 = 4⇡a✏0 V0 = Qtot . At large r: V ⇡
=
=
=
.X
i da =
a
r
4⇡✏0 r
4⇡✏0
r
r
Problem 3.44
dr'
r' q'=q
Use multipole expansion (Eq. 3.95): ⇢ d⌧ 0 ! dr0 ;
Q
=
; the r0 integral breaks into two pieces:
2a
r
q'=p-q
2 a
3
Z
Za
1
1 X 1 4
V (r) =
(r0 )n Pn (cos ✓0 ) dr0 + (r0 )n Pn (cos ✓0 ) dr0 5 .
4⇡✏0 n=0 rn+1
0
0
In the first integral ✓ = ✓ (see diagram); in the second integral ✓ = ⇡ ✓, so cos ✓0 =
( 1)n Pn (z), so the integrals cancel when n is odd, and add when n is even.
0
0
cos ✓. But Pn ( z) =
Z
1
X
1 Q
1
V (r) = 2
Pn (cos ✓) xn dx.
4⇡✏0 2a n=0,2,4,... rn+1
a
0
The integral is
an+1
, so
n+1

Q 1 X
1 ⇣ a ⌘n
V =
Pn (cos ✓) .
4⇡✏0 r n=0,2,4,... n + 1 r
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79
CHAPTER 3. POTENTIAL
Problem 3.45
Use separation of variables in cylindrical coordinates (Prob. 3.24):
V (s, ) = a0 + b0 ln s +
1
X
⇥ k
s (ak cos k + bk sin k ) + s
k
k=1
⇤
(ck cos k + dk sin k ) .
P1
s < R : V (s, ) = k=1 sk (ak cos k + bk sin k ) (ln s and s k blow up at s = 0);
P1
s > R : V (s, ) = k=1 s k (ck cos k + dk sin k ) (ln s and sk blow up as s ! 1).
(We
X may as well pick constantsXso V ! 0 as s ! 1, and hence a0 = 0.) Continuity at s = R )
Rk (ak cos k + bk sin k ) =
R k (ck cos k + dk sin k ), so ck = R2k ak , dk = R2k bk . Eq. 2.36 says:
@V
@V
1
=
. Therefore
@s R+
@s R
✏0
X
or:
X
k
R
(c cos k + dk sin k )
k+1 k
X
2kRk
1
kRk
1
(ak cos k + bk sin k ) =
⇢
1
,
✏0
(ak cos k + bk sin k ) =
0 /✏0
0 /✏0
(0 <
(⇡ <
< ⇡)
< 2⇡)
.
Fourier’s trick: multiply by (cos l ) d and integrate from 0 to 2⇡, using
Z2⇡
Z2⇡
sin k cos l d = 0;
0
Then
2lRl
1
⇡al =
0
✏0
cos k cos l d =
0
2 ⇡
Z
4 cos l d
0
Z2⇡
⇡
Multiply by (sin l ) d and integrate, using
3
cos l d 5 =
2⇡
R
0
✏0
(
sin k sin l d =
0
2lR
l 1
0
2 ⇡
Z
4 sin l d
Z2⇡
3
0
(
⇢
cos l
l
sin l d 5 =
✏0
✏0
⇡
0
⇢
⇢
0,
0,
if l is even
=
) bl =
4 0 /l✏0 , if l is odd
2 0 /⇡✏0 l2 Rl
⇡bl =
Conclusion:
V (s, ) =
2 0R
⇡✏0
X
k=1,3,5,...
1
sin k
k2
⇢
0
0, k 6= l
⇡, k = l
sin l
l
.
2⇡
⇡
⇡
0
cos l
+
l
2⇡
⇡
)
=
.
(s/R)k (s < R)
(R/s)k (s > R)
.
In =
)
= 0;
al = 0.
0, k 6= l
:
⇡, k = l
if l is even
, if l is odd
Problem 3.46
1
1 X Pn (cos ✓)
Use Eq. 3.95, in the form V (r) =
In ;
4⇡✏0 n=0 rn+1
⇡
sin l
l
⇢
1
Za
z n (z) dz.
a
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reproduced, in any form or by any means, without permission in writing from the publisher.
0
l✏0
(2
2 cos l⇡)
Contents
1
Contents
Problem 3.40
80
Contents
Z
⇣ ⌘
a
⇡z
Problem
(a) I0 = k 3.40
cos
2a
dz = k
a
CHAPTER 3. POTENTIAL

λ(z)
✻
⇣ ⇡z ⌘
2a
sin
⇡
2a
V (r, ✓) ⇠
=
1
4⇡✏0
a
=
a
✓
2ak h ⇣ ⇡ ⌘
sin
⇡
2
4ak
⇡
◆
1
.
r
sin
⇣ ⇡ ⌘i 4ak
=
. Therefore:
2
⇡
(Monopole.)
✲z
Problem 3.40
−a
a
(b)
I0 = 0.
a
⇢⇣ ⌘
⇣ ⇡z ⌘ az
⇣ ⇡z ⌘ a
λ(z) Z
a 2
✻
I1 = kλ(z)z sin(⇡z/a) dz = k
sin
cos
⇡
a
⇡
a
✻
a
a
✲
z
⇢
⌘2 z
−a
a ⇣ a✲
a2
a2
2a2
−a = k
a
[sin(⇡) sin( ⇡)]
cos(⇡)
cos( ⇡) = k
;
⇡
⇡
⇡
⇡
λ(z)
λ(z)
✻✻
✓ 2 ◆
1
2a k 1
⇠
λ(z)
V
(r,
✓)
cos ✓. (Dipole.)
=
✲✲
✻
z
4⇡✏0
⇡
r2
−a−a
aa z
−a
a✲
z
(c)
I0 = I1 =
0.
λ(z)
a
⇢
Z✻
⇣ ⇡z ⌘
⇣ ⇡z ⌘
2z cos(⇡z/a) (⇡z/a)2 2
2
I2 =λ(z)
k z cos
+
sin
✲ z a dz = k
(⇡/a)2
(⇡/a)3
a
−a ✻ a a
⇣✲
−a
a
a ⌘z2
4a3 k
= 2k
[a cos(⇡) + a cos( ⇡)] =
.
⇡
⇡2
λ(z)
✻
a✲
−a
z
V (r, ✓) ⇠
=
1
4⇡✏0
✓
4a3 k
⇡2
◆
1
3 cos2 ✓
2r3
1 .
a
a
2
Problem 3.47
c
⃝2005
protected
under
all
currently
exist. qNo
(a)
The
average
field laws
dueastothey
a point
charge
atportion
r is of this material may be
Problem
3.41
reproduced, in any form or by any means, without permission in writing from the publisher.
r
✮
r
q
Eave =
1
4
3
3 ⇡R
1
Z
E d⌧,
1
Z
where E =
r̂
r
1
q
4⇡✏0 r 2
r̂
,
dτ
c
⃝2005
Pearson Education, Inc., Upper
This material is
so E
q 2 d⌧.
ave =River,
4
3 No
4⇡✏portion
protected under all copyright laws as they currently⇡R
exist.
of this material may be
0
3
reproduced, in any form or by any means, without permission in writing from the publisher.
(Here r is the source point, d⌧ is the field point, so r goes from r to d⌧ .) The field at r due to uniform
Z
r̂
1
charge ⇢ over the sphere is E⇢ =
⇢ 2 d⌧. This time d⌧ is the source point and r is the field
r
4⇡✏
0
c
⃝2005
protected
under
as they
currently
exist.
No portion
of this material
r allgoes
point, so
from laws
d⌧ to
r, and
hence
carries
the opposite
sign.may
So be
with ⇢ = q/ 43 ⇡R3 , the two
reproduced, in any form or by any means, without permission in writing from the publisher.
expressions agree: Eave = E⇢ .
(b) From Prob. 2.12:
E⇢ =
1
⇢r =
3✏0
q r
=
4⇡✏0 R3
p
.
4⇡✏0 R3
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81
CHAPTER 3. POTENTIAL
(c) If there are many charges inside the sphere, Eave is the sum of the individual averages, and ptot is the
p
sum of the individual dipole moments. So Eave =
. qed
4⇡✏0 R3
(d) The same argument, only with q placed at r outside the sphere, gives
Eave = E⇢ =
1
4⇡✏0
4
3
3 ⇡R ⇢
r2
r̂ (field at r due to uniformly charged sphere) =
1
q
r̂.
2
4⇡✏0 r
But this is precisely the field produced by q (at r) at the center of the sphere. So the average field (over
the sphere) due to a point charge outside the sphere is the same as the field that same charge produces
at the center. And by superposition, this holds for any collection of exterior charges.
Problem 3.48
(a)
p
ˆ
(2 cos ✓ r̂ + sin ✓ ✓)
4⇡✏0 r3
p
=
[2 cos ✓(sin ✓ cos x̂ + sin ✓ sin ŷ + cos ✓ ẑ)
4⇡✏0 r3
+ sin ✓(cos ✓ cos x̂ + cos ✓ sin ŷ sin ✓ ẑ)]
2
Edip =
=
Eave =
=
But
Z2⇡
0
3
p 6
7
2
2
43 sin ✓ cos ✓ cos x̂ + 3 sin ✓ cos ✓ sin ŷ + 2 cos ✓ sin ✓ ẑ5 .
4⇡✏0 r3
|
{z
}
1
4
3
3 ⇡R
1
4
3
⇡R
3
cos d =
Z2⇡
Z
✓
=3 cos2 ✓ 1
Edip d⌧
p
4⇡✏0
◆Z
1 ⇥
3 sin ✓ cos ✓(cos x̂ + sin ŷ) + 3 cos2 ✓
r3
sin d = 0, so the x̂ and ŷ terms drop out, and
2⇡
R
⇤
1 ẑ r2 sin ✓ dr d✓ d .
d = 2⇡, so
0
0
Eave =
1
4
3
⇡R
3
✓
p
4⇡✏0
◆
2⇡
ZR
Z⇡
1
dr
r
0
0
(
|
3 cos2 ✓
1 sin ✓ d✓ .
{z
}
cos3 ✓+cos ✓)|⇡
0 =1 1+1 1=0
Z
1
dr,
r
blows up, since ln r ! 1 as r ! 0. If, as suggested, we truncate the r integral at r = ✏, then it is finite, and
the ✓ integral gives Eave = 0.]
(b) We want E within the ✏-sphere to be a delta function: E = A 3 (r), with A selected so that the average
field is consistent with the general theorem in Prob. 3.47:
Z
1
A
p
p
p 3
Eave = 4 3
A 3 (r) d⌧ = 4 3 =
)A=
, and hence E =
(r).
3
4⇡✏0 R
3✏0
3✏0
3 ⇡R
3 ⇡R
Evidently Eave = 0, which contradicts the result of Prob. 3.47. [Note, however, that the r integral,
0
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R
82
CHAPTER 3. POTENTIAL
Problem 3.49 We need to show that the field inside the sphere approaches a delta-function with the right
coefficient (Eq. 3.106) in the limit as R ! 0. From Eq. 3.86, the potential inside is
V =
k
k
r cos ✓ =
z,
3✏0
3✏0
so
E=
rV =
k
ẑ.
3✏0
From Prob. 3.30, the dipole moment of this configuration is p = (4⇡R3 k/3) ẑ, so k ẑ = 3p/(4⇡R3 ), and hence
the field inside is
1
E=
p.
4⇡✏0 R3
Clearly E ! 1 as R ! 0 (if p is held constant); its volume integral is
Z
1
4
1
E d⌧ =
p ⇡R3 =
p,
4⇡✏0 R3 3
3✏0
which matches the delta-function term in Eq. 3.106. X
Problem 3.50
Z
(a) I = (rV1 ) · (rV2 ) d⌧ . But r·(V1 rV2 ) = (rV1 ) · (rV2 ) + V1 (r2 V2 ), so
I=
Problem 3.41
Z
2
r·(V1 rV2 ) d⌧
Z
I
1
V1 (r V2 ) =
V1 (rV2 ) · da +
✏
0
S
2
Z
V1 ⇢2 d⌧.
Z
1
But the surface integral is over a huge sphere “at infinity”, where V1 and V2 ! 0. So I =
V1 ⇢2 d⌧ . By
✏0
Z
Z
Z
r
1
the same
with 1 and 2 reversed, I =
V2 ⇢1 d⌧ . So V1 ⇢2 d⌧ = V2 ⇢1 d⌧ . qed
q
✮ argument,
✏0 R
9
R
dτ 8
< Situation (1 ) : Qa = a ⇢1 d⌧ = Q; Qb = b ⇢1 d⌧ = 0; V1b ⌘ Vab . =
(b)
R
R
:
;
Situation (2 ) : Qa = a ⇢2 d⌧ = 0; Qb = b ⇢2 d⌧ = Q; V2a ⌘ Vba .
8R
9
R
R
< V1 ⇢2 d⌧ = V1a a ⇢2 d⌧ + V1b b ⇢2 d⌧ = Vab Q. =
:R
V2 ⇢1 d⌧ = V2a
R
a
⇢1 d⌧ + V2b
R
b
⇢1 d⌧ = Vba Q.
Green’s reciprocity theorem says QVab = QVba , so Vab = Vba . qed
Problem 3.51
;
(a) Situation (1): actual. Situation (2): right plate at V0 , left plate at V = 0, no charge at x.
Z
V =0
V =0
x
V1 ⇢2 d⌧ = Vl1 Ql2 + Vx1 Qx2 + Vr1 Qr2 .
✲x
0 q
d
R
But Vl1 = Vr1 = 0 and Qx2 = 0, so V1 ⇢2 d⌧ = 0.
Z
V2 ⇢1 d⌧ = Vl2 Ql1 + Vx2 Qx1 + Vr2 Qr1 .
But Vl2 = 0 Qx1 = q, Vr2 = V0 , Qr1 = Q2 , and Vx2 = V0 (x/d). So 0 = V0 (x/d)q + V0 Q2 , and hence
Q2 =
qx/d.
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83
CHAPTER 3. POTENTIAL
Situation (1): actual. Situation (2): left plate at V0 , right plate at V = 0, no charge at x.
Z
Z
V1 ⇢2 d⌧ = 0 = V2 ⇢1 d⌧ = Vl2 Ql1 + Vx2 Qx1 + Vr2 Qr1 = V0 Q1 + qVx2 + 0.
⇣
But Vx2 = V0 1
x⌘
, so
d
Q1 =
q(1
x/d).
(b) Situation (1): actual. Situation (2): inner sphere at V0 , outer sphere at zero, no charge at r.
Z
V1 ⇢2 d⌧ = Va1 Qa2 + Vr1 Qr2 + Vb1 Qb2 .
R
But Va1 = Vb1 = 0, Qr2 = 0. So V1 ⇢2 d⌧ = 0.
Z
V2 ⇢1 d⌧ = Va2 Qa1 + Vr2 Qr1 + Vb2 Qb1 = Qa V0 + qVr2 + 0.
But Vr2 is the potential at r in configuration 2: V (r) = A + B/r, with V (a) = V0 ) A + B/a = V0 , or
aA + B = aV0 , and V (b) = 0 ) A + B/b = 0, or bA + B = 0. Subtract: (b a)A = aV0 ) A =
aV0 /(b a); B a1 1b = V0 = B (baba) ) B = abV0 /(b a). So V (r) = (baV0a) rb 1 . Therefore
aV0
Qa V0 + q
(b a)
✓
b
r
◆
1 = 0;
qa
Qa =
(b
a)
✓
b
r
◆
1 .
Now let Situation (2) be: inner sphere at zero, outer at V0 , no charge at r.
Z
Z
V1 ⇢2 d⌧ = 0 = V2 ⇢1 d⌧ = Va2 Qa1 + Vr2 Qr1 + Vb2 Qb1 = 0 + qVr2 + Qb V0 .
B
This time V (r) = A +
with V (a) = 0 ) A + B/a = 0; V (b) = V0 ) A + B/b = V0 , so
r
⇣
⌘
bV0
a
bV0 ⇣
a⌘
qb ⇣
a⌘
V (r) =
1
. Therefore q
1
+ Qb V0 = 0;
Qb =
1
.
(b a)
r
(b a)
r
(b a)
r
Problem 3.52
8
Z
3
3
3
X
1 < X 0X
(a)
r̂i r̂j Qij =
3
r̂i ri
r̂j rj0
:
2
i,j=1
i=1
j=1
But
3
X
i=1
r̂i ri0 = r̂ · r0 = r0 cos ↵ =
3
X
j=1
r̂j rj0 ;
X
i,j
(r0 )2
X
r̂i r̂j
i,j
r̂i r̂j
ij
=
X
ij
9
=
;
⇢ d⌧ 0
r̂j r̂j = r̂ · r̂ = 1.
Z
⌘
1 1
1 ⇣ 02
2
02
3r
cos
↵
r
⇢ d⌧ 0 = the third term in Eq. 3.96.
4⇡✏0 r3
2
⇥
(b) Because x2 = y 2 = (a/2)2 for all four charges, Qxx = Qyy = 12 3(a/2)2
p
Because z = 0 for all four charges, Qzz = 12 [ ( 2a/2)2 ](q q q + q) = 0 and
This leaves only
Qxy = Qyx =
So
X
p
⇤
( 2a/2)2 (q q q + q) = 0.
Qxz = Qyz = Qzx = Qzy = 0.
⇣a⌘ ⇣ a⌘
⇣ a⌘ ⇣a⌘
⇣ a⌘ ⇣ a⌘ i
3 h⇣ a ⌘ ⇣ a ⌘
3
q+
( q) +
( q) +
q = a2 q.
2 2
2
2
2
2
2
2
2
2
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84
CHAPTER 3. POTENTIAL
(c)
2Qij =
=
Z
⇥
3(ri
Z
di )(rj
dj )
(r
⇥
⇤
3ri rj r2 ij ⇢ d⌧
Z
2
d ij ⇢ d⌧ = Qij
d)2
3di
Z
ij
⇤
(I0 ll drop the primes, for simplicity.)
Z
Z
Z
3dj ri ⇢ d⌧ + 3di dj ⇢ d⌧ + 2d · r⇢ d⌧
⇢ d⌧
rj ⇢ d⌧
3(di pj + dj pi ) + 3di dj Q + 2
So if p = 0 and Q = 0 then Qij = Qij . qed
(d) Eq. 3.95 with n = 3:
Z
1 1
Voct =
(r0 )3 P3 (cos ↵)⇢ d⌧ 0 ;
4⇡✏0 r4
ij d
P3 (cos ✓) =
·p
d2
ij
ij Q.
1
5 cos3 ✓
2
3 cos ✓ .
1 1 X
r̂i r̂j r̂k Qijk .
4⇡✏0 r4
Voct =
i,j,k
Define the “octopole moment” as
1
2
Qijk ⌘
Problem 3.53
⇢ ✓
1
1
V =
q
r1
4⇡✏0
r
r
r
r
1
=
2
=
3
=
1
=
1
r
2
1
r
1
3
But q 0 =
+q
0
⇥ 0 0 0
5ri rj rk
✓
1
r
r
4
◆
✓
1
r
jk
+ rj0
ik
+ rk0
⇤
ij )
Problem 3.46
r
3
4
⇢(r0 ) d⌧ 0 .
◆
r
✯
❃
✒
✼▼
r4
r2
r
−q′
!
r
2
◆
r3
b
"#
a
2r
⇠
= 2 cos ✓ (we want a
a
r1
θ
+q
!"#\$ !"#\$+q′
−q
1
1
(r0 )2 (ri0
1
p
r2 + a2 2ra cos ✓,
p
r2 + a2 + 2ra cos ✓,
p
r2 + b2 2rb cos ✓,
p
r2 + b2 + 2rb cos ✓.
Expanding as in Ex. 3.10:
✓
◆
Z
b
\$!
r, not r
"#
a
\$
a, this time).
2b
⇠
= 2 cos ✓ (here we want b ⌧ r, because b = R2 /a, Eq. 3.16)
r
2 R2
=
cos ✓.
a r2
R
q (Eq. 3.15), so
a
V (r, ✓) ⇠
=

1
2r
q 2 cos ✓
4⇡✏0 a
R 2 R2
1
q
cos ✓ =
2
a ar
4⇡✏0
✓
2q
a2
◆✓
r
R3
r2
◆
cos ✓.
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85
CHAPTER 3. POTENTIAL
Set E0 =
1 2q
(field in the vicinity of the sphere produced by ±q):
4⇡✏0 a2
✓
E0 r
V (r, ✓) =
R3
r2
◆
cos ✓
(agrees with Eq. 3.76).
Problem 3.54
The boundary conditions are
9
(i) V = 0 when y = 0, >
>
=
(ii) V = V0 when y = a,
(iii) V = 0 when x = b, >
>
;
(iv) V = 0 when x = b.
Go back to Eq. 3.26 and examine the case k = 0: d2 X/dx2 = d2 Y /dy 2 = 0, so X(x) = Ax + B, Y (y) = Cy + D.
But this configuration is symmetric in x, so A = 0, and hence the k = 0 solution is V (x, y) = Cy + D. Pick
D = 0, C = V0 /a, and subtract o↵ this part:
V (x, y) = V0
y
+ V̄ (x, y).
a
The remainder (V̄ (x, y)) satisfies boundary conditions similar to Ex. 3.4:
9
(i) V̄ = 0 when y = 0,
>
>
=
(ii) V̄ = 0 when y = a,
(iii) V̄ = V0 (y/a) when x = b, >
>
;
(iv) V̄ = V0 (y/a) when x = b.
(The point of peeling o↵ V0 (y/a) was to recover (ii), on which the constraint k = n⇡/a depends.)
The solution (following Ex. 3.4) is
V̄ (x, y) =
1
X
Cn cosh(n⇡x/a) sin(n⇡y/a),
n=1
and it remains to fit condition (iii):
V̄ (b, y) =
Invoke Fourier’s trick:
X
Cn cosh(n⇡b/a)
Z
X
a
Cn cosh(n⇡b/a) sin(n⇡y/a) =
sin(n⇡y/a) sin(n0 ⇡y/a) dy =
0
a
Cn cosh(n⇡b/a) =
2
V0
a
Z
a
V0 (y/a).
V0
a
Z
a
y sin(n0 ⇡y/a) dy,
0
y sin(n⇡y/a) dy.
0
⇣ ⌘
⇣ ay ⌘
2V0
a 2
Cn =
sin(n⇡y/a)
cos(n⇡y/a)
a2 cosh(n⇡b/a)
n⇡
n⇡
✓ 2◆
2V0
a
2V0
( 1)n
= 2
cos(n⇡) =
.
a cosh(n⇡b/a) n⇡
n⇡ cosh(n⇡b/a)
a
0
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86
CHAPTER 3. POTENTIAL
V (x, y) = V0
"
#
1
y
2 X ( 1)n cosh(n⇡x/a)
+
sin(n⇡y/a) .
a ⇡ n=1 n cosh(n⇡b/a)
V (x, y) = (C sin kx + D cos kx) Aeky + Be
ky
.
Note that the configuration is symmetric in x, so C = 0, and V (x, 0) = 0 ) B =
constants)
A, so (combining the
V (x, y) = A cos kx sinh ky.
But V (b, y) = 0, so cos kb = 0, which means that kb = ±⇡/2, ±3⇡/2, . . . , or k = (2n 1)⇡/2b ⌘ ↵n , with
n = 1, 2, 3, . . . (negative k does not yield a di↵erent solution—the sign can be absorbed into A). The general
linear combination is
1
X
V (x, y) =
An cos ↵n x sinh ↵n y,
n=1
and it remains to fit the final boundary condition:
V (x, a) = V0 =
1
X
An cos ↵n x sinh ↵n a.
n=1
Use Fourier’s trick, multiplying by cos ↵n0 x and integrating:
V0
Z
V0
b
b
cos ↵n0 x dx =
1
X
An sinh ↵n a
n=1
1
2 sin ↵n0 b X
=
An sinh ↵n a(b
↵n0
n=1
2V0 sin ↵n b
So An =
. But sin ↵n b = sin
b ↵n sinh ↵n a
V (x, y) =
✓
2n
1
2
⇡
◆
Z
b
b
n0 n )
=
cos ↵n0 x cos ↵n x dx,
= bAn0 sinh ↵n0 a.
( 1)n , so
1
2V0 X
sinh ↵n y
( 1)n
cos ↵n x.
b n=1
↵n sinh ↵n a
Problem 3.55
(a) Using Prob. 3.15b (with b = a):
V (x, y) =
4V0 X sinh(n⇡x/a) sin(n⇡y/a)
.
⇡
n sinh(n⇡)
n odd
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87
CHAPTER 3. POTENTIAL
(y) =
✏0
@V
@x
x=0
=
✏0
4V0 X ⇣ n⇡ ⌘ cosh(n⇡x/a) sin(n⇡y/a)
⇡
a
n sinh(n⇡)
n odd
x=0
4✏0 V0 X sin(n⇡y/a)
=
.
a
sinh(n⇡)
n odd
Z a
Z a
4✏0 V0 X
1
=
(y) dy =
sin(n⇡y/a) dy.
a
sinh(n⇡) 0
0
n odd
Z a
a
a
2a
a
But
sin(n⇡y/a) dy =
cos(n⇡y/a) 0 =
[1 cos(n⇡)] =
(since n is odd).
n⇡
n⇡
n⇡
0
8✏0 V0 X
1
✏0 V0
=
=
ln 2.
⇡
n sinh(n⇡)
⇡
n odd
[Summing the series numerically (using Mathematica) gives 0.0866434, which agrees precisely with ln 2/8.
The series can be summed analytically, by manipulation of elliptic integrals—see “Integrals and Series, Vol. I:
Elementary Functions,” by A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (Gordon and Breach, New
York, 1986), p. 721. I thank Ram Valluri for calling this to my attention.]
Using Prob. 3.54 (with b = a/2):
"
#
y
2 X ( 1)n cosh(n⇡x/a) sin(n⇡y/a)
V (x, y) = V0
+
.
a ⇡ n
n cosh(n⇡/2)
(x) =
=
=
=
=
"
#
@V
1
2 X ⇣ n⇡ ⌘ ( 1)n cosh(n⇡x/a) cos(n⇡y/a)
✏0
= ✏0 V0
+
@y y=0
a ⇡ n
a
n cosh(n⇡/2)
y=0
"
#
"
#
X ( 1)n cosh(n⇡x/a)
1 2 X ( 1)n cosh(n⇡x/a)
✏0 V0
✏0 V0
+
=
1+2
.
a a n
cosh(n⇡/2)
a
cosh(n⇡/2)
n
"
#
Z a/2
X ( 1)n Z a/2
✏0 V0
(x) dx =
a+2
cosh(n⇡x/a) dx .
a
cosh(n⇡/2) a/2
a/2
n
Z a/2
a/2
a
2a
cosh(n⇡x/a) dx =
sinh(n⇡x/a)
=
sinh(n⇡/2).
But
n⇡
n⇡
a/2
a/2
"
#
"
#
✏0 V0
4a X ( 1)n tanh(n⇡/2)
4 X ( 1)n tanh(n⇡/2)
a+
= ✏0 V0 1 +
a
⇡ n
n
⇡ n
n
✏0 V0
ln 2.
⇡
[The numerical value is -0.612111, which agrees with the expected value (ln 2 ⇡)/4.]
(b) From Prob. 3.24:
◆
1 ✓
X
1
k
V (s, ) = a0 + b0 ln s +
ak s + bk k [ck cos(k ) + dk sin(k )].
s
k=1
In the interior (s < R) b0 and bk must be zero (ln s and 1/s blow up at the origin). Symmetry ) dk = 0. So
V (s, ) = a0 +
1
X
ak sk cos(k ).
k=1
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4
Problem 3.48
88
y
CHAPTER 3. POTENTIAL
✻
V0
✲x
At the surface:
V (R, ) =
X
ak R cos(k ) =
k
k=0
⇢
V0 , if
⇡/4 <
0, otherwise.
< ⇡/4,
Fourier’s trick: multiply by cos(k 0 ) and integrate from ⇡ to ⇡:
Problem 3.49
8
⇡/4
Z ⇡
Z ⇡/4
1
<
X
V0 sin(k 0 )/k 0
= (V0 /k 0 ) sin(k 0 ⇡/4), if k 0 6= 0,
k
0
0
ak R
cos(k ) cos(k ) d = V0
cos(k ) d =
⇡/4
: V ⇡/2, if k 0 = 0.
⇡
⇡/4
k=0
0
z
But
✻
θ
φ
Z
l
T❦
8
< 0, if k 6= k 0
0
cos(k ) cos(k ) d = 2⇡, if k = k 0 = 0,
:
⇡
⇡, if k = k 0 6= 0.
⇡
mg
❄
So 2⇡a0 = V0 ⇡/2 ) a0 = V0 /4; ⇡ak Rk = (2V0 /k) sin(k⇡/4) ) ak = (2V0 /⇡kRk ) sin(k⇡/4) (k 6= 0); hence
V (s, ) = V0
"
#
1
1
2 X sin(k⇡/4) ⇣ s ⌘k
+
cos(k ) .
4 ⇡
k
R
k=1
Using Eq. 2.49, and noting that in this case n̂ =
( ) = ✏0
@V
@s
s=R
= ✏0 V0
ŝ:
1
2 X sin(k⇡/4) k
ks
⇡
kRk
1
cos(k )
k=1
s=R
We want the net (line) charge on the segment opposite to V0 ( ⇡ <
=
=
Z
( )R d = 2R
4✏0 V0
⇡
Z
⇡
( )d =
3⇡/4
1
X
k=1
sin(k⇡/4)

sin(k )
k
=
1
2✏0 V0 X
sin(k⇡/4) cos(k ).
⇡R
k=1
<
3⇡/4 and 3⇡/4 <
< ⇡):
Z ⇡
1
4✏0 V0 X
sin(k⇡/4)
cos(k ) d
⇡
3⇡/4
k=1
⇡
3⇡/4
=
1
4✏0 V0 X sin(k⇡/4) sin(3k⇡/4)
.
⇡
k
k=1
k sin(k⇡/4) sin(3k⇡/4) product
p
p
1
1/ 2
1/ 2
1/2
2
1p
-1
-1
p
3
1/ 2
1/ 2
1/2
4
0p
0p
0
5
-1/ 2
-1/ 2
1/2
c
6Pearson Education,
-1p
⃝2005
Inc., Upper
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7
-1/
2
-1/
2
1/2
reproduced, in any form or by any means, without permission in writing from the publisher.
8
0
0
0
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89
CHAPTER 3. POTENTIAL
"
#
4✏0 V0 1 X 1 1 X 1
=
=
= 0.
⇡
2 1,3,5... k 2 1,3,5,... k
P
Ouch! What went wrong? The problem is that the series (1/k) is divergent, so the “subtraction” 1 1
is suspect. One way to avoid this is to go back to V (s, ), calculate ✏0 (@V /@s) at s 6= R, and save the limit
s ! R until the end:
1
@V
2✏0 V0 X sin(k⇡/4) ksk 1
( , s) ⌘ ✏0
=
cos(k )
4
@s
⇡
k
Rk
4✏0 V0
⇡
"
1 X 1
2 1,3,5... k
X
1
k
2,6,10,...
#
k=1
1
2✏0 V0 X k
x
Problem =3.48
⇡R
1
sin(k⇡/4) cos(k )
k=1
(x) ⌘
( , s)R d =
1
4✏0 V0 X 1 k
x
⇡
k
1
(where x ⌘ s/R ! 1 at the end).
sin(k⇡/4) sin(3k⇡/4)
k=1
3
5
 ✓
◆
✓
◆
✻ 4✏0 V0 1
x
x
1 x2
x6
x10
=
x+
+
+ ···
+
+
+ ···
⇡ V 2x
3
5
x 2
6
10
0
✓
◆ ✓
◆
2✏0 V✲
x3
x5
x6
x10
0
x x+
=
+
+ ···
x2 +
+
+ ···
.
⇡x
3
5
3
5
✓
◆
✓
◆
1+x
x3
x5
But (see math tables) : ln
=2 x+
+
+ ··· .
1 x
3
5

✓
◆
✓
◆
✓
◆✓
◆
2
2✏0 V0 1
1+x
1
1+x
✏0 V0
1+x
1 + x2
=
ln
ln
=
ln
⇡x
2
1 x
2
1 x2
⇡x
1 x
1 x2

Problem 3.49
✏0 V0
(1 + x)2
✏0 V0
=
ln
;
= lim (x) =
ln 2.
2
x!1
⇡x
1+x
⇡
y
Problem 3.56
z
✻
θ
φ
F = qE =
l
T❦
qp
ˆ
(2 cos ✓ r̂ + sin ✓ ✓).
4⇡✏0 r3
mg
❄
Now consider the pendulum: F = mg ẑ T r̂, where T
mg cos = mv 2 /l and (by conservation of
2
2
energy) mgl cos = (1/2)mv ) v = 2gl cos (assuming it started from rest at = 90 , as stipulated). But
cos = cos ✓, so T = mg( cos ✓) + (m/l)( 2gl cos ✓) = 3mg cos ✓, and hence
F=
mg(cos ✓ r̂
ˆ + 3mg cos ✓ r̂ = mg(2 cos ✓ r̂ + sin ✓ ✓).
ˆ
sin ✓ ✓)
This total force is such as to keep the pendulum on a circular arc, and it is identical to the force on q in the
field of a dipole, with mg \$ qp/4⇡✏0 l3 . Evidently q also executes semicircular motion, as though it were on a
tether of fixed length l.
Problem 3.57 Symmetry suggests that the plane of the orbit is perpendicular to the z axis, and since we need
a centripetal force, pointing in toward the axis, the orbit must lie at the bottom of the field loops (Fig. 3.37a),
where the z component of the field is zero. Referring to Eq. 3.104,
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90
CHAPTER 3. POTENTIAL
p
p
E·ẑ = 0 ) 3(p·r̂)(r̂·ẑ) p·ẑ = 0, or 3 cos2 ✓ 1 = 0. So cos2 ✓ = 1/3, cos ✓ = 1/ 3, sin ✓ = 2/3, z/s =
p
tan ✓ ) z =
2 s.
The field at the orbit is (Eq. 3.103)
!
r
1
p
2ˆ
E=
2 p r̂ +
✓
4⇡✏0 r3
3
3
r h
i
p
p
2
=
2(sin
✓
cos
x̂
+
sin
✓
sin
ŷ
+
cos
✓
ẑ)
+
(cos
✓
cos
x̂
+
cos
✓
sin
ŷ
sin
✓
ẑ)
4⇡✏0 r3 3
r h⇣
⌘
⇣ p
⌘
⇣ p
⌘ i
p
p
2
=
2 sin ✓ + cos ✓ cos x̂ +
2 sin ✓ + cos ✓ sin ŷ +
2 cos ✓ sin ✓ ẑ
3
4⇡✏0 r
3
!
r "
r
r ! #
p
p 1
p
2
2
1
2
p
=
2
(cos x̂ + sin ŷ) +
2p
ẑ
4⇡✏0 r3 3
3
3
3
3
r h
i
p
p
2
p p
p
p
=
3
(cos
x̂
+
sin
ŷ)
=
2 ŝ =
ŝ.
3
3
4⇡✏0 r
3
4⇡✏0 r
3 3⇡✏0 s3
p
(I used s = r sin ✓ = r 2/3, in the last step.)
The centripetal force is
F = qE =
qp
=
3 3⇡✏0 s3
p
mv 2
s
qp
v2 = p
3 3⇡✏0 ms2
)
The angular momentum is
L = smv =
r
v=
)
1
s
r
qp
p
.
3 3⇡✏0 m
qpm
p
,
3 3⇡✏0
the same for all orbits, regardless of their radius (!), and the energy is
W =
Problem 3.58
Potential of q:
1
1
qp
q p cos ✓
qp
p
= p
mv 2 + qV =
+
2
2
2
2 3 3⇡✏0 s
4⇡✏0 r
6 3⇡✏0 s2
Vq (r) =
1 q
,
4⇡✏0 r
where
Equation 3.94, with r0 ! a and ↵ ! ✓:
= a2 + r2 2ar cos ✓.
1
1
1 X ⇣ a ⌘n
=
r r n=0 r Pn (cos ✓). So
Vq (r, ✓) =
Meanwhile, the potential of
is (Eq. 3.79)
r
qp
p
= 0.
4⇡✏0 3(3/2)s2
2
1
q 1 X ⇣ a ⌘n
Pn (cos ✓).
4⇡✏0 r n=0 r
V (r, ✓) =
1
X
Bl
Pl (cos ✓).
rl+1
l=0
Comparing the two (Vq = V ) we see that Bl = (q/4⇡✏0 )al , and hence (Eq. 3.81) Al = (q/4⇡✏0 )al /R2l+1 . Then
(Eq. 3.83)
" 1
#
1
1 ⇣ ⌘l
⇣ a ⌘l
X ⇣ a ⌘l
X
q X
q
a
(✓) =
(2l + 1)
Pl (cos ✓) =
2
l
Pl (cos ✓) +
Pl (cos ✓) .
4⇡R2
R
4⇡R2
R
R
l=0
l=0
l=0
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91
CHAPTER 3. POTENTIAL
Now (second line above, with r ! R)
p
1
a2 + R2
2aR cos ✓
=
1 X ⇣ a ⌘l
Pl (cos ✓).
R
R
1
l=0
Di↵erentiating with respect to a:
d
da
Thus
✓
p
1
a2 + R2
2aR cos ✓
◆
=
(a R cos ✓)
1 X ⇣ a ⌘l
=
l
Pl (cos ✓).
aR
R
(a2 + R2 2aR cos ✓)3/2
l=0
1

q
(a R cos ✓)
R
2aR 2
+ 2
4⇡R2
(a + R2 2aR cos ✓)3/2
(a + R2 2aR cos ✓)1/2
⇥
⇤
2a(a R cos ✓) + (a2 + R2 2aR cos ✓)
q
q
(R2 a2 )
=
=
.
2
2
3/2
2
4⇡R
4⇡R (a + R2 2aR cos ✓)3/2
(a + R
2aR cos ✓)
(✓) =
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92
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Chapter 4
Electric Fields in Matter
Problem 4.1
E = V /x = 500/10 3 = 5⇥105 . Table 4.1: ↵/4⇡✏0 = 0.66⇥10 30 , so ↵ = 4 ⇡(8.85⇥10 12 )(0.66⇥10
7.34 ⇥ 10 41 . p = ↵E = ed ) d = ↵E/e = (7.34 ⇥ 10 41 )(5 ⇥ 105 )/(1.6 ⇥ 10 19 ) = 2.29 ⇥ 10 16 m.
d/R = (2.29 ⇥ 10
Rex/↵ = (0.5 ⇥ 10
Problem 4.2
16
)/(0.5 ⇥ 10
10
10
)(1.6 ⇥ 10
) = 4.6 ⇥ 10
19
)(10
3
6
30
)=
. To ionize, say d = R. Then R = ↵E/e = ↵V /ex ) V =
)/(7.34 ⇥ 10
First find the field, at radius r, using Gauss’ law:
R
41
) = 108 V.
E·da =

Z
4⇡q r 2r/a 2
4q
a
=
⇢ d⌧ =
e
r dr = 3
e
3
⇡a 0
a
2
0

✓
◆

2q
a2
a2
2r/a
2
=
e
r
+
ar
+
=
q
1
a2
2
2
1
✏0
1 1
4⇡✏0 r 2 Qenc .
◆ r
a2
Qenc
r + ar +
2
0
✓
◆
2
r
r
e 2r/a 1 + 2 + 2 2
.
a
a
h
⇣
⌘i
1 q
r
r2
2r/a
[Note: Qenc (r ! 1) = q.] So the field of the electron cloud is Ee = 4⇡✏
1
e
1
+
2
+
2
. The
2
2
a
a
0 r
proton will be shifted from r = 0 to the point d where Ee = E (the external field):

✓
◆
1 q
d
d2
2d/a
E=
1 e
1+2 +2 2
.
4⇡✏0 d2
a
a
Z
r
2r/a
✓
Qenc , or E =
2
Expanding in powers of (d/a):
e
1
e
2d/a
✓
2d/a
=1
◆
=1
d
d2
1+2 +2 2
a
a
✓
2d
a
1
◆
◆2
2d
a
✓ ◆2
d
d
2 +2
a
a
1
+
2
✓
◆3
✓ ◆2
2d
d
d
+ ··· = 1 2 + 2
a
a
a
!✓
✓ ◆3
◆
4 d
d
d2
+ ···
1+2 +2 2
3 a
a
a
1
3!
✓
d
d2
d
d2
d3
1/
2/
2 /2 + 2/ + 4 /2 + 4 /3
a
a
a
a
a
✓ ◆3
4 d
=
+ higher order terms.
3 a
= 1/
d2
2 /2
a
4
3
✓ ◆3
d
+ ···
a
d3
4 d3
4 /3 +
+ ···
a
3 a3
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1
93
CHAPTER 4. ELECTRIC FIELDS IN MATTER
E=
1 q
4⇡✏0 d2
✓
4 d3
3 a3
◆
=
1 4
1
(qd) =
p.
4⇡✏0 3a3
3⇡✏0 a3
↵ = 3⇡✏0 a3 .
[Not so di↵erent from the uniform sphere model of Ex. 4.1 (see Eq. 4.2). Note that this result predicts
1
3 3
3
10 3
= 0.09 ⇥ 10 30 m3 , compared with an experimental value (Table 4.1) of
4⇡✏0 ↵ = 4 a = 4 0.5 ⇥ 10
0.66 ⇥ 10 30 m3 . Ironically the “classical” formula (Eq. 4.2) is slightly closer to the empirical value.]
Contents
Problem 4.3
H
Rr
⇢(r) = Ar. Electric field (by Gauss’s Law): E·da = E 4⇡r2 = ✏10 1Qenc = ✏10 0 Ar 4⇡r2 dr, or E =
1 4⇡A r4
Ar2
=
. This “internal” field balances the external field E when nucleus is “o↵-center” an amount
2
4⇡r ✏0 4
4✏0
p
p
p
Problem
4✏0 E/A. So the induced dipole moment is p = ed = 2e ✏0 /A E. Evidently
0 = E ) d =
p is proportional to E 1/2 .
For Eq. 4.1 to hold in the weak-field limit, E must be proportional to r, for small r, which means that ⇢
must go to a constant (not zero) at the origin: ⇢(0) 6= 0 (nor infinite).
Problem 4.4
Contents
r
1 q
Field of q: 4⇡✏
2 r̂. Induced dipole moment of atom: p = ↵ E =
0 r
↵q
r̂.
2
4⇡✏0 r
✓
◆
q
1 1
2↵ q
Hello
Field of this dipole, at location of q (✓ = ⇡, in Eq. 3.103): E =
(to the right).
4⇡✏0 r3 4⇡✏0 r2
✓
◆2
q
1
Problem
Force on 4.6
q due to this field: F = 2↵
(attractive).
5
4⇡✏
r
θ+
0
Problem 4.4✸
p
− 4.5
Problem
z
p1 ˆ
Field of p1 at p2 (✓ = ⇡/2 in Eq. 3.103): E1 =
✓ (points down).
4⇡✏0 r3
z
p1 p2
Torque
(points into the page).
+ ❦on p2 : N2 = p2 ⇥ E1 = p2 E1 sin 90 = p2 E1 =
4⇡✏0 r3
pi −
θ
p2
Field of p2 at p1 (✓ = ⇡ in Eq. 3.103): E2 =
( 2 r̂) (points to the right).
r
4⇡✏0 r3
q
2p1 p2
Torque on p1 : N1 = p1 ⇥ E2 =
(points into the page).
4⇡✏0 r3
Problem 4.6
Problem 4.6
(a)
θ+
✸
p
−
z
+❦
pi
z
θ
(b)
θ
pi ✻
✲θ
p
Use image dipole as shown in Fig. (a).
Redraw, placing pi at the origin, Fig. (b).
2z
−
Ei =
p
ˆ
(2 cos ✓ r̂ + sin ✓ ✓);
4⇡✏0 (2z)3
ˆ
p = p cos ✓ r̂ + p sin ✓ ✓.
h
i
p2
ˆ ⇥ (2 cos ✓ r̂ + sin ✓ ✓)
ˆ
(cos
✓
r̂
+
sin
✓
✓)
4⇡✏0 (2z)3
h
i
2
p
ˆ + 2 sin ✓ cos ✓( ˆ)
=
cos
✓
sin
✓
4⇡✏0 (2z)3
c
⃝2005
Pearson Education, Inc.,
( ) (out of the page).
protected under all copyright laws as
they
currently exist. No portion of this material may be
4⇡✏
0 (2z)
N = p ⇥ Ei =
reproduced, in any form or by any means, without permission in writing from the publisher.
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reproduced, in any form or by any means, without permission in writing from the publisher.
c
⃝2005
1
94
CHAPTER 4. ELECTRIC FIELDS IN MATTER
p2 sin 2✓
(out of the page).
4⇡✏0 (16z 3 )
For 0 < ✓ < ⇡/2, N tends to rotate p counterclockwise; for ⇡/2 < ✓ < ⇡, N rotates p clockwise. Thus the
stable orientation is perpendicular to the surface—either " or #.
But sin ✓ cos ✓ = (1/2) sin 2✓, so N =
Problem 4.7
If the potential is zero at infinity, the energy of a point charge Q is (Eq. 2.39) W = QV (r). For a physical
dipole, with q at r and +q at r + d,
" Z
#
U = qV (r + d)
qV (r) = q[V (r + d)
r+d
V (r)] = q
r
E · dl .
For an ideal dipole the integral reduces to E · d, and
U=
qE · d =
p · E,
since p = qd. If you do not (or cannot) use infinity as the reference point, the result still holds, as long as
you bring the two charges in from the same point, r0 (or two points at the same potential). In that case
W = Q[V (r) V (r0 )], and
U = q[V (r + d)
V (r0 )]
q[V (r)
V (r0 )] = q[V (r + d)
V (r)],
as before.
Problem 4.8
U=
p1 ·E2 , but E2 =
Problem 4.9
1 1
4⇡✏0 r 3
[3 (p2 ·r̂) r̂
p2 ]. So U =
1 1
4⇡✏0 r 3
[p1 ·p2
3 (p1 ·r̂) (p2 ·r̂)]. qed
1 q
q
x x̂ + y ŷ + z ẑ
r̂ =
.
4⇡✏0 r2
4⇡✏0 (x2 + y 2 + z 2 )3/2
✓
◆
@
@
@
q
x
Fx = px
+ py
+ pz
2
2
@x
@y
@z 4⇡✏0 (x + y + z 2 )3/2
⇢ 

q
1
3
2x
3
2y
=
px
x 2
+ py
x 2
2
2
2
3/2
2
2
5/2
2
4⇡✏0
2 (x + y + z )
2 (x + y + z 2 )5/2
(x + y + z )



3
2z
q
px
3x
q
p
+ pz
x 2
=
(px x + py y + pz z) =
3
5
2
2
5/2
2 (x + y + z )
4⇡✏0 r
r
4⇡✏0 r3
(a) F = (p · r)E (Eq. 4.5); E =
F=
1 q
[p
4⇡✏0 r3
3r(p · r)
r5
.
x
3(p · r̂) r̂] .
1 1
1 1
{3 [p · ( r̂)] ( r̂) p} =
[3(p · r̂) r̂
3
4⇡✏0 r
4⇡✏0 r3
are because r points toward p, in this problem.)
(b) E =
F = qE =
p] . (This is from Eq. 3.104; the minus signs
1 q
[3(p · r̂) r̂
4⇡✏0 r3
p] .
[Note that the forces are equal and opposite, as you would expect from Newton’s third law.]
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
✲θ
p
θ
pi ✻
z
Problem 4.7
CHAPTER 4. ELECTRIC FIELDS IN MATTER
95
Problem 4.10
(a)
b
✻
y
= P·n̂ = kR; ⇢b =
E
✻
r·P =
1 @ 2
(r kr) =
r2 @r
1
3kr2 =
r2
3k.
(b) For r < R, E = 3✏10θ⇢r r̂ (Prob. 2.12), so E = (k/✏0 ) r.
✲
✲
✒p
p
x
For r > R, same as if all charge at center; but Qtot = (kR)(4⇡R2 ) + ( 3k)( 43 ⇡R3 ) = 0, so E = 0.
Problem 4.11
⇢b = 0; b = P·n̂ = ±P (plus sign at one end—the one P points toward ; minus sign at the other—the one
P points
away 4.11
from).
Problem
(i) L
a. Then the ends look like point charges, and the whole thing is like a physical dipole, of length L and
charge P ⇡a2 . See Fig. (a).
(ii) L ⌧ a. Then it’s like a circular parallel-plate capacitor. Field is nearly uniform inside; nonuniform “fringing
field” at the edges. See Fig. (b).
(iii) L ⇡ a. See Fig. (c).
✛
✛
✛
−
✛
✛
✛
(a) Like a dipole
+
✲
P
✛
✻✻
✛
✛
✛
✛
✛
✛
✛
✛
✛
✛
✛
− +❄ ❄
✛
✛
✛
✛
✛
✛
✛
✛
✛
✲
P
(b) Like a parallel-plate capacitor
✲
P
(c)
Problem 4.12
n
o
R P· r̂
R r̂
1
1
V = 4⇡✏
. But the term in curly brackets is precisely the field of a uniformly
2 d⌧ = P·
2 d⌧
4⇡✏
0
0
r
r
charged sphere, divided by ⇢. The integral was done explicitly in Probs. 2.7 and 2.8:
1
4⇡✏0
Z
8
9
1 (4/3)⇡R3 ⇢
>
>
>
r̂, (r > R), >
>
>
=
r2
1 < 4⇡✏0
r̂
r 2 d⌧ = ⇢ >
>
>
Problem 4.13
>
3
>
: 1 (4/3)⇡R ⇢ r, (r < R). >
;
3
4⇡✏0
R
So V (r, ✓) =
8
9
>
>
R3
R3 P cos ✓
>
>
>
P·r̂ =
, (r > R), >
>
>
2
2
>
>
3✏0 r
< 3✏0 r
=
>
>
>
1
P r cos ✓
>
>
P·r =
,
:
3✏0
3✏0
>
>
>
>
(r < R). >
;
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
Think of it as two cylinders of opposite uniform charge density ±⇢. Inside, the field at a distance s from
the axis of a uniformly charge cylinder is given by Gauss’s law: E2⇡s` = ✏10 ⇢⇡s2 ` ) E = (⇢/2✏0 )s. For
two such cylinders, one plus and one minus, the net field (inside) is E = E+ + E = (⇢/2✏0 ) (s+ s ). But
s+ s = d, so E = ⇢d/(2✏0 ), where d is the vector from the negative axis to positive axis. In this case
the total dipole moment of a chunk of length ` is P ⇡a2 ` = ⇢⇡a2 ` d. So ⇢d = P, and E =
s < a.
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P/(2✏0 ), for
96
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Outside, Gauss’s law gives E2⇡s` =
⇣
⌘
2
ŝ+
ŝ
E+ + E = ⇢a
, where
2✏0
s+
s
✓
ŝ+
s+
1
2
✏0 ⇢⇡a `
)E=
⇢a2 ŝ
2✏0 s ,
for one cylinder. For the combination, E =
d
s± = s ⌥ ;
2
✓
◆✓
◆ 1
✓
◆✓
◆ 1
✓
◆✓
◆
s±
d
d2
1
d
s·d
1
d
s·d
2
⇠
⇠
= s⌥
s +
⌥s·d
1⌥ 2
1± 2
= 2 s⌥
= 2 s⌥
s2±
2
4
s
2
s
s
2
s
✓
◆
1
(s · d) d
= 2 s±s 2 ⌥
(keeping only 1st order terms in d).
s
s
2
◆
✓
◆ ✓
◆
✓
◆
ŝ
1
(s · d) d
(s · d) s
1
s(s · d)
= 2
s+s 2
s s 2 +
= 2 2
d .
s
s
s
2
s
2
s
s2
E(s) =
a2 1
[2(P · ŝ) ŝ
2✏0 s2
P] ,
for s > a.
Problem 4.14
H
H
R
Total charge
is Qtot = S b da + V ⇢b d⌧ = S P · da
H on the dielectric
R
theorem says S P · da = V r·P d⌧ , so Qenc = 0. qed
Problem 4.15
V
r·P d⌧ . But the divergence
◆
⇢
k
k
+P · r̂ = k/b (at r = b),
=
;
=
P·n̂
=
b
P · r̂ = k/a (at r = a).
r
r2
1 Qenc
Gauss’s law ) E = 4⇡✏
r̂.
For
r
<
a,
Q
=
0,
so
E
=
0.
For r > b, Qenc = 0 (Prob. 4.14), so E = 0.
2
enc
0 r
R
r
k
k
2
For a < r < b, Qenc = a
4⇡a2 + a r2 4⇡r dr = 4⇡ka 4⇡k(r a) = 4⇡kr; so E = (k/✏0 r) r̂.
H
(b) D·da = Qfenc = 0 ) D = 0 everywhere. D = ✏0 E + P = 0 ) E = ( 1/✏0 )P, so
E = 0 (for r < a and r > b);
E = (k/✏0 r) r̂ (for a < r < b).
(a) ⇢b =
r·P =
1 @
r2 @r
✓
R
r2
Problem 4.16
(a) Same as E0 minus the field at the center of a sphere with uniform polarization P. The latter (Eq. 4.14)
1
is P/3✏0 . So E = E0 +
P. D = ✏0 E = ✏0 E0 + 13 P = D0 P + 13 P, so D = D0 23 P.
3✏0
(b) Same as E0 minus the field of ± charges at the two ends of the “needle”—but these are small, and far
away, so E = E0 . D = ✏0 E = ✏0 E0 = D0 P, so D = D0 P.
(c) Same as E0 minus the field of a parallel-plate capacitor with upper plate at
(1/✏0 )P , so E = E0 +
1
✏0 P.
= P . The latter is
D = ✏0 E = ✏0 E0 + P, so D = D0 .
protected under all copyright laws as they currently exist. No portion of this material may be
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97
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Problem 4.17
✻✻✻✻✻✻✻
❄ ❄ ❄❄ ❄ ❄❄ ❄ ❄
❄ ❄❄
❄❄ ❄
✻✻✻
✻✻✻
6
P
(uniform)
Problem 4.18
E
(field of two circular plates)
D
(same as E outside, but lines
continuous, since ∇·D = 0)
For more detailed figures see the solution to Problem 6.14, reading P for M, E for H, and D for B.
Problem 4.18
R
(a) Apply D · da = Qfenc to the gaussian surface shown. DA = A ) D = . (Note: D = 0 inside the
metal plate.) This is true in both slabs; D points down.
6
+σ
❖
(b)
D
=
✏E
)
E
=
/✏
in
slab
1,
E
=
/✏
in
slab
2.
But
✏ = ✏0 ✏r , so ✏1 = 2✏0 ; ✏2 = 32 ✏0 . E1 = /2✏0 ,
1
2
Problem 4.18
E2 = 2 /3✏0 .
(c) P = ✏0
e E,
so P = ✏0
e d/(✏0 ✏r )
=(
e /✏r )
;
e
(d) V = E1 a + E2 a = ( a/6✏0 )(3 + 4) = 7 a/6✏0 .
= ✏r
1 ) P = (1
✏r 1 ) .
P1 = /2, P2 = /3.
+σ
−σ/2
⃝
1
= +P1 at bottom of slab (1) = /2,
bottom of slab (2) = /3,
b = +P
2 at +σ/2
+σ
P1 at top of slab (1) =
/2;
top of slab (2) =
/3.
b =
b = ❖ P2 at −σ/3
⃝
2
⇢
+σ/3
total surface charge above:
( /2) = /2,
−σ
(f) In slab 1:
=) E1 =
.X
total surface charge below : ( /2) ( /3) + ( /3)
=
/2,
2✏0
⇢
2
total surface charge above:
( /2) + ( /2) ( /3) = 2 /3,
In slab 2:
=) E2 =
.X
total surface charge below : ( /3)
= 2 /3,
3✏0
(e) ⇢b = 0;
b
+σ
−σ/2
⃝
1
c
⃝2005
+σ/2
Problem
4.22
protected
under
all copyright laws as they currently exist. No portion of this material may
be
−σ/3
reproduced, in any form or by any means, without permission in writing from the publisher.
⃝
2
+σ/3
−σ
Problem 4.19
E0 ✻
✛
φ
x
✻
With no dielectric, C0 = A✏0 /d (Eq. 2.54).
y
In configuration (a), with + on upper plate,
on lower, D = between the plates.
✙
z
E = /✏0 (in air) and E = /✏ (in dielectric). So V = ✏0 d2 + ✏ d2 = 2✏Qd
1 + ✏✏0 .
0A
Problem 4.22
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
y
E0 ✻
✛
✙
z
φ
x
✻
98
Ca =
Q
V
=
✏0 A
d
⇣
2
1+1/✏r
⌘
=)
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Ca
2✏r
=
.
C0
1 + ✏r
In configuration (b), with potential di↵erence V : E = V /d, so = ✏0 E = ✏0 V /d (in air).
P = ✏0 e E = ✏0 e V /d (in dielectric), so b = ✏0 e V /d (at top surface of dielectric).
✏0 e V /d, so f = ✏0 V (1 + e )/d = ✏0 ✏r V /d (on top plate above dielectric).
tot = ✏0 V /d = f + b = f
✓
◆
✓
◆
✓
◆
Q
1
A
A
A
V
V
A✏0 1 + ✏r
Cb
1 + ✏r
=) Cb =
=
+ f
=
✏0 + ✏0 ✏r =
.
=
.
V
V
2
2
2V
d
d
d
2
C0
2
Cb
[Which is greater? C
0
If the x axis points down:
Ca
C0
=
1+✏r
2
2✏r
1+✏r
=
(1+✏r )2 4✏r
2(1+✏r )
E
(a) air
(a) dielectric
2
V
(✏r +1) d
V
d x̂
V
d x̂
(b) dielectric
b
(a)
(b)
V =
Z
0
1
E · dl =
x̂
2✏r ✏0 V
(✏r +1) d
x̂
x̂
2✏r ✏0 V
(✏r +1) d
✏0 V
d x̂
✏0 V
✏r d x̂
x̂
(top surface)
2(✏r 1) ✏0 V
(✏r +1) d
(✏r 1) ✏0dV
Problem 4.20
R
D·da = Qfenc ) D4⇡r2 = ⇢ 43 ⇡r3 ) D =
⇢R3 /3r2 ) E = (⇢R3 /3✏0 r2 ) r̂, for r > R.
⇢R3 1
3✏0 r
R
1
1+2✏r +4✏2r 4✏r
2(1+✏r )
D
2✏r V
(✏r +1) d
(b) air
=
f
=
(1 ✏r )2
2(1+✏r )
> 0. So Cb > Ca .]
P
0
2(✏r 1) ✏0 V
(✏r +1) d
x̂
0
(✏r
1) ✏0dV x̂
(top plate)
2✏r ✏0 V
(✏r +1) d
✏r ✏0dV (left); ✏0dV
(right)
1
3 ⇢r
) E = (⇢r/3✏) r̂, for r < R; D4⇡r2 = ⇢ 43 ⇡R3 ) D =
Z
rdr =
⇢
3✏
0
R
⇢R2
⇢ R2
⇢R2
+
=
3✏0
3✏0
3✏ 2
✓
1+
1
2✏r
◆
.
Problem 4.21
Let Q be the charge on a length ` of the inner conductor.
I
Q
Q
Q
D · da = D2⇡s` = Q ) D =
; E=
(a < s < b), E =
(b < r < c).
2⇡s`
2⇡✏0 s`
2⇡✏s`
◆
◆
 ✓ ◆
Z a
Z b✓
Z c✓
Q
ds
Q
ds
Q
b
✏0 ⇣ c ⌘
V =
E · dl =
+
=
ln
+ ln
.
2⇡✏0 ` s
2⇡✏` s
2⇡✏0 `
a
✏
b
c
a
b
C
Q
2⇡✏0
=
=
.
`
V`
ln(b/a) + (1/✏r ) ln(c/b)
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+σ/2
−σ/3
⃝
2
+σ/3
−σ
99
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Problem
4.22
Problem
4.22
Same method as Ex. 4.7: solve Laplace’s equation for Vin (s, ) (s < a) and Vout (s, ) (s > a), subject to
x
the boundary conditions
✻
8
φ
at s = a,
< (i) Vin = Vout
✻
@Vin
@Vout
E
0
(ii) ✏ @s = ✏0 @s
at s = a,
:
✛
(iii) Vout ! E0 s cos for s
a.
y
✙
z
From Prob. 3.24 (invoking boundary condition (iii)):
Vin (s, ) =
1
X
sk (ak cos k + bk sin k ),
Vout (s, ) =
1
X
E0 s cos +
k=1
s
k
(ck cos k + dk sin k ).
k=1
(I eliminated the constant terms by setting V = 0 on the y z plane.) Condition (i) says
X
X
ak (ak cos k + bk sin k ) = E0 a cos +
a k (ck cos k + dk sin k ),
while (ii) says
✏r
X
kak
1
X
1
(ak cos k⃝2005
sin k Education,
) = E0 cos
ka kRiver,
(ckNJ.
cosAll
k rights
+ dkreserved.
sin k ).This material is
c + bkPearson
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in any form or by any means, without permission in writing from the publisher.
Evidently bk = dk = 0 for all k, areproduced,
k = ck = 0 unless k = 1, whereas for k = 1,
aa1 =
E0 a + a
1
c1 ,
✏r a1 =
E0
a
2
c1 .
Solving for a1 ,
a1 =
and hence Ein (s, ) =
E0
,
(1 + e /2)
so Vin (s, ) =
E0
s cos
(1 + e /2)
=
E0
x,
(1 + e /2)
@Vin
E0
x̂ =
. As in the spherical case (Ex. 4.7), the field inside is uniform.
@x
(1 + e /2)
Problem 4.23
P0 = ✏0 e E0 ; E1 =
⇣
⌘n
e
En =
E0 , so
3
1
P0 =
3✏0
e
3
E0 ; P1 = ✏0
e E1
E = E0 + E1 + E2 + · · · =
The geometric series can be summed explicitly:
1
X
n=0
xn =
1
1
x
,
so
✏0 2e
E0 ; E2 =
3
=
"
1 ⇣
X
e
3
n=0
E=
⌘n
1
(1 +
e /3)
#
2
1
P1 = e E0 ; . . .. Evidently
3✏0
9
E0 .
E0 ,
which agrees with Eq. 4.49. [Curiously, this method formally requires that e < 3 (else the infinite series
diverges), yet the result is subject to no such restriction, since we can also get it by the method of Ex. 4.7.]
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100
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Problem 4.24
Potentials:
8
> Vout (r, ✓) =
<
P l
E0 r cos ✓ + ⌘rB
(r > b);
l+1 Pl (cos ✓),
P⇣
l
Vmed (r, ✓) =
Al rl + rB̄
P
(cos
✓),
(a
<
r < b);
l
l+1
>
:
Vin (r, ✓) = 0,
(r < a).
Boundary Conditions:
8
< (i) Vout = Vmed , (r = b);
med
(ii) ✏ @[email protected]
= ✏0 @[email protected] , (r = b);
:
(iii) Vmed = 0,
(r = a).
◆
X Bl
X✓
B̄l
l
(i) ) E0 b cos ✓ +
Pl (cos ✓) =
Al b + l+1 Pl (cos ✓);
bl+1
b
X
X
B̄
Bl
l
(ii) ) ✏r
lAl bl 1 (l + 1) l+2 Pl (cos ✓) = E0 cos ✓
(l + 1) l+2 Pl (cos ✓);
b
b
B̄
l
(iii) ) Al al + l+1 = 0 ) B̄l = a2l+1 Al .
a
For l 6= 1 :
✓
◆
Bl
a2l+1 Al
l
(i) l+1 = Al b
) Bl = Al b2l+1 a2l+1 ;
b
bl+1

✓
◆
a2l+1 Al
Bl
l
(ii) ✏r lAl bl 1 + (l + 1) l+2
= (l + 1) l+2 ) Bl = ✏r Al
b2l+1 + a2l+1
b
b
l+1
) Al = Bl = 0.
For l = 1 :
B1
a3 A1
E0 b + 2 = A1 b
) B1 E0 b3 = A1 b3 a3 ;
2
b
b
✓
◆
a3 A1
B1
(ii) ✏r A1 + 2 3
= E0 2 3 ) 2B1 E0 b3 = ✏r A1 b3 + 2a3 .
b
b
(i)
So
⇥
3E0 b3 = A1 2 b3
Vmed (r, ✓) =
E(r, ✓) =
2[1
a3 + ✏r b3 + 2a3
⇤
;
3E0
3
(a/b) ] + ✏r [1 + 2(a/b)3 ]
rVmed =
2[1
A1 =
✓
r
3E0
.
(a/b)3 ] + ✏r [1 + 2(a/b)3 ]
2[1
a3
r2
◆
cos ✓,
3E0
(a/b)3 ] + ✏r [1 + 2(a/b)3 ]
⇢✓
◆
2a3
1 + 3 cos ✓ r̂
r
✓
1
a3
r3
◆
sin ✓ ✓ˆ .
Problem 4.25
There are four charges involved: (i) q, (ii) polarization charge surrounding q, (iii) surface charge ( b ) on
the top surface of the lower dielectric, (iv) surface charge ( b0 ) on the lower surface of the upper dielectric.
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101
CHAPTER 4. ELECTRIC FIELDS IN MATTER
In view of Eq. 4.39, the bound charge (ii) is qp =
qt = q + qp = q/(1 + 0e ) = q/✏0r . As in Ex. 4.8,
(a)
b
= ✏0
(b)
0
b
=
Solve for
b,
e
✏0 0e
0
b:
"
"
1
qd/✏0r
4⇡✏0 (r2 + d2 ) 32
1
qd/✏0r
4⇡✏0 (r2 + d2 ) 32
first divide by
0
b
0
e
b
e
Plug this into (a) and solve for
b
0
b
b,
=
2✏0
b
0
b
2✏0
2✏0
#
+
0
e ),
so the total (point) charge at (0, 0, d) is
(here
b
= P·n̂ = +Pz = ✏0
(here
b
=
Pz =
✏0
e Ez );
0
e Ez ).
0
e
using ✏0r = 1 +
0
e:
1 qd/✏0r
1
qd
b
e
0
( e + 0e ), so b =
;
e (1 + e )
3
4⇡ (r2 + d2 ) 2
2
4⇡ (r2 + d2 ) 32 [1 + ( e + 0e )/2]
(
)
1
qd
1
1
qd/✏0r
1
qd
✏r 0e /✏0r
0
= e
+
, so b0 =
.
3
3
3
0
4⇡ (r2 + d2 ) 2 [1 + ( e + e )/2] 2⇡ (r2 + d2 ) 2
4⇡ (r2 + d2 ) 2 [1 + ( e + 0e )/2]
=
The total bound surface charge is
0
e
2✏0
#
0
e /(1
(respectively) and subtract:
"
#
1
qd/✏0r
1
qd/✏0r
b
0
0
=
) b= e
+
.
2⇡ (r2 + d2 ) 32
2⇡ (r2 + d2 ) 32
e
e
and
0
b
b
q(
e ).
t
=
b+
0
b
=
( 0e
qd
e)
1
4⇡ (r 2 +d2 ) 32 ✏0r [1+( e + 0e )/2]
(which vanishes, as it should, when
The total bound charge is (compare Eq. 4.51):
✓ 0
◆
( 0e
✏r ✏r q
e )q
qt = 0
=
, and hence
2✏r [1 + ( e + 0e )/2]
✏0r + ✏r ✏0r
1
V (r) =
4⇡✏0
Meanwhile, since
(
q/✏0r
p
x2 + y 2 + (z
qt
d)2
+p
x2 + y 2 + (z + d)2

q
q
✏0 ✏r
2q
+
q
=
1 + r0
= 0
,
t
0
0
✏r
✏r
✏r + ✏r
✏r + ✏r
V (r) =
Problem 4.26
From Ex. 4.5:
)
(for z > 0).
1
[2q/(✏0r + ✏r )]
p
(for z < 0).
4⇡✏0 x2 + y 2 + (z d)2
9
(r < a) >
>
>
=
0,
(r < a)
r̂,
(a
<
r
<
b)
D=
, E = 4⇡✏r2
.
Q
>
>
r̂, (r > a)
>
>
Q
2
>
>
4⇡r
:
r̂,
(r > b) ;
4⇡✏0 r2
( Z
)
( ✓
◆
Z
Z
1
1 Q2
1 b 1 1 2
1 1 1
Q2 1
1
W =
D·E d⌧ =
4⇡
r dr +
dr =
2
2 (4⇡)2
✏ a r2 r2
✏0 b r2
8⇡ ✏
r
⇢
✓
◆
✓
◆
Q2
1
1 1
1
Q2
1
e
=
+
=
+
.
8⇡✏0 (1 + e ) a b
b
8⇡✏0 (1 + e ) a
b
(
)
8
0,
>
>
> Q
<
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b
1
+
✏
0
a
✓
1
r
◆
1
b
)
102
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Problem 4.27
R
Using Eq. 4.55: W = ✏20 E 2 d⌧ . From Ex. 4.2 and Eq. 3.103,
8
9
1
>
P ẑ,
(r < R) >
<
=
3✏0
E=
, so
3
>
ˆ (r > R) >
: R P (2 cos ✓ r̂ + sin ✓ ✓),
;
3
3✏0 r
✓
◆2
✏0 P
4 3
2⇡ P 2 R3
Wr<R =
⇡R =
.
2 3✏0
3
27 ✏0
✓
◆2 Z
✏0 R3 P
1
Wr>R =
4 cos2 ✓ + sin2 ✓ r2 sin ✓ dr d✓ d
2
3✏0
r6
Z ⇡
Z 1
(R3 P )2
1
⇡(R3 P )2
=
2⇡
(1 + 3 cos2 ✓) sin ✓ d✓
dr =
( cos ✓
4
18✏0
9✏0
0
R r
✓
◆
⇡(R3 P )2
4
4⇡R3 P 2
=
=
.
9✏0
3R3
27✏0
cos3 ✓)
⇡
0
✓
1
3r3
◆
1
R
2⇡R3 P 2
.
9✏0
Wtot =
This is the correct electrostatic energy of the configuration, but it is not the “total work necessary to assemble
the system,” because it leaves out the mechanical energy involved in polarizing the molecules.
R
Using Eq. 4.58: W = 12 D·E d⌧ . For r > R, D = ✏0 E, so this contribution is the same as before.
✏0 2
1
2
1
For r < R,
⇣ D = ⌘✏0 E + P = 3 P + P = 3 P = 2✏0 E, so 2 D·E = 2 2 E , and this contribution is
2
3
3
2
P R
R P
now ( 2) 2⇡
= 4⇡
27 ✏0
27 ✏0 , exactly cancelling the exterior term. Conclusion: Wtot = 0. This is not
surprising, since the derivation in Sect. 4.4.3 calculates the work done on the free charge, and in this problem
there is no free charge in sight. Since this is a nonlinear dielectric, however, the result cannot be interpreted as
the “work necessary to assemble the configuration”—the latter would depend entirely on how you assemble it.
Problem 4.28
First find the capacitance, as a function of h:
Air part: E =
Oil part: D =
Q=
0
h + (`
2
4⇡✏0 s
2 0
4⇡s
=) V =
=) E =
h) = ✏r h
2
4⇡✏0
2 0
4⇡✏s
ln(b/a),
=) V =
2 0
4⇡✏
h + ` = [(✏r
C=
ln(b/a),
9
=
;
=)
1)h + `] = (
✏0
=
eh
0
✏
;
0
=
✏
✏0
= ✏r .
+ `), where ` is the total height.
Q
( e h + `)
( e h + `)
=
4⇡✏0 = 2⇡✏0
.
V
2 ln(b/a)
ln(b/a)
1 2 2⇡✏0 e
The net upward force is given by Eq. 4.64: F = 12 V 2 dC
dh = 2 V ln(b/a) .
2
2
The gravitational force down is F = mg = ⇢⇡(b
a )gh.
h=
⇢(b2
✏0 e V 2
.
a2 )g ln(b/a)
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7
Problem 4.29
z
✻
p1
✻
✰
x
CHAPTER 4. ELECTRIC FIELDS IN MATTER
p
✲ 2✲
❄ y
103 E1
7
Problem 4.29
z
@
✻
(a) Eq. 4.5 )Problem
F2 = (p2 4.29
· r) E1 = p2
(E1 );
@y
p1
p
p1 ˆ
p1
✲ 2✲
✻
Eq. 3.103 ) E1 =
✓
=
ẑ.
Therefore
❄ y
4⇡✏0 r3
4⇡✏0 y 3
✰
E
1
x
 ✓ ◆
z✻
p1 p2 d
1
3p1 p2
3p1 p2
F2 =
ẑ =
ẑ, or F2 =
ẑ (upward).
4⇡✏0 dy y 3
4⇡✏0 y 4
4⇡✏0 r4
✲
✻
7
p2
y
To calculate F1 , put p2 at thez origin, pointing in the z direction; then p1
✰
✛
✻
x
is at r ẑ, and it points in the ŷ direction. So F1 = (p1 · r) E2 =
p1
Problem 4.29
p
p
1
@E2
✲ 2✲of x, y, and z.
✻ a function
z✻
p1
; we need E2 as
❄ y
@y x=y=0, z= r
✰
E1
x

1 ✻1 3(p
✲2 · r)r
From
Eq.
3.104:
E
=
p2 , where r = x x̂+y ŷ+z ẑ, p2 = p2 ẑ, and hence p2 ·r = p2 z.
2
p
y r2
Problem 4.30
2
3
4⇡✏0 r
✰ ✛

2
2
2
2
2
p2x 3z(x
x̂
2z 2 ) ẑ
p1+ y ŷ + z ẑ) (x + y + z ) ẑ = p2 3xz x̂ + 3yz ŷ (x + y
E2 =
2
2
2
5/2
2
2
2
5/2
4⇡✏0
4⇡✏0
(x + y + z )
(x + y + z )
⇢
@E2
p2
5 2y
1
=
[3xz x̂ + 3yz ŷ (x2 + y 2 2z 2 ) ẑ] + 5 (3z ŷ 2y ẑ) ;
@y
4⇡✏0
2 r7
r
z✻
✓
◆
@E2
p2 3z 4.30
p2
3r
3p1 p2
Problem
=
ŷ; F1 = p1
ŷ =
ŷ.
@y (0,0)
4⇡✏0 r5
4⇡✏0 r5
4⇡✏0 r4
✲
✻
p
y
+
But 2ŷ in these coordinates corresponds to ẑ in the original
system, so these results are consistent with
F
◆
✰
E
Newton’s
third
law:
F
=
F
.
1
2
✛
x
p(b)
1 From the remark following Eq. 4.5,
✌
N2 = (p2−⇥✍FE1 ) + (r ⇥ F2 ). The first term was calculated in
✌ r ŷ:
Prob. 4.5; the second we get from (a), using r =
✌
✌
✓
◆
✌
p1 p2
3p1 p2
3p1 p2
2p1 p2
( x̂); r ⇥ F2 = (r ŷ) ⇥
ẑ =
x̂; so N2 =
x̂.
p2 ⇥ E1 =
4⇡✏0 r3
4⇡✏0 r4
4⇡✏0 r3
4⇡✏0 r3
Problem 4.30
+
F
◆ center of E
This is equal and opposite to the torque on p1 due to p2 , with respect to the
p1 (see Prob. 4.5).
✍F
✌
−
Problem 4.30
✌
Net force is to the right (see diagram). Note that the field lines must
bulge
to the right, as shown, because
✌
✌
E is perpendicular to the surface of each conductor.
✌
+
✌
F
◆
✍F
E
−
c
⃝2005
Pearson Education, Inc., Upper Saddle River,
protected under all copyright laws as they currently exist. No portion of this material may be
✌
reproduced, in any form or by any means, without permission in writing from the publisher.
✌
✌
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
104
CHAPTER 4. ELECTRIC FIELDS IN MATTER
1 @
1 Q
Problem 4.31 In cylindrical coordinates (in the z = 0 plane), p = p ˆ, p · r = p
, and E =
ŝ, so
s@
4⇡✏0 s2
✓
◆
p @
1 Q
pQ @ŝ
pQ ˆ
Q
F = (p · r)E =
ŝ =
=
=
p. X
s@
4⇡✏0 s2
4⇡✏0 s3 @
4⇡✏0 s3
4⇡✏0 R3
Qualitatively, the forces on the negative and positive ends, though equal in magnitude, point in slightly di↵erent
directions, and they combine to make a net force in the “forward” direction:
y
F+
p
F_
f
R
x
To keep the dipole going in a circle, there must be a centripetal force exerted by the track (we may as well
take it to act at the center of the dipole, and it is irrelevant to the problem), and to keep it aiming in the
tangential direction there must be a torque (which we could model by radial forces of equal magnitude acting at
the two ends). Indeed, if the dipole has the orientation indicated in the figure, and is moving in the ˆ direction,
the torque exerted by Q is clockwise, whereas the rotation is counterclockwise, so these constraint forces must
actually be larger than the forces exerted by Q, and the net force will be in the “backward” direction—tending
to slow the dipole down. [If the motion is in the ˆ direction, then the electrical forces will dominate, and the
net force will be in the direction of p, but this again will tend to slow it down.]
Problem 4.32
(a) According to Eqs. 4.1 and 4.5, F = ↵(E · r)E. From the product rule,
rE 2 = r(E · E) = 2E ⇥ (r ⇥ E) + 2(E · r)E.
But in electrostatics r ⇥ E = 0, so (E · r)E = 12 r(E 2 ), and hence
F=
1
↵r(E 2 ). X
2
[It is tempting to start with Eq. 4.6, and write F = rU = r(p · E) = ↵r(E · E) = ↵r(E 2 ). The
error occurs in the third step: p should not have been di↵erentiated, but after it is replaced by ↵E we are
di↵erentiating both E’s.]
(b) Suppose E 2 has a local maximum at point P . Then there is a sphere (of radius R) about P such that
E (P 0 ) < E 2 (P ), and hence |E(P 0 )| < |E(P )|, for all points on the surface. But if there is no charge inside the
2
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105
CHAPTER 4. ELECTRIC FIELDS IN MATTER
sphere, then Problem 3.4a says the average field over the spherical surface is equal to the value at the center:
Z
1
E da = E(P ),
4⇡R2
or, choosing the z axis to lie along E(P ),
1
4⇡R2
Z
Ez da = E(P ).
But if E 2 has a maximum at P , then
Z
Z
Z
Ez da  |E| da < |E(P )| da = 4⇡R2 E(P ),
and it follows that E(P ) < E(P ), a contradiction. Therefore, E 2 cannot have a maximum in a charge-free
region. [It can have a minimum, however; at the midpoint between two equal charges the field is zero, and this
is obviously a minimum.]
Problem 4.33
P = kr = k(x x̂ + y ŷ + z ẑ) =) ⇢b =
Total volume bound charge: Qvol =
b
r·P =
k(1 + 1 + 1) =
3k.
= ka/2. Clearly,
b
3ka3 .
= P·n̂. At top surface, n̂ = ẑ, z = a/2; so
b
= ka/2 on all six surfaces.
Total surface bound charge: Qsurf = 6(ka/2)a2 = 3ka3 . Total bound charge is zero. X
Problem 4.34
Say the high voltage is connected to the bottom plate, so the electric field points in the x direction, while
the free charge density ( f ) is positive on the lower plate and negative on the upper plate. (If you connect
x
the battery the other way, all the signs will switch.) The susceptibility is e = , and the permittivity is
d
⇣
x⌘
✏ = ✏0 1 +
. Between the plates
d
Z 0
Z d
⇣
1
1
x⌘ d
f
f
f
fd
D = f x̂, E = D =
x̂; V =
E · dl =
dx =
d ln 1 +
=
ln 2.
✏
✏0 (1 + x/d)
✏
(1
+
x/d)
✏
d
✏
0
0 0
0
0
d
So
f
=
✏0 V
,
d ln 2
E=
V
1
x̂,
d ln 2 (1 + x/d)
P = ✏0
eE
=
✏0 V
x
x̂.
d2 ln 2 (1 + x/d)
The bound charges are therefore
⇢b =
r·P=

✏0 V
1
2
d ln 2 (1 + x/d)
The total bound charge is
Z
Z
Qb =
⇢b d⌧ +
b da =
✏0 V A
=
d ln 2
✓
1
2
1
1+
2
◆
x/d
=
(1 + x/d)2
✏0 V
d2 ln 2
= 0,
Z
0
d
✏0 V
1
;
2
d ln 2 (1 + x/d)2
b
= P · n̂ =
1
✏0 V
✏0 V A
A dx +
A=
(1 + x/d)2
2d ln 2
d ln 2
X
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
"
8
>
<0
(x = 0),
✏0 V
>
:
2d ln 2
1
d
d (1 + x/d)
d
(x = d).
1
+
2
0
#
106
CHAPTER 4. ELECTRIC FIELDS IN MATTER
where A is the area of the plates.
Problem 4.35
I
q
1
q
r̂
q e
r̂
D·da = Qfenc ) D =
r̂; E = D =
; P = ✏0 e E =
.
2
2
4⇡r
✏
4⇡✏0 (1 + e ) r
4⇡(1 + e ) r2
✓
◆
q e
r̂
q e
e
3
⇢b = r·P =
r· 2 = q
(r) (Eq. 1.99); b = P·r̂ =
;
4⇡(1 + e )
r
1+ e
4⇡(1 + e )R2
Qsurf =
b (4⇡R
2
)= q
e
1+
The compensating negative charge is at the center:
.
e
Z
⇢b d⌧ =
q e
1+ e
Z
3
(r) d⌧ =
q
e
1+
.
e
Problem 4.36
Ek is continuous (Eq. 4.29); D? is continuous (Eq. 4.26, with
✏1 Ey1 = ✏2 Ey2 , and hence
f
= 0). So Ex1 = Ex2 , Dy1 = Dy2 )
tan ✓2
Ex2 /Ey2
Ey 1
✏2
=
=
= . qed
tan ✓1
Ex1 /Ey1
Ey 2
✏1
If 1 is air and 2 is dielectric, tan ✓2 / tan ✓1 = ✏2 /✏0 > 1, and the field lines bend away from the normal. This is
the opposite of light rays, so a convex “lens” would defocus the field lines.
Problem 4.37
In view of Eq. 4.39, the net dipole moment at the center is p0 = p 1+e e p = 1+1 e p =
potential produced by p0 (at the center) and b (at R). Use separation of variables:
8
9
1
X
>
>
B
l
>
>
>
Pl (cos ✓)
(Eq. 3.72) >
>
>
< Outside: V (r, ✓) =
=
rl+1
l=0
1
>
1 p cos ✓ X
>
+
Al rl Pl (cos ✓) (Eqs. 3.66, 3.102) >
>
2
;
4⇡✏0 ✏r r
l=0
9
Bl
l
2l+1
>
= Al R ,
or Bl = R
Al (l 6= 1) >
>
=
Rl+1
.
>
B1
1
p
p
>
3>
=
+ A1 R, or B1 = 4⇡✏0 ✏r + A1 R ;
R2
4⇡✏0 ✏r R2
>
>
>
>
: Inside:
V continuous at R )
@V
@r
R+
@V
@r
=
R
=
(l + 1)
For l = 1:
Bl
Rl+2
2
8
>
>
>
<
>
>
>
:
X
We want the
.
V (r, ✓) =
(l + 1)
1
P · r̂ =
✏0
lAl Rl
1
✏r p.
1
B1
1
2p
+
3
R
4⇡✏0 ✏r R3
=
Bl
1 2p cos ✓
Pl (cos ✓) +
l+2
R
4⇡✏0 ✏r R3
1
(✏0
✏0
e lAl R
A1 =
l 1
e
✓
e E·r̂)
=
e
(l 6= 1); or
@V
@r
=
R
X
e
⇢
(2l + 1)Al Rl
1
2p
+ A1
4⇡✏0 ✏r R3
◆
B1 +
lAl Rl
1
Pl (cos ✓) =
1
✏0
b
1 2p cos ✓ X
+
lAl Rl
4⇡✏0 ✏r R3
1
=
p
4⇡✏0 ✏r
e lAl R
l 1
A1 R3
=
2
1
Pl (cos ✓) .
) Al = 0 (` 6= 1).
1
ep
+
4⇡✏0 ✏r
e
A1 R3
;
2
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107
CHAPTER 4. ELECTRIC FIELDS IN MATTER
A1 R3
A1 R3
1
ep
)
(3 + e ) =
.
2
2
4⇡✏0 ✏r

1
2 ep
1
2(✏r 1)p
p
2(✏r 1)
p
3✏r
) A1 =
=
; B1 =
1+
=
.
3
3
4⇡✏0 R ✏r (3 + e )
4⇡✏0 R ✏r (✏r + 2)
4⇡✏0 ✏r
(✏r + 2)
4⇡✏0 ✏r ✏r + 2
✓
◆✓
◆
p cos ✓
3
V (r, ✓) =
(r R).
4⇡✏0 r2
✏r + 2
p
4⇡✏0 ✏r
A1 R3 +
A1 R3
=
2
p
4⇡✏0 ✏r
Meanwhile, for r  R, V (r, ✓) =
1
ep
+
4⇡✏0 ✏r
e
1 p cos ✓
1 pr cos ✓ 2(✏r 1)
+
2
4⇡✏0 ✏r r
4⇡✏0 R3 ✏r (✏r + 2)
=

✓
◆
p cos ✓
✏r 1 r3
1+2
4⇡✏0 r2 ✏r
✏r + 2 R3
(r  R).
Problem 4.38
Given two solutions, V1 (and E1 = rV1 , D1 = ✏E1 ) and V2 (E2 = rV2 , D2 = ✏E2 ), define V3 ⌘ V2 V1
(E3 = E2 E1 , D3 = D2 D1 ).
R
R
R
R
r·(V3 D3 ) d⌧ = S V3 D3 · da = 0, (V3 = 0 on S), so (rV3 ) · D3 d⌧ + V3 (r·D3 ) d⌧ = 0.
V
R
But r·D3 = r·D2 r·D1 = ⇢f ⇢f = 0,Rand rV3 = rV2 rV1 = E2 + E1 = E3 , so E3 · D3 d⌧ = 0.
But D3 = D2 D1 = ✏E2 ✏E1 = ✏E3 , so ✏(E3 )2 d⌧ = 0. But ✏ > 0, so E3 = 0, so V2 V1 = constant. But
at surface, V2 = V1 , so V2 = V1 everywhere. qed
Problem 4.39
R
R
r̂, in which case P = ✏0 e V0 2 r̂,
2
r
r
R
✏0 e V0
in the region z < 0. (P = 0 for z > 0, of course.) Then b = ✏0 e V0 2 (r̂·n̂) =
. (Note: n̂ points out
R
R
of dielectric ) n̂ = r̂.) This b is on the surface at r = R. The flat surface z = 0 carries no bound charge,
since n̂ = ẑ ? r̂. Nor is there any volume bound charge (Eq. 4.39). If V is to have the required spherical
symmetry, the net charge must be uniform:
2
tot 4⇡R = Qtot = 4⇡✏0 RV0 (since V0 = Qtot /4⇡✏0 R), so tot = ✏0 V0 /R. Therefore
(a) Proposed potential: V (r) = V0
f
=
⇢
R
. If so, then E =
r
rV = V0
(✏0 V0 /R), on northern hemisphere
(✏0 V0 /R)(1 + e ), on southern hemisphere
.
(b) By construction, tot = b + f = ✏0 V0 /R is uniform (on the northern hemisphere b = 0, f = ✏0 V0 /R;
on the southern hemisphere b = ✏0 e V0 /R, so f = ✏V0 /R). The potential of a uniformly charged sphere is
V0 =
Qtot
=
4⇡✏0 r
tot (4⇡R
4⇡✏0 r
2
)
=
✏0 V0 R2
R
= V0 . X
R ✏0 r
r
(c) Since everything is consistent, and the boundary conditions (V = V0 at r = R, V ! 0 at 1) are met,
Prob. 4.38 guarantees that this is the solution.
(d) Figure (b) works the same way, but Fig. (a) does not: on the flat surface, P is not perpendicular to n̂,
so we’d get bound charge on this surface, spoiling the symmetry.
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108
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Problem 4.40
✏0 e
Eext (Ex. 4.7).
1 + e /3
◆✓
◆
✓
◆
✓
◆✓
◆2 ✓ ◆ ✓
◆ Z
✏0 e
d
✏0 e
1
1
F=
ŝ d⌧ =
ŝ d⌧
1 + e /3
2⇡✏0 s ds 2⇡✏0 s
1 + e /3
2⇡✏0
s
s2
✓ 2 ◆
✓
◆ 2 3
1 4 3
R
e
e
=
⇡R
ŝ
=
ŝ.
2
3
1 + e /3 4⇡ ✏0 s 3
3 + e ⇡✏0 s3
Eext =
2⇡✏0 s
Z ✓
ŝ. Since the sphere is tiny, this is essentially constant, and hence P =
Problem 4.41
1
The density of atoms is N = (4/3)⇡R
3 . The macroscopic field E is Eself + Eelse , where Eself is the average
field over the sphere due to the atom itself.
p = ↵Eelse ) P = N ↵Eelse .
[Actually, it is the field at the center, not the average over the sphere, that belongs here, but the two are in
fact equal, as we found in Prob. 3.47d.] Now
Eself =
(Eq. 3.105), so
E=
✓
1 ↵
Eelse + Eelse =
4⇡✏0 R3
So
P=
e
e
or
✏0
↵=
N (1 +
N↵
3✏0
e
3✏0
e
=
/3)
N
(3
+
e
=
e
=
◆
Eelse =
✓
1
N↵
3✏0
◆
Eelse .
e E,
N ↵/✏0
.
(1 N ↵/3✏0 )
⌘
N↵
N↵ ⇣
e
)
1+
=
✏0
✏0
3
. But
e
↵
4⇡✏0 R3
N↵
E = ✏0
N ↵/3✏0 )
(1
and hence
Solving for ↵:
1
1 p
4⇡✏0 R3
e
= ✏r
e,
3✏0
1, so ↵ =
N
✓
✏r 1
✏r + 2
◆
. qed
Problem 4.42
For an ideal gas, N = Avagadro’s number/22.4 liters = (6.02 ⇥ 1023 )/(22.4 ⇥ 10 3 ) = 2.7 ⇥ 1025 . N ↵/✏0 =
(2.7 ⇥ 1025 )(4⇡✏0 ⇥ 10 30 ) /✏0 = 3.4 ⇥ 10 4 , where is the number listed in Table 4.1.
9
H:
= 0.667, N ↵/✏0 = (3.4 ⇥ 10 4 )(0.67) = 2.3 ⇥ 10 4 , e = 2.5 ⇥ 10 4 >
>
=
He: = 0.205, N ↵/✏0 = (3.4 ⇥ 10 4 )(0.21) = 7.1 ⇥ 10 5 , e = 6.5 ⇥ 10 5
agreement is quite good.
Ne: = 0.396, N ↵/✏0 = (3.4 ⇥ 10 4 )(0.40) = 1.4 ⇥ 10 4 , e = 1.3 ⇥ 10 4 >
>
;
Ar: = 1.64, N ↵/✏0 = (3.4 ⇥ 10 4 )(1.64) = 5.6 ⇥ 10 4 , e = 5.2 ⇥ 10 4
protected under all copyright laws as they currently exist. No portion of this material may be
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109
CHAPTER 4. ELECTRIC FIELDS IN MATTER
Problem 4.43
(a) Doing the (trivial)
hui =
R pE
ue
pE
R pE
pE
= kT
= kT
e
(⇥
e
integral, and changing the remaining integration variable from ✓ to u (du = pE sin ✓ d✓),
u/kT
u/kT
du
du
=
(kT )2 e
u/kT
[ (u/kT )
kT e
pE
pE
1]|
u/kT pE
pE
⇤ ⇥
⇤)
epE/kT + (pE/kT )e pE/kT + (pE/kT )epE/kT
e pE/kT epE/kT
 pE/kT
✓
◆
e
+ e pE/kT
pE
pE pE/kT
= kT pE coth
.
kT
e
e pE/kT
pE/kT
⇢
✓
◆
hui
pE
kT
= N p coth
.
pE
kT
pE
⇣
⌘
8
3
1
x
x3
Let y ⌘ P/N p, x ⌘ pE/kT . Then y = coth x 1/x. As x ! 0, y = x1 + x3 x45 + · · ·
x = 3
45 +· · · !
0, so the graph starts at the origin, with an initial slope of 1/3. As x ! 1, y ! coth(1) = 1, so the graph
Problem 4.40 goes asymptotically to y = 1 (see Figure).
P = N hpi; p = hp cos ✓iÊ = hp · Ei(Ê/E) =
P
Np
1
hui(Ê/E); P = N p
✻
✲
pE/kT
(b) For small x, y ⇡ 13 x, so
P
Np
⇡
pE
3kT
, or P ⇡
For water at 20 = 293 K, p = 6.1 ⇥ 10
30
N p2
3kT E
= ✏0
6.0 ⇥ 1023
= 2.11 ⇥ 1025 ;
2.85 ⇥ 10 2
Table 4.2 gives 5.9 ⇥ 10
3
e
) P is proportional to E, and
C m; N =
molecules
molecules
⇥ moles
volume =
mole
gram
(0.33⇥1029 )(6.1⇥10 30 )2
(3)(8.85⇥10 12 )(1.38⇥10 23 )(293)
1
N = 6.0 ⇥ 1023 ⇥ 18
⇥ 106 = 0.33 ⇥ 1029 ; e =
experimental value of 79, so it’s pretty far o↵.
For water vapor at 100 = 373 K, treated as an ideal gas,
N=
eE
=
volume
mole
= (22.4 ⇥ 10
(2.11 ⇥ 1025 )(6.1 ⇥ 10
(3)(8.85 ⇥ 10 12 )(1.38 ⇥ 10
, so this time the agreement is quite good.
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
3
⇥
e
=
N p2
.
3✏0 kT
grams
volume .
= 12. Table 4.2 gives an
)⇥
)
30 2
23 )(373)
373
293
= 2.85 ⇥ 10
= 5.7 ⇥ 10
3
.
2
m3 .
2
CHAPTER 5. MAGNETOSTATICS
110
CHAPTER 5. MAGNETOSTATICS
Chapter 1
Chapter 5
Magnetostatics
Magnetostatics
Problem 5.1
Problem 5.1
Since v ⇥ B points upward, and that is also the direction of the force, q must be positive. To find R, in
terms of a and d, use the pythagorean theorem:
(R
2Rd + d2 + a2 = R2 ) R =
a2 + d2
.
2d
!
d)2 + a2 = R2 ) R2
R
R
\$
(a2 + d2 )
p = qBR = qB
.
2d
"#
The cyclotron formula then gives
a
}d
Problem 5.2
The general solution is (Eq. 5.6):
Problem 5.2
E
y(t) = C1 cos(!t) + C2 sin(!t) + t + C3 ;
B
z(t) = C2 cos(!t)
C1 sin(!t) + C4 .
(a) y(0) = z(0) = 0; ẏ(0) = E/B; ż(0) = 0. Use these to determine C1 , C2 , C3 , and C4 .
y(0) = 0 ) C1 + C3 = 0; ẏ(0) = !C2 + E/B = E/B ) C2 = 0; z(0) = 0 ) C2 + C4 = 0 ) C4 = 0;
ż(0) = 0 ) C1 = 0, and hence also C3 = 0. So y(t) = Et/B; z(t) = 0. Does this make sense? The magnetic
force is q(v ⇥ B) = q(E/B)B ẑ = qE, which exactly cancels the
z electric force; since there is no net force,
✻
the particle moves
in
a
straight
line
at
constant
speed.
X
z
β
✻
(b) Assuming it starts from the origin, so C3 = C1 , C4 = C2 , we have ż(0) = 0 ) C1✸
= 0 ) C3 = 0;
β
✲y
E
E
E
E
E
E
ẏ(0) =
) C2 ! +
=
) C2 = ◆
= C4 ; y(t) =
sin(!t) + t;
2B
B
2B
2!B
B
✲ y 2!B
E
E
E
E
−β
z(t) =
cos(!t) + (b) , or y(t) =
[2!t sin(!t)] ; z(t) =
[1 (c)cos(!t)] . Let ⌘ E/2!B.
2!B
2!B
2!B
2!B
Then y(t) = [2!t sin(!t)] ; z(t) = [1 cos(!t)] ; (y 2 !t) =
sin(!t), (z
)=
cos(!t) )
(y 2 !t)2 + (z
)2 = 2 . This
is
a
circle
of
whose
center
moves
to
the
right
at
constant
speed:
c
⃝2005
y0 = 2 !t; z0 = .
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced,
in any form or by any
in writing from the publisher.
E
E
E means, without
E permission
E
(c) ż(0) = ẏ(0) =
) C1 ! =
) C1 = C3 =
; C2 ! +
=
) C2 = C4 = 0.
B
B
!B
B
B
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
R
"#
R
\$
Problem
5.2
CHAPTER
5. MAGNETOSTATICS
a
}d
111
E
E
E
E
E
E
cos(!t) + t +
; z(t) =
sin(!t). y(t) =
[1 + !t cos(!t)] ; z(t) =
sin(!t).
!B
B
!B
!B
!B
!B
Let ⌘ E/!B; then [y
(1 + !t)] =
cos(!t), z = sin(!t); [y
(1 + !t)]2 + z 2 = 2 . This is a circle
of radius whose center is at y0 = (1 + !t), z0 = 0.
z
✻
z
β
✻
✸
β
✲y
◆
✲y
y(t) =
(b)
−β
(c)
Problem 5.3
c
⃝2005
(a) From Eq. 5.2, F = q[E +
(v ⇥ B)]
= all
0)
E = vB
v = currently
.
protected
under
laws)
as they
exist. No portion of this material may be
B
reproduced, in any form or by any means, without permission in writing from the publisher.
q
v
E
(b) From Eq. 5.3, mv = qBR )
=
=
.
m
BR
B2R
Problem 5.4
Suppose I flows counterclockwise (if not, change the sign of the answer). The force on the left side (toward
the left) cancels the force on the right side (toward the right); the force on the top is IaB = Iak(a/2) =
Ika2 /2, (pointing upward), and the force on the bottom is IaB = Ika2 /2 (also upward). So the net force is
F = Ika2 ẑ.
Problem 5.5
I
, because the length-perpendicular-to-flow is the circumference.
2⇡a
Z
Z
Z
↵
1
I
I
(b) J = ) I = J da = ↵
s ds d = 2⇡↵ ds = 2⇡↵a ) ↵ =
;J =
.
s
s
2⇡a
2⇡as
(a) K =
Problem 5.6
(a) v = !r, so K = !r.
(b) v = !r sin ✓ ˆ ) J = ⇢!r sin ✓ ˆ, where ⇢ ⌘ Q/(4/3)⇡R3 .
Problem 5.7Z
Z ✓ ◆
Z
dp
d
@⇢
=
⇢r d⌧ =
r d⌧ =
(r · J)r d⌧ (by the continuity equation). Now product rule #5
dt
dt V
@t
R
says r · (xJ) = x(r · J) + J · (rx). But rx = x̂, so r · (xJ) = x(r · J) + Jx . Thus V (r · J)x d⌧ =
Z
Z
R
r · (xJ) d⌧
Jx d⌧ . The first term is S xJ · da (by the divergence theorem), and since J is entirely
V
V
R
R
inside V, it is zero on the surface S. Therefore V (r · J)x d⌧ =
J d⌧ , or, combining
this with the y and
V x
Z
R
R
dp
z components, V (r · J)r d⌧ =
J d⌧. Or, referring back to the first line,
= J d⌧ . qed
V
dt
Here’s a quicker method, if the distribution consists of a collection of point charges. Use Eqs. 5.30 and
3.100:
Z
X
d X
dp
J d⌧ =
qi vi =
qi ri =
.
dt
dt
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112
CHAPTER 5. MAGNETOSTATICS
Problem 5.8
(a) Use Eq. 5.37, with z = R, ✓2 =
p
2µ0 I
.
⇡R
✓1 = 45 , and four sides: B =
⇡
nµ0 I
, and n sides: B =
sin(⇡/n).
n
2⇡R
nµ0 I ⇣ ⇡ ⌘
µ0 I
(c) For small ✓, sin ✓ ⇡ ✓. So as n ! 1, B !
=
(same as Eq. 5.41, with z = 0).
2⇡R n
2R
(b) z = R, ✓2 =
✓1 =
Problem 5.9
✓
◆
µ0 I 1 1
(a) The straight segments produce no field at P . The two quarter-circles give B =
(out).
8
a b
µ0 I
µ0 I
(b) The two half-lines are the same as one infinite line:
; the half-circle contributes
.
2⇡R
4R
✓
◆
µ0 I
2
So B =
1+
(into the page).
4R
⇡
Problem 5.10
(a) The forces on the two sides cancel. At the bottom, B =
top, B =
µ0 I
)F =
2⇡s
✓
µ0 I
2⇡s
µ0 I
µ0 I 2 a
µ0 I 2 a2
)F =
(down). The net force is
(up).
2⇡(s + a)
2⇡(s + a)
2⇡s(s + a)
◆
Ia =
µ0 I 2 a
(up). At the
2⇡s
"#
\$
µ0 I
(b) The force on the bottom is the same as before, µ0 I 2 a/2⇡s (up). On the left side, B =
ẑ;
2⇡y
✓
◆
µ0 I
µ0 I 2
CHAPTER
3
dF = I(dl ⇥ B)
= I(dx x̂ +5.dyMAGNETOSTATICS
ŷ + dz ẑ) ⇥
ẑ =
( dx ŷ + dy x̂). But the x component
cancels the
2⇡y
2⇡y p
Z
p
µ0 I 2 (s/ 3+a/2) 1
corresponding Problem
term from 5.10
the right side, and Fy =
dx. Here y = 3x, so
p
2⇡ s/ 3
y
!
p
p !
2
2
µ0 I
s/ 3 + a/2
µ0 I
3a
p ln
p
p ln 1 +
Fy =
=
. The force on the right side is the same, so the net
2s
2 3⇡
s/ 3
2 3⇡
y
"
p !#
✻
µ0 I 2 a
2
3a
p ln 1 +
force on the triangle is
.
a
2⇡
s
2s
3
a
s
◦
✲x
!
60
√s
3
✰
z
Problem 5.11
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
η
✲z
θ
a
}
❄
Problem 5.11
113
CHAPTER 5. MAGNETOSTATICS
}
Problem 5.11
Use Eq. 5.41
of width dz, with I ! nI dz:
r
Z for a ring
µ0 nI
a2
θ
B=
dz. But z = a cot ✓,
3/2
2
a
(a2 + z 2 )
❄
a
1
sin3 ✓
!
"#
\$
so dz =
d✓, and
=
.
dz
3/2
z
a3
sin2 ✓
(a2 + z 2 )
So
Z 2 3
Z
µ0 nI
a sin ✓
µ0 nI
µ0 nI
µ0 nI
✓2
B=
sin ✓ d✓ =
cos ✓ ✓1 =
(cos ✓2 cos ✓1 ).
2 ( a d✓) =
3
2
2
2
2
a sin ✓
For an infinite solenoid, ✓2 = 0, ✓1 = ⇡, so (cos ✓2
Problem 5.12
cos ✓1 ) = 1
( 1) = 2, and B = µ0 nI. X
z
Rsinq
Rcosq
q
R
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
Field (at center of sphere) due to the ring at ✓ (see figure) is (Eq. 5.41):
dB =
µ0 dI
(R sin ✓)2
µ0
=
sin2 ✓ dI.
2 [(R sin ✓)2 + (R cos ✓)2 ]3/2
2R
dI = K R d✓,
so
K = v,
=
Q
,
4⇡R2
v = !R sin ✓,
Q
Q!
!R sin ✓R d✓ =
sin ✓ d✓.
4⇡R2
4⇡
✓ ◆
Z
µ0 Q! ⇡ 3
µ0 Q! 4
µ0 Q!
B=
sin ✓ d✓ =
. B=
ẑ.
2R 4⇡ 0
8⇡R 3
6⇡R
dI =
Problem 5.13
µ0 2 v 2
Magnetic attraction per unit length (Eqs. 5.40 and 5.13): fm =
.
2⇡ d
1
Electric field of one wire (Eq. 2.9): E =
. Electric repulsion per unit length on the other wire:
2⇡✏0 s
2
1
1
1
fe =
. They balance when µ0 v 2 = , or v = p
. Putting in the numbers,
2⇡✏0 d
✏0
✏0 µ0
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
✲z
114
CHAPTER 5. MAGNETOSTATICS
1
v=p
= 3.00 ⇥ 108 m/s. This is precisely the speed of light(!), so in fact you could
(8.85 ⇥ 10 12 )(4⇡ ⇥ 10 7 )
never get the wires going fast enough; the electric force always dominates.
Problem 5.14
(
)
I
0,
for s < a;
(a) B · dl = B 2⇡s = µ0 Ienc ) B = µ0 I ˆ
, for s > a.
2⇡s
Z a
Z a
2⇡ka3
(b) J = ks; I =
J da =
ks(2⇡s) ds =
)k =
3
0
0
8
µ0 Is2 ˆ
>
>
>
< 2⇡a3 ,
2⇡ks3
s3
= I 3 , for s < a; Ienc = I, for s > a. So B =
>
3
a
>
4
>
: µ0 I ˆ,
2⇡s
Problem 5.15Problem 5.14
By the right-hand-rule, the field points in the
At
z
= 0, B = 0. Use the amperian loop shown:
I
B · dl = Bl = µ0 Ienc = µ0 lzJ ) B =
so B =
⇢
3I
. Ienc =
2⇡a3
9
>
for s < a; >
>
=
Z
s
J da =
0
Z
s
ks̄(2⇡s̄) ds̄ =
0
>
>
>
for s > a. ;
CHAPTER 5. MAGNETOSTAT
ŷ direction for z > 0, and in the +ŷ direction for z < 0.
µ0 Jz ŷ ( a < z < a). If z > a, Ienc = µ0 laJ,
µ0 Ja ŷ, for z > +a;
+µ0 Ja ŷ, for z > a.
z
✛
z{
!
✻
✲
"#
l
amperian loop
✠
✲y
\$
Problem 5.16
z to the right (ẑ),
The field inside a solenoid is µ0 nI, and outside it is zero. The outer solenoid’s field points
✻
whereas the inner one points to the left ( ẑ). So: (i) B = µ0 I(n2 n1 ) ẑ, (ii) B = µ0 In2 ẑ, (iii) B = 0.
loop 1
Problem 5.17
I
✿ it, and into
From Ex. 5.8, the top plate produces a field µ0 K/2 (aiming out of the page, for points above
the page, for points below). The bottom plate produces a field µ0 K/2 (aiming into the page, for points ηabove
1
it, and out of the page, for points below). Above and below both plates the two fields cancel; between the plates
❥r
✣
µ0 K, pointing
5.17in.
(a) B = µ0 v (in) betweem the plates, B = 0 elsewhere.
R
✰
η
(b) The Lorentz force law says F = (K ⇥ B) da, so the force per unit area xis f = K ⇥ B. Here K =2 v,
to the right, and B (the field of the lower plate) is µ0 v/2, into the page. So fm = µ0
v /2 (up).
2 2
(c) The electric field of the lower plate is
fe =
2
/2✏0 ; the electric force per unit area on the upper plate
loop is
2
I
p
✿in Prob. 5.13.
/2✏0 (down). They balance if µ0 v = 1/✏0 , or v = 1/ ✏0 µ0 = c (the speed of light), as
2
Problem 5.18
We might as well orient the axes so the field point r lies on the y axis: r = (0, y, 0). Consider a source point
at (x0 , y 0 , z 0 ) on loop #1:
r
=
x0 x̂ + (y
y 0 ) ŷ
z 0 ẑ; dl0 = dx0 x̂ + dy 0 ŷ;
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
✲y
4
CHAPTER 5. MAGNETOSTATICS
Problem 5.14
115
CHAPTER 5. MAGNETOSTATICS
dl0 ⇥
r
z
✻
amperian loop
x̂
ŷ
ẑ
✠ 0
0
0
0
0 ✛
0
0
dy
0 = ( z dy ) x̂ + (z zdx
= dx
[(y y ) dx0 + x0 dy 0 ] ẑ.
{ ) ŷ +
✲
✲y
0
0
0
x (y y ) z
dB1 =
"# \$y 0 ) dx0 + x0 dy 0 ] ẑ
µ0 I dl0 ⇥ r
µ0 I ( z 0 dy 0 ) x̂ + (z 0 dx0 ) ŷ! + [(y
=
.
l
3
3/2
r
4⇡
4⇡
[(x0 )2 + (y y 0 )2 + (z 0 )2 ]
Now consider the symmetrically placed source element on
loop #2, at (x0 , y 0 , z 0 ). Since z 0 changes sign, while everything else is the same, the x̂ and ŷ components from dB1 and
dB2 cancel, leaving only a ẑ component. qed
With this, Ampére’s law yields immediately:
z
✻
loop 1
I
✿
(
µ0 nI ẑ, inside the solenoid;
B=
0,
outside
Problem 5.17
r
1
❥r
✣
(the same as for a circular solenoid—Ex. 5.9).
For the toroid, N/2⇡s = n (the number of turns per unit
length), so Eq. 5.60 yields B = µ0 nI inside, and zero outside,
consistent with the solenoid. [Note: N/2⇡s = n applies only
if the toroid is large in circumference, so that s is essentially
constant over the cross-section.]
✰
x
r
I
✿
✲y
2
loop 2
Problem 5.19
R
It doesn’t matter. According to Theorem 2, in Sect. 1.6.2, J · da is independent of surface, for any given
boundary line, provided that J is divergenceless, which it is, for steady currents (Eq. 5.33).
Problem 5.20
✓ ◆
charge
charge atoms moles grams
1
(a) ⇢ =
=
·
·
·
= (e)(N )
(d), where
volume
atom
mole gram volume
M
e
N
M
d
=
=
=
=
charge of electron
atomic mass of copper
density of copper
=
=
=
=
1.6 ⇥ 10 19 C,
6.0 ⇥ 1023 mole,
64 gm/mole,
3
9.0 gm/cm .
◆
9.0
= 1.4 ⇥ 104 C/cm3 .
64
I
I
1
(b) J = 2 = ⇢v ) v = 2 =
= 9.1 ⇥ 10 3 cm/s, or about 33 cm/hr. This
⇡s
⇡s ⇢
⇡(2.5 ⇥ 10 3 )(1.4 ⇥ 104 )
is astonishingly small—literally ✓
slower ◆than a snail’s pace.
µ0 I1 I2
(4⇡ ⇥ 10 7 )
(c) From Eq. 5.40, fm =
=
= 2 ⇥ 10 7 N/cm.
2⇡ ✓ d ◆
2⇡ ✓
◆ ✓ 2◆
✓
◆
1
1
1 1
I1 I2
c
µ0 I1 I2
c2
1 2
(d) E =
; fe =
= 2
=
= 2 fm , where
2
2⇡✏0 d
2⇡✏0
d
v 2⇡✏0
d
v
2⇡
d
v
✓
◆
c
⃝2005
Pearson2 Education,
Inc., Upper
fe under
3.0 ⇥ laws
10 as they currently exist.
p
protected
No portion of this material may be
c ⌘ 1/ ✏0 µ0 = 3.00 ⇥ 108 m/s. Here
=
=
=without
1.1 ⇥ permission
1025 .
reproduced,
or by any 3means,
in writing from the publisher.
fm inv 2any form9.1
⇥ 10
⇢ = (1.6 ⇥ 10
19
)(6.0 ⇥ 1023 )
fe = (1.1 ⇥ 1025 )(2 ⇥ 10
7
✓
) = 2 ⇥ 1018 N/cm.
protected under all copyright laws as they currently exist. No portion of this material may be
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116
CHAPTER 5. MAGNETOSTATICS
Problem 5.21
Ampére’s law says r ⇥ B = µ0 J. Together with the continuity equation (5.29) this gives r · (r ⇥ B) =
µ0 r · J = µ0 @⇢/@t, which is inconsistent with div(curl)=0 unless ⇢ is constant (magnetostatics). The other
Maxwell equations are OK: r ⇥ E = 0 ) r · (r ⇥ E) = 0 (X), and as for the two divergence equations, there
is no relevant vanishing second derivative (the other one is curl(grad), which doesn’t involve the divergence).
Problem 5.22
At this stage I’d expect no changes in Gauss’s law or Ampére’s law. The divergence of B would take the
form r · B = ↵0 ⇢m , where ⇢m is the density of magnetic charge, and ↵0 is some constant (analogous to ✏0
and µ0 ). The curl of E becomes r ⇥ E = 0 Jm , where Jm is the magnetic current density (representing the
flow of magnetic charge), and 0 is another constant. Presumably magnetic charge is conserved, so ⇢m and Jm
satisfy a continuity equation: r · Jm = @⇢m /@t.
As for the Lorentz force law, one might guess something of the form qm [B + (v ⇥ E)] (where qm is the
magnetic charge). But this is dimensionally impossible, since E has the sameCHAPTER
units as vB.
Evidently we
5. MAGNETOSTATICS
need to divide (v ⇥ E) by something with the dimensions of velocity-squared. The natural candidate is

1
c2 = 1/✏0 µ0 : F = qe [E + (v ⇥ B)] + qm B
(v ⇥ E) . In this form the magnetic
analog
Problem
5.22 to Coulomb’s
c2
↵0 qm1 qm2
r̂, so to determine ↵0 we would first introduce (arbitrarily) a unit of magnetic charge,
4⇡ r2
then measure the force between unit charges at a given separation. [For further details, and an explanation of
Problem 5.23
r
µ0 I
dz =
ẑ
4⇡
z1
z2
z1
dz
p
z 2 + s2
dl=dz
s
"
#
p
z2 + (z2 )2 + s2
µ0 I
p
=
ln
ẑ
4⇡
z1 + (z1 )2 + s2
}
I ẑ
Z
}
µ0
A=
4⇡
Z
✲z
r
!
"#
\$
⌘i z2
p
µ0 I h ⇣
z2
ẑ ln z + z 2 + s2
4⇡
z1
"
#
@A ˆ
µ0 I
1
s
1
s
ˆ
p
p
p
p
B = r⇥A=
=
@s
4⇡ z2 + (z2 )2 + s2 (z2 )2 + s2
z1 + (z1 )2 + s2 (z1 )2 + s2
"
#
p
p
(z2 )2 + s2
z1
(z1 )2 + s2
µ0 Is z2
1
1 Problem
ˆ 5.26
p
p
=
4⇡ (z2 )2 [(z2 )2 + s2 ] (z2 )2 + s2
z12 [(z1 )2 + s2 ] (z1 )2 + s2
#
"
#
✓
◆"
µ0 Is
1
z2
z1
µ0 I
z2
z1
ˆ
ˆ,
p
p
p
=
1 p
+1
=
4⇡
s2
4⇡s
(z2 )2 + s2
(z1 )2 + s2
(z2 )2 + s2
(z1 )2 + s2
z1
z2
or, since sin ✓1 = p
and sin ✓2 = p
,
2
2
(z1 ) + s
(z2 )2 + s2
=
=
µ0 I
(sin ✓2
4⇡s
Problem 5.24
sin ✓1 ) ˆ (as in Eq. 5.37).
1 @
k
1
1
A =k )B=r⇥A=
(sk) ẑ = ẑ; J =
(r ⇥ B) =
s @s
s
µ0
µ0

@
@s
✓ ◆
k
s
ˆ=
k ˆ
.
µ0 s2
Problem 5.36
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reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 5.37
117
CHAPTER 5. MAGNETOSTATICS
Problem 5.25
1
1
r·A =
r · (r ⇥ B) =
[B · (r ⇥ r) r · (r ⇥ B)] = 0, since r ⇥ B = 0 (B is uniform) and
2
2
1
1
r ⇥ r = 0 (Prob. 1.63). r ⇥ A =
r ⇥ (r ⇥ B) =
[(B · r)r (r · r)B + r(r · B) B(r · r)]. But
2
2
@x
@y
@z
(r · r)B = 0 and r · B = 0 (since B is uniform), and r · r =
+
+
= 1 + 1 + 1 = 3. Finally,
@x
@y
@z
✓
◆
@
@
@
1
(B·r)r = Bx
+ By
+ Bz
(x x̂+y ŷ+z ẑ) = Bx x̂+By ŷ+Bz ẑ = B. So r⇥A =
(B 3B) = B.
@x
@y
@z
2
qed
}
}
Problem 5.26
MAGNETOSTATICS
5 wire). In cylindrical
(a) A isCHAPTER
parallel (or5.antiparallel)
to I, and is a function only of s (the distance from the
@A ˆ
µ0 I ˆ
coordinates, then, A = A(s) ẑ, so B = r ⇥ A =
=
(the field of an infinite wire). Therefore
@s
2⇡s
5.22
@A
µProblem
µ0 I
0I
=
, and A(r) =
ln(s/a) ẑ (the constant a is arbitrary; you could use 1, but then the units
@s
2⇡s
2⇡
@Az
@Az ˆ µ0 I ˆ
look fishy). r · A =
= 0. X r ⇥ A =
=
= B. X
@z
@s
2⇡s
I
I
µ0 Is2
(b) Here Ampére’s law gives
B · dl = B 2⇡s = µ0 Ienc = µ0 J ⇡s2 = µ0
⇡s2 =
.
2
⇡R
R2
z1
dl=dz
µ0 Is ˆ @A
µ0 I s ✲
µ0 I 2
B=
.
=
) zA =
(s
b2 ) ẑ. Here b is again arbitrary, except that since A
2
2⇡ R
@s
2⇡ R2
4⇡R2
s η µ0 I
µ0 I
must be continuous at R,
ln(R/a) =
(R2 b2 ), which means that we must pick a and b such that
2
2⇡
4⇡R
!
"#
\$
8 µ I
9
z
0
2
2
>
>
(s
R
)
ẑ,
for
s

R;
>
>
< 4⇡R2
=
2
2 ln(R/b) = 1 (b/R) . I’ll use a = b = R. Then A =
>
>
>
>
: µ0 I ln(s/R) ẑ,
for s R. ;
2⇡
Problem Problem
5.27
5.26
µ0 K
K = K x̂ ) B = ±
ŷ (plus for z < 0, minus for z > 0).
2
A is parallel (or antiparallel) to K, and depends only on z, so A = A(z) x̂.
x̂
ŷ
ẑ
@A
µ0 K
B = r ⇥ A = @/@x @/@y @/@z =
ŷ = ±
ŷ.
@z
2
A(z) 0
0
A=
µ0 K
|z| x̂ will do the job—or this plus any constant.
2
Problem 5.28Z
✓ ◆
µ0
J
0
(a) r·A =
r·
r d⌧ ;
4⇡
✓
J
◆
1
r
r
(r·J)+J·r
✲y
✠ ✠ ✠ ✠ ✠ ✠K
✠
x
✓
1
◆
. But the first term is zero, because J(r0 )
✓ ◆
✓ ◆
1
1
0
0
is a function of the source coordinates, not the field coordinates. And since r = r r , r
= r
.
r
r
✓ ◆
✓ ◆
✓ ◆
✓ ◆
J
1
J
1
1
0
Problem
5.36
So r ·
= J·r
. But r0 ·
=
(r0 · J) + J · r0
, and r0 · J = 0 in magnetostatics
r
r
r
r
r
✓ ◆
✓ ◆
✓ ◆
Z
J
J
µ0
J
0
(Eq. 5.33). So r·
= r0 ·
, and hence, by the divergence theorem, r·A =
r0 ·
r
r
r d⌧ =
4⇡
r·
=
z
✻
r
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 5.37
118
CHAPTER 5. MAGNETOSTATICS
I
µ0
J
0
r · da , where the integral is now over the surface surrounding all the currents. But J = 0 on this
4⇡
surface, so r · A = 0. X
✓ ◆
✓ ◆
Z
Z 
µ0
J
µ0
1
1
0
(b) r ⇥ A =
r⇥
d⌧ =
(r ⇥ J) J ⇥ r
d⌧ 0 . But r ⇥ J = 0 (since J is
r
r
r
4⇡
4⇡
✓ ◆
Z
r̂
1
µ0
J ⇥ r̂
0
not a function of r), and r
=
(Eq.
1.101),
so
r
⇥
A
=
2
r
r
r 2 d⌧ = B. X
4⇡
✓ ◆
✓ ◆
✓ ◆
Z
µ0
J
J
1
(c) r2 A =
r2
d⌧ 0 . But r2
= Jr2
r
r ✓ ◆ r (once again, J is a constant, as far as
4⇡
1
di↵erentiation with respect to r is concerned), and r2
= 4⇡ 3 ( r ) (Eq. 1.102). So
r
Z
⇥
⇤
µ0
r2 A =
J(r0 ) 4⇡ 3 ( r ) d⌧ 0 = µ0 J(r). X
4⇡
Problem I5.29
µ0 I =
B · dl =
Z
a
b
rU · dl =
For an infinite straight wire, B =
[U (b)
µ0 I ˆ
.
2⇡s
U (a)] (by the gradient theorem), so U (b) 6= U (a). qed
U=
µ0 I
2⇡
would do the job, in the sense that
µ0 I
µ0 I 1 @ ˆ
r( ) =
2⇡
2⇡ s @
value; it works (say) for 0  < 2⇡, but at 2⇡ it “jumps” back to zero.
rU =
Problem 5.30
Use Eq. 5.69, with R ! r̄ and
! ⇢ dr̄:
Z
Z R
µ0 !⇢ sin ✓ ˆ r 4
µ0 !⇢
ˆ
r̄
dr̄
+
r
sin
✓
r̄ dr̄
3
r2
3
0
r
 ✓ 5◆
✓ 2
◆
⇣ µ !⇢ ⌘
1 r
r
r2 ˆ
0
ˆ = µ0 !⇢ r sin ✓ R
=
sin ✓ 2
+
R2 r 2
.
3
r
5
2
2
3
5
⇢

✓ 2
◆

✓ 2
µ0 !⇢
1
@
R
r2
1 @
R
2
B = r⇥A=
sin ✓ r sin ✓
r̂
r sin ✓
2
r sin ✓ @✓
3
5
r @r
3
✓ 2
◆
✓
◆
R
r2
R2
2r2
= µ0 !⇢
cos ✓ r̂
sin ✓ ✓ˆ .
3
5
3
5
A =
r2
5
◆
✓ˆ
Problem 5.31
8
9
Rx
@Wz
>
>
0
0
>
>
=
F
)
W
(x,
y,
z)
=
F
(x
,
y,
z)
dx
+
C
(y,
z).
>
>
y
z
y
1
0
<
=
@x
(a)
>
>
Rx
>
>
@Wy
>
:
;
= Fz ) Wy (x, y, z) = + 0 Fz (x0 , y, z) dx0 + C2 (y, z). >
@x
These satisfy (ii) and (iii), for any C1 and C2 ; it remains to choose these functions so as to satisfy (i):
Z x
Z x
@Fy (x0 , y, z) 0
@C1
@Fz (x0 , y, z) 0
@C2
@Fx
@Fy
@Fz
dx +
dx
= Fx (x, y, z). But
+
+
= 0, so
@y
@y
@z
@zZ
@x
@y
@z
0
Z x0
x
@Fx (x0 , y, z) 0
@C1
@C2
@Fx (x0 , y, z) 0
dx +
= Fx (x, y, z). Now
dx = Fx (x, y, z) Fx (0, y, z), so
0
@x
@y
@z
@x0
0
0
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119
CHAPTER 5. MAGNETOSTATICS
@C1
@y
Z y
@C2
= Fx (0, y, z). We may as well pick C2 = 0, C1 (y, z) =
Fx (0, y 0 , z) dy 0 , and we’re done, with
@z
0
Z x
Z y
Z x
Wx = 0; Wy =
Fz (x0 , y, z) dx0 ; Wz =
Fx (0, y 0 , z) dy 0
Fy (x0 , y, z) dx0 .
0
✓
◆
✓
◆
0
✓
◆
0
@Wz
@Wy
@Wx
@Wz
@Wy
@Wx
x̂ +
ŷ +
ẑ
@y
@z
@z
@x
@x
@y
Z
x
x
@Fy (x0 , y, z) 0
@Fz (x0 , y, z) 0
dx
dx x̂ + [0 + Fy (x, y, z)] ŷ + [Fz (x, y, z) 0] ẑ.
@y
0
0

Z [email protected]
0
@Fx (x , y, z) 0
But r · F = 0, so the x̂ term is Fx (0, y, z) +
dx = Fx (0, y, z) + Fx (x, y, z) Fx (0, y, z),
@x0
0
so r ⇥ W = F. X
Z x
Z y
Z x
@Wx @Wy @Wz
@Fz (x0 , y, z) 0
@Fx (0, y 0 , z) 0
@Fy (x0 , y, z) 0
r·W =
+
+
= 0+
dx +
dy
dx 6= 0,
@x
@y
@z
@y
@z
@z
0
0
0
in general. Z
Z y
Z x
x
x2
y2
(c) Wy =
x0 dx0 =
; Wz =
y 0 dy 0
z dx0 =
zx.
2
2
0
0
0
(b) r ⇥ W =

Z
= Fx (0, y, z)
W=
x2
ŷ +
2
✓
y2
2
◆
zx ẑ.
x̂
ŷ
ẑ
@/@z
r ⇥ W = @/@x @/@y
= y x̂ + z ŷ + x ẑ = F. X
0 x2 /2 (y 2 /2 zx)
Problem 5.32
(a) At the surface of the solenoid, Babove = 0, Bbelow = µ0 nI ẑ = µ0 K ẑ; n̂ = ŝ; so K ⇥ n̂ = K ẑ.
Evidently Eq. 5.76 holds. X
(b) In Eq. 5.69, ✓
both expressiions
reduce to (µ0 R2 ! /3) sin ✓ ˆ at the surface, so Eq. 5.77 is satisfied.
◆
4
@A
µ0 R !
2 sin ✓ ˆ
2µ0 R!
@A
µ0 R!
=
=
sin ✓ ˆ;
=
sin ✓ ˆ. So the left side of
@r R+
3
r3
3
@r
3
R
R
! ⇥ r) = !R sin ✓ ˆ, so the right side of Eq. 5.78 is
Eq. 5.78 is µ0 R! sin ✓ ˆ. Meanwhile K = v = (!
ˆ
µ0 !R sin ✓ , and the equation is satisfied.
Problem 5.33
@A
@A
Because Aabove = Abelow at every point on the surface, it follows that
and
are the same above
@x
@y
and below; any discontinuity
is confined to the
✓
◆ normal
✓ derivative.
◆
@Ayabove
@Aybelow
@Axabove
@Axbelow
Babove Bbelow =
x̂ +
ŷ. But Eq. 5.76 says this equals
+
@z
@z
@z
@z
@Ayabove
@Aybelow
@Axabove @Axbelow
µ0 K( ŷ). So
=
, and
= µ0 K. Thus the normal derivative of the [email protected]
@z
@z
@z
@Aabove
@Abelow
ponent of A parallel to K su↵ers a discontinuity µ0 K, or, more compactly:
= µ0 K.
@n
@n
Problem 5.34
ˆ ✓ˆ = m cos ✓ r̂ m sin ✓ ✓ˆ (Fig. 5.54). Then
(Same idea as Prob. 3.36.) Write m = (m · r̂) r̂ + (m · ✓)
ˆ and Eq. 5.89 , Eq. 5.88. qed
3(m · r̂) r̂ m = 3m cos ✓ r̂ m cos ✓ r̂ + m sin ✓ ✓ˆ = 2m cos ✓ r̂ + m sin ✓ ✓,
Problem 5.35
(a) m = Ia = I⇡R2 ẑ.
⌘
µ0 I⇡R2 ⇣
ˆ .
(b) B ⇡
2
cos
✓
r̂
+
sin
✓
✓
4⇡ r3
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120
CHAPTER 5. MAGNETOSTATICS
µ0 IR2
ẑ (for z < 0, ✓ = ⇡, r̂ = ẑ, so the field
2z 3
is the same, with |z|3 in place of z 3 ). The exact answer (Eq. 5.41) reduces (for z
R) to B ⇡ µ0 IR2 /2|z|3 ,
so they agree.
(c) On the z axis, ✓ = 0, r = z, r̂ = ẑ (for z > 0), so B ⇡
Problem 5.36
The field of one side is given by Eq. 5.37, with s !
p
(w/2)
z 2 + (w/2)2 and sin ✓2 = sin ✓1 = p
;
z 2 + w2 /2
µ0 I
w
p
p
B=
. To pick o↵ the vertical
2
2
4⇡
z + (w /4) z 2 + (w2 /2)
(w/2)
component, multiply by sin = p
; for all four
z 2 + (w/2)2
sides, multiply by 4: B =
µ0 I
w2
p
ẑ.
2⇡ (z 2 + w2 /4) z 2 + w2 /2
µ0 Iw2
ẑ. The field of a dipole m = Iw2 ,
2⇡z 3
µ0 m
for points on the z axis (Eq. 5.88, with r ! z, r̂ ! ẑ, ✓ = 0) is B =
ẑ. X
2⇡ z 3
Problem 5.37
RR
(a) For a ring, m = I⇡r2 . Here I ! v dr = !r dr, so m = 0 ⇡r2 !r dr = ⇡ !R4 /4.
For z
w, B ⇡
(b) The total charge on the shaded ring is dq =
(2⇡R sin ✓)R d✓. The time for one revolution is dt = 2⇡/!.
dq
So the current in the ring is I =
= !R2 sin ✓ d✓. The
dt
area of the ring is ⇡(R sin ✓)2 , so the magnetic moment of the
ring is dm = ( !R2 sin ✓ d✓)⇡R2 sin2 ✓, and the total dipole
moment of the shell is
R⇡
4⇡
m = !⇡R4 0 sin3 ✓ d✓ = (4/3) !⇡R4 , or m =
!R4 ẑ.
3
µ0 4⇡
sin ✓
µ0 !R4 sin ✓ ˆ
!R4 2 ˆ =
,
4⇡ 3
r
3
r2
which is also the exact potential (Eq. 5.69); evidently a spinning sphere produces a perfect dipole field, with
no higher multipole contributions.
The dipole term in the multipole expansion for A is therefore Adip =
Problem 5.38
(a) I dl ! J d⌧, so
A=
Z
1
µ0 X 1
(r0 )n Pn (cos ✓)J d⌧.
4⇡ n=0 rn+1
Z
µ0
µ0 dp
J d⌧ =
(Prob. 5.7), where p is the total electric dipole moment. In mag4⇡r
4⇡r dt
netostatics, p is constant, so dp/dt = 0, and hence Amon = 0. qed
H
R
(c) m = Ia = 12 I (r ⇥ dl) ! m = 12 (r ⇥ J) d⌧. qed
(b) Amon =
Problem 5.39
(a) Yes. (Magnetic forces do no work.)
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121
CHAPTER 5. MAGNETOSTATICS
µ0 ˆ
(b) B =
;
2⇡s
ŝ ˆ ẑ
⇣µ ⌘
0
˙
ˆ
v = ṡ ŝ + s
+ ż ẑ; (v ⇥ B) = ṡ s ˙ ż =
( ż ŝ + ṡ ẑ).
2⇡s
µ0
0 2⇡s 0
F = q(v ⇥ B) =
(c) a = (s̈
0
2⇡s
( ż r̂ + ṡ ẑ).
s ˙ 2 ) ŝ + (s ¨ + 2ṡ ˙ ) ˆ + z̈ ẑ (see, for example J. R. Taylor, Classical Mechanics, Eq. 1.47).
F = ma
where ↵ ⌘
⇣ qµ ⌘
⇣ qµ ⌘
0
. Thus
2⇡m
)
s̈
(s̈
s ˙2 =
↵
s ˙ 2 ) ŝ + (s ¨ + 2ṡ ˙ ) ˆ + z̈ ẑ = ( ż ŝ + ṡ ẑ),
s
ż
↵ ,
s
s ¨ + 2ṡ ˙ = 0,
ṡ
z̈ = ↵ .
s
(d) If ż is constant, then z̈ = 0, and it follows from the third equation that ṡ = 0, and hence s is constant.
1p
Then the first equation says ˙ 2 = ↵(ż/s2 ), so ˙ = ±
↵ż is also constant (the second equation holds
s
automatically). The charge moves in a helix around the wire.
Problem 5.40
The mobile charges do pull in toward the axis, but the resulting concentration of (negative) charge sets up
an electric field that repels away further accumulation. Equilibrium is reached when the electric repulsion on a
mobile charge q balances the magnetic attraction: F = q[E + (v ⇥ B)] = 0 ) E = (v ⇥ B). Say the current
is in the z direction: J = ⇢ v ẑ (where ⇢ and v are both negative).
I
µ0 ⇢ vs ˆ
B · dl = B 2⇡s = µ0 J⇡s2 ) B =
;
2
Z
1
1
E · da = E 2⇡sl = (⇢+ + ⇢ )⇡s2 l ) E =
(⇢+ + ⇢ )s ŝ.
✏0
2✏0
✓ 2◆
h
⇣ µ ⇢ vs ⌘i µ
1
v
0
0
2
2
ˆ
(⇢+ + ⇢ )s ŝ =
(v ẑ) ⇥
=
⇢ v s ŝ ) ⇢+ + ⇢ = ⇢ (✏0 µ0 v ) = ⇢
.
2✏0
2
2
c2
✓
◆
v2
⇢
2
Evidently ⇢+ = ⇢
1
= 2 , or ⇢ =
⇢+ . In this naive model the mobile negative charges fill a
c2
smaller inner cylinder, leaving a shell of positive (stationary) charge at the outside. But since v ⌧ c, the e↵ect
is extremely small.
Problem 5.41
(a) If positive charges flow to the right, they are deflected down, and the bottom plate acquires a positive
charge.
(b) qvB = qE ) E = vB ) V = Et = vBt, with the bottom at higher potential.
(c) If negative charges flow to the left, they are also deflected down, and the bottom plate acquires a negative
charge. The potential di↵erence is still the same, but this time the top plate is at the higher potential.
Problem 5.42
R
RFrom Eq. 5.17,R F = I (dl ⇥ B). But B is constant, in this case, so it comes outside the integral: F =
I
dl ⇥ B, and dl = w, the vector displacement from the point at which the wire first enters the field to
the point where it leaves. Since w and B are perpendicular, F = IBw, and F is perpendicular to w.
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122
CHAPTER 5. MAGNETOSTATICS
Problem 5.43
The angular momentum acquired by the particle as it moves out from the center to the edge is
L=
Z
dL
dt =
dt
Z
N dt =
Z
(r ⇥ F) dt =
Z
r ⇥ q(v ⇥ B) dt = q
Z
r ⇥ (dl ⇥ B) = q
Z
(r · B) dl
Z
B(r · dl) .
But r is perpendicular to B, so r · B = 0, and r · dl = r · dr = 12 d(r · r) = 12 d(r2 ) = r dr = (1/2⇡)(2⇡r dr).
Z R
Z
R
q
q
q
So L =
B 2⇡r dr =
B da. It follows that L =
, where = B da is the total flux.
2⇡ 0
2⇡
2⇡
In particular, if = 0, then L = 0, and the charge emerges with zero angular momentum, which means it is
going along a radial line. qed
Problem 5.44
R
From Eq. 5.24, F = (K ⇥ Bave ) da. Here K = v, v = !R sin ✓ ˆ, da = R2 sin ✓ d✓ d , and
Bave = 12 (Bin + Bout ). From Eq. 5.70,
Bin =
Bout =
=
Bave =
K ⇥ Bave =
2
2
ˆ From Eq. 5.69,
µ0 R! ẑ = µ0 R!(cos ✓ r̂ sin ✓ ✓).
3
3
✓
◆

✓
◆
✓
◆
µ0 R4 ! sin ✓ ˆ
µ0 R4 !
1
@ sin2 ✓
1 @ sin ✓ ˆ
r⇥A=r⇥
=
r̂
✓
3
r2
3
r sin ✓ @✓
r2
r @r
r
µ0 R4 !
ˆ = µ0 R! (2 cos ✓ r̂ + sin ✓ ✓)
ˆ (since r = R).
(2 cos ✓ r̂ + sin ✓ ✓)
3r3
3
µ0 R!
ˆ
(4 cos ✓ r̂ sin ✓ ✓).
6
✓
◆h
i
µ0 R!
ˆ ⇥ (4 cos ✓ r̂ sin ✓ ✓)
ˆ = µ0 ( !R)2 (4 cos ✓ ✓ˆ + sin ✓ r̂) sin ✓.
( !R sin ✓)
6
6
Picking out the z component of ✓ˆ (namely,
µ0
(K ⇥ Bave )z =
( !R)2 sin2 ✓ cos ✓, so
2
Fz =
µ0
( !R)2 R2
2
Z
sin ✓) and of r̂ (namely, cos ✓), we have
sin ✓ cos ✓ d✓ d =
3
µ0
( !R2 )2 2⇡
2
✓
sin4 ✓
4
◆
⇡/2
, or F =
0
µ0 ⇡
( !R2 )2 ẑ.
4
Problem 5.45
µ0 qe qm
µ0 qe qm
(v ⇥ r̂); a =
(v ⇥ r).
2
4⇡ r
4⇡ mr3
1 d
1 d 2
dv
dv
(b) Because a ? v, a · v = 0. But a · v =
(v · v) =
(v ) = v . So
= 0. qed
2 dt
2 dt
dt
dt
✓
◆
⇣
⌘
dQ
µ0 qe qm d r
µ0 qe qm
µ0 qe qm v
r dr
(c)
= m(v ⇥ v) + m(r ⇥ a)
=0+
[r ⇥ (v ⇥ r]
dt
4⇡ dt r
4⇡r3
4⇡
r
r2 dt
⇢

µ0 qe qm 1 2
v
r d p
µ0 qe qm v (r̂ · v)
v
r̂ 2(r · v)
[r v (r · v)r]
= 0. X
=
+ 2 ( r · r) =
r̂
+
4⇡
r3
r
r dt
4⇡
r
r
r
2r
r
µ0 qe qm
(d) (i) Q · ˆ = Q(ẑ · ˆ) = m(r ⇥ v) · ˆ
(r̂ · ˆ). But ẑ · ˆ = r̂ · ˆ = 0, so (r ⇥ v) · ˆ = 0. But
4⇡
(a) F = ma = qe (v ⇥ B) =
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123
CHAPTER 5. MAGNETOSTATICS
r = r r̂, and v =
dl
= ṙ r̂ + r✓˙ ✓ˆ + r sin ✓ ˙ ˆ (where dots denote di↵erentiation with respect to time), so
dt
ˆ
r̂ ✓ˆ
˙ ˆ.
0
r⇥v = r 0
= ( r2 sin ✓ ˙ ) ✓ˆ + (r2 ✓)
˙
˙
ṙ r✓ r sin ✓
Therefore (r ⇥ v) · ˆ = r2 ✓˙ = 0, so ✓ is constant. qed
µ0 qe qm
(ii) Q · r̂ = Q(ẑ · r̂) = m(r ⇥ v) · r̂
(r̂ · r̂). But ẑ · r̂ = cos ✓, and (r ⇥ v) ? r ) (r ⇥ v) · r̂ = 0, so
4⇡
µ0 qe qm
µ0 qe qm
Q cos ✓ =
, or Q =
. And since ✓ is constant, so too is Q. qed
4⇡
4⇡ cos ✓
ˆ = m(r ⇥ v) · ✓ˆ µ0 qe qm (r̂ · ✓).
ˆ But ẑ · ✓ˆ = sin ✓, r̂ · ✓ˆ = 0, and (r ⇥ v) · ✓ˆ = r2 sin ✓ ˙
(iii) Q · ✓ˆ = Q(ẑ · ✓)
4⇡
Q
k
Q
µ0 qe qm
(from (i)), so Q sin ✓ = mr2 sin ✓ ˙ ) ˙ =
= 2 , with k ⌘
=
.
mr2
r
m
4⇡m cos ✓
k
k2
k 2 sin2 ✓
(e) v 2 = ṙ2 + r2 ✓˙2 + r2 sin2 ✓ ˙ 2 , but ✓˙ = 0 and ˙ = 2 , so ṙ2 = v 2 r2 sin2 ✓ 4 = v 2
.
r
r
r2
r⇣ ⌘
✓ ◆2
⇣ ⌘
dr
ṙ2
v 2 (k sin ✓/r)2
vr 2
dr
vr 2
2
2
=
=
=
r
sin
✓
;
=
r
sin2 ✓.
˙2
d
(k 2 /r4 )
k
d
k
Z
Z
⇣ vr ⌘
dr
1
vr
q
(f)
= d )
sec 1
; sec[(
, or
0 =
0 ) sin ✓] =
sin
✓
k
sin
✓
k
sin
✓
2
r (vr/k)2 sin ✓
r( ) =
A
cos[(
0 ) sin ✓]
, where A ⌘
µ0 qe qm tan ✓
.
4⇡mv
Problem 5.46
Put the field point on the x axis, so r = (s, 0, 0). Then B =
Z
µ0
(K ⇥ r̂ )
ˆ
r 2 da; da = R d dz; K = K = K( sin x̂ +
4⇡
cos ŷ); r = (s R cos ) x̂ R sin ŷ z ẑ.
x̂
ŷ
ẑ
sin
cos
0
K ⇥ r
=
K
=
(s R cos ) ( R sin ) ( z)
K [( z cos ) x̂ + ( z sin ) ŷ + (R s cos ) ẑ] ;
r 2 = z 2 + R2 + s2 2Rs cos . The x and y components
integrate to zero (z integrand is odd, as in Prob. 5.18).
⇢Z 1
Z
(R s cos )
µ0 KR 2⇡
dz
d
dz
=
(R
s
cos
)
d ,
2 + d2 )3/2
4⇡
(z 2 + R2 + s2 2Rs cos )3/2
(z
0
1
Z 1
1
dz
2z
2
where d2 ⌘ R2 + s2 2Rs cos . Now
= p
= 2.
2
2
3/2
2 z 2 + d2 0
d
(z
+
d
)
d
1
Z
⇤
µ0 KR 2⇡
(R s cos )
1 ⇥ 2
=
d ; (R s cos ) =
(R
s2 ) + (R2 + s2 2Rs cos ) .
2
2
2⇡
(R + s
2Rs cos )
2R
0

Z 2⇡
Z 2⇡
µ0 K
d
=
(R2 s2 )
+
d .
4⇡
(R2 + s2 2Rs cos )
0
0
Bz =
µ0
KR
4⇡
Z
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124
Z
2⇡
0
d
a + b cos
=2
Z
2Rs, so a2
d
a + b cos
0
= p
b=
⇡
CHAPTER 5. MAGNETOSTATICS
4
a2
b2
tan
=p
1
"p
4
a2
a2
b2
tan
1
"p
a2
#
b2 tan( /2)
a+b
tan(⇡/2)
4
=p
2
a+b
a
b2
b2
#
⇣⇡⌘
⇡
0
=p
2
2⇡
a2
b2 = R4 + 2R2 s2 + s4 4R2 s2 = R4 2R2 s2 + s4 = (R2 s2 )2 ;

✓
◆
µ0 K (R2 s2 )
µ0 K R2 s2
Bz =
2⇡
+
2⇡
=
+
1
.
4⇡
|R2 s2 |
2
|R2 s2 |
b2
p
a2
. Here a = R2 + s2 ,
b2 = |R2
d2 |.
µ0 K
µ0 K
(1+1) = µ0 K. Outside the solenoid, s > R, so Bz =
( 1+1) = 0.
2
2
Here K = nI, so B = µ0 nI ẑ(inside), and 0(outside) (as we found more easily using Ampére’s law, in Ex. 5.9).
Inside the solenoid, s < R, so Bz =
Problem 5.47
µ0 IR2
(a) From Eq. 5.41, B =
2
@B
µ0 IR2
=
@z
2
@B
@z
z=0
(
(
1
3/2
[R2 + (d/2 + z)2 ]
( 3/2)2(d/2 + z)
+
1
3/2
[R2 + (d/2
( 3/2)2(d/2
z)( 1)
z)2 ]
)
.
)
+
5/2
5/2
[R2 + (d/2 + z)2 ]
[R2 + (d/2 z)2 ]
)
(
3µ0 IR2
(d/2 + z)
(d/2 z)
=
+
.
5/2
5/2
2
[R2 + (d/2 + z)2 ]
[R2 + (d/2 z)2 ]
(
)
3µ0 IR2
d/2
d/2
=
+
= 0. X
5/2
5/2
2
[R2 + (d/2)2 ]
[R2 + (d/2)2 ]
(b) Di↵erentiating again:
@2B
3µ0 IR2 n
1
(d/2 + z)( 5/2)2(d/2 + z)
=
+
2
5/2
7/2
2
2
@z
2
[R + (d/2 + z) ]
[R2 + (d/2 + z)2 ]
1
(d/2 z)( 5/2)2(d/2 z)( 1) o
+
+
.
5/2
7/2
[R2 + (d/2 z)2 ]
[R2 + (d/2 z)2 ]
(
)
✓
@2B
3µ0 IR2
2
2(5/2)2(d/2)2 2
3µ0 IR2
=
+
=
R2
5/2
7/2
7/2
@z 2 z=0
2
[R2 + (d/2)2 ]
[R2 + (d/2)2 ]
[R2 + (d/2)2 ]
=
3µ0 IR2
d2
5d2
+
4
4
◆
d2 R2 . Zero if d = R, in which case
7/2
[R2 + (d/2)2 ]
(
)
µ0 IR2
1
1
1
8µ0 I
B(0) =
+
= µ0 IR2
= 3/2 .
3/2
3/2
2
3/2
2
(5R /4)
5 R
[R2 + (R/2)2 ]
[R2 + (R/2)2 ]
Problem 5.48
The total charge on the shaded ring is dq = (2⇡r) dr. The
time for one revolution is dt = 2⇡/!. So the current in the
dq
ring is I =
= !r dr. From Eq. 5.41, the magnetic field of
dt
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125
CHAPTER 5. MAGNETOSTATICS
µ0
r2
!r 2
dr ẑ, and the total field of the disk is
2
(r + z 2 )3/2
this ring (for points on the axis) is dB =
Z
R
r3 dr
ẑ. Let u ⌘ r2 , so du = 2r dr. Then
2
2 3/2
0 (r + z )
 ✓
◆ R2

Z 2
µ0 ! R
u du
µ0 !
u + 2z 2
µ0 ! (R2 + 2z 2 )
p
=
=
2 p
=
4
4
2
(u + z 2 )3/2
u + z2
R2 + z 2
0
0
B =
µ0 !
2
When z
so
R, the term in square brackets is
✓
◆✓
◆
2z 2 (1 + R2 /2z 2 )
R2
1 R2
3 R4
[]=
2z ⇡ 2z 1 + 2
1
+
2z
2 z2
8 z4
z[1 + (R/z)2 ] 1/2
✓
◆
✓
◆
2
4
2
4
4
R
3R
R
R
1R
R4
⇡ 2z 1
+
+
1
=
2z
=
,
2z 2
8 z4
2z 2
4z 4
8 z4
4z 4
2z ẑ.
B⇡
1
µ0 ! R4
µ0 !R4
ẑ =
ẑ.
4
2 4z
8z 3
Meanwhile, from Eq. 5.88, the dipole field is
Bdip =
⌘
µ0 m ⇣
ˆ ,
2
cos
✓
r̂
+
sin
✓
✓
4⇡r3
and for points on the +z axis ✓ = 0, r = z, r̂ = ẑ, so Bdip =
µ0 m
ẑ. In this case (Problem 5.37a) m = ⇡ !R4 /4,
2⇡z 3
µ0 !R4
ẑ, in agreement with the approximation. X
8z 3
Problem 5.49
Z
µ0 I
dl0 ⇥ r
B =
r 3 ; r = R cos x̂ + (y R sin ) ŷ + z ẑ. (For simplicity I’ll drop the prime on
4⇡
.) r 2 = R2 cos2 + y 2 2Ry sin + R2 sin2 + z 2 = R2 + y 2 + z 2 2Ry sin . The source coordinates
(x0 , y 0 , z 0 ) satisfy x0 = R cos ) dx0 = R sin d ; y 0 = R sin ) dy 0 = R cos d ; z 0 = 0 ) dz 0 = 0. So
dl0 = R sin d x̂ + R cos d ŷ.
x̂
ŷ
ẑ
0
r
R
sin
d
R
cos
d
0 = (Rz cos d ) x̂ + (Rz sin d ) ŷ + ( Ry sin d + R2 d ) ẑ.
dl ⇥
=
R cos
(y R sin ) z
so Bdip =
Bx =
µ0 IRz
4⇡
Z
cos
2⇡
0
d
3/2
(R2 + y 2 + z 2
2Ry sin )
=
µ0 IRz 1
1
p
4⇡ Ry R2 + y 2 + z 2
2⇡
2Ry sin
= 0,
0
since sin = 0 at both limits. The y and z components are elliptic integrals, and cannot be expressed in terms
of elementary functions.
Bx = 0;
µ0 IRz
By =
4⇡
Z
0
sin
2⇡
(R2
+
y2
+
z2
d
µ0 IR
; Bz =
3/2
4⇡
2Ry sin )
Z
(R
2⇡
0
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reproduced, in any form or by any means, without permission in writing from the publisher.
(R2
+
y2
y sin ) d
+ z2
3/2
2Ry sin )
.
126
CHAPTER 5. MAGNETOSTATICS
Problem 5.50
From the Biot-Savart law, the field of loop #1 is B =
F = I2
I
2
dl2 ⇥ B =
µ0
I1 I2
4⇡
I I
1
dl2 ⇥ (dl1 ⇥
r
2
µ0
I1 I2
4⇡
F=
r̂
)
2
⇢I I
µ0 I1
4⇡
I
r
1
I
dl1
; the force on loop #2 is
2
r̂
. Now dl2 ⇥ (dl1 ⇥
r̂
r 2 (dl1 · dl2 )
r̂
dl1 ⇥
I
) = dl1 (dl2 ·
(dl2 ·
r
r̂
r̂
)
r̂
(dl1 · dl2 ), so
)
2
The first term is what we want. It remains to show that the second term is zero:
⇥
⇤
r = (x2 x1 ) x̂ + (y2 y1 ) ŷ + (z2 z1 ) ẑ, so r2 (1/ r ) = @ (x2 x1 )2 + (y2 y1 )2 + (z2 z1 )2 1/2 x̂
@x2
⇤ 1/2
⇤ 1/2
@ ⇥
@ ⇥
+
(x2 x1 )2 + (y2 y1 )2 + (z2 z1 )2
ŷ +
(x2 x1 )2 + (y2 y1 )2 + (z2 z1 )2
ẑ
@y2
@z2
✓ ◆
I
I
r
r̂
r̂
(x2 x1 )
(y2 y1 )
(z2 z1 )
1
=
x̂
ŷ
ẑ =
=
. So
· dl2 =
r2
· dl2 = 0 (by
3
3
3
3
2
2
r
r
r
Corollary 2 in Sect. 1.3.3). qed
Problem 5.51 I
µ0 I
dl ⇥ r̂
(a) B =
r2 =
I4⇡
µ0 I
d✓
so B =
. X
4⇡
r
µ0 I
4⇡
r
I
r
dl ⇥ r̂
, dl ⇥ r̂ =
r2
µ0 I 1
(b) For a circular loop, r = R is constant; B =
4⇡ R
(c)
r
r
dl sin ẑ, dl sin
I
d✓ =
= r d✓ (see diagram in the book),
µ0 I
, which agrees with Eq. 5.41 (z = 0).
2R
0.8
0.6
0.4
0.2
-0.5
0.5
1.0
1.5
-0.2
-0.4
µ0 I
B=
4⇡a
(d) B =
µ0 I
4⇡p
Problem 5.52
(a)
Z
2⇡
Z
0
µ0
4⇡
p
0
(1 + e cos ✓) d✓ =
B=
2⇡
Z

µ0 I 2 3/2
✓ d✓ =
✓
4⇡a 3
µ0 I
µ0 I
(2⇡) =
.
4⇡p
2p
J⇥
r
r̂
2
d⌧
)
2⇡
0
p
µ0 I 2⇡
=
.
3a
X
A=
1
4⇡
Z
B⇥
r
r̂
2
d⌧.
1
⇢. For dielectrics (with no free charge), ⇢b = r · P
✏0Z
1
P(r0 ) · r̂
0
(Eq. 4.12), and the resulting potential is V (r) =
r 2 d⌧ . In general, ⇢ = ✏0 r · E (Gauss’s law),
4⇡✏0
(b) Poisson’s equation (Eq. 2.24) says r2 V =
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127
CHAPTER 5. MAGNETOSTATICS
Z
1
E(r0 ) · r̂
0
r 2 d⌧ . qed
4⇡
[There are many other ways to obtain this result. For example, using Eq. 1.100:
✓
◆
✓
◆
r̂
r̂
0
r·
= r ·
= 4⇡ 3 ( r ) = 4⇡ 3 (r r0 ),
2
2
so the analogy is P !
✏0 E, and hence V (r) =
r
V (r) =
Z
V (r ) (r r ) d⌧ =
0
0
3
0
(Eq. 1.59). But r0 V (r0 ) =
1
4⇡
r
Z
0
V (r )r ·
0
✓
r̂
r2
◆
d⌧ 0 =
1
4⇡
Z
r̂ ⇥ 0 0 ⇤ 0
r 2 · r V (r ) d⌧
E(r0 ), and the surface integral ! 0 at 1, so V (r) =
before. You can also check the result, by computing its gradient—but it’s not easy.]
1
4⇡
1
4⇡
Z
I
V (r0 )
E(r0 ) ·
r
2
r̂
r̂
0
r 2 ·da
d⌧ 0 , as
Problem 5.53
Rr
Rr
(a) For uniform B, 0 (B ⇥ dl) = B ⇥ 0 dl = B ⇥ r 6= A = 12 (B ⇥ r).
✓
◆
✓
◆
I
µ0 I ˆ
µ0 I
µ0 I
µ0 Iw 1 1
(b) B =
, so B ⇥ dl =
ŝ
ŝ w =
ŝ 6= 0.
2⇡s
2⇡a
2⇡b
2⇡
a b
R1
1
(c) A = r ⇥ B 0 d =
2 (r ⇥ B).
Z 1
µ0 I ˆ
µ0 I ˆ
µ0 I
1
µ0 I
(d) B =
; B( r) =
; A =
(r ⇥ ˆ)
d =
(r ⇥ ˆ). But r here is the
2⇡s
2⇡ s
2⇡s
2⇡s
0
i
µ0 I h
vector from the origin—in cylindrical coordinates r = s ŝ + z ẑ. So A =
s(ŝ ⇥ ˆ) + z(ẑ ⇥ ˆ) , and
2⇡s
µ0 I
ˆ
ˆ
(ŝ ⇥ ) = ẑ, (ẑ ⇥ ) = ŝ. So A =
(z ŝ s ẑ.
2⇡s
The examples in (c) and (d) happen to be divergenceless, but this is not in general the case. For (letting
R1
L ⌘ 0 B( r) d , for short) r · A = r · (r ⇥ L) = [L · (r ⇥ r) r · (r ⇥ L)] = r · (r ⇥ L), and
R1
R1
R1
R1
r ⇥ L = 0 [r ⇥ B( r)] d = 0 2 [r ⇥ B( r)] d = µ0 0 2 J( r) d , so r · A = µ0 r · 0 2 J( r) d , and
it vanishes in regions where J = 0 (which is why the examples in (c) and (d) were divergenceless). To construct
an explicit counterexample, we need the field at a point where J 6= 0—say, inside a wire with uniform current.
µ0 J ˆ
Here Ampére’s law gives B 2⇡s = µ0 Ienc = µ0 J⇡s2 ) B =
s , so
2
Z
1
✓
µ0 J
2
◆
s ˆd =
r⇥
0

µ0 J 1 @ 2
@
r·A =
(s z) +
( s2 ) =
6
s @s
@z
A =
µ0 J
µ0 Js
s(r ⇥ ˆ) =
(z ŝ sẑ).
6
6
✓
◆
µ0 J 1
µ0 Jz
2sz =
6= 0.
6
s
3
Conclusion: (ii) does not automatically yield r · A = 0.
Problem 5.54
(a) Exploit the analogy with the electrical case:
1 1
1 p · r̂
[3(p · r̂) r̂ p] (Eq. 3.104) = rV, with V =
4⇡✏0 r3
4⇡✏0 r2
µ0 1
B =
[3(m · r̂) r̂ m] (Eq. 5.89) = rU, (Eq. 5.67).
4⇡ r3
E=
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(Eq. 3.102).
128
CHAPTER 5. MAGNETOSTATICS
µ0 m · r̂
.
4⇡ r2
(b) Comparing Eqs. 5.69 and 5.87, the dipole moment of the shell is m = (4⇡/3)! R4 ẑ (which we also got
Evidently the prescription is p/✏0 ! µ0 m : U (r) =
µ0 ! R4 cos ✓
for r > R.
3
r2
Inside the shell, the field is uniform (Eq. 5.70): B = 23 µ0 !R ẑ, so U (r) = 23 µ0 !Rz + constant. We may
in Prob. 5.37). Using the result of (a), then, U (r) =
as well pick the constant to be zero, so U (r) =
2
3 µ0
!Rr cos ✓ for r < R.
[Notice that U (r) is not continuous at the surface (r = R): Uin (R) = 23 µ0 !R2 cos ✓ 6= Uout (R) =
!R2 cos ✓. As I warned you on p. 245: if you insist on using magnetic scalar potentials, keep away from
places where there is current!]
(c)
✓
◆
✓
◆
µ0 !Q
3r2
6r2
@U
1 @U ˆ
1 @U ˆ
B =
1
cos
✓
r̂
1
sin ✓ ✓ˆ = rU =
r̂
✓
.
4⇡R
5R2
5R2
@r
r @✓
r sin ✓ @
@U
= 0 ) U (r, ✓, ) = U (r, ✓).
@
✓
◆✓
◆
✓
◆✓
◆
1 @U
µ0 !Q
6r2
µ0 !Q
6r2
=
1
sin ✓ ) U (r, ✓) =
1
r cos ✓ + f (r).
r @✓
4⇡R
5R2
4⇡R
5R2
✓
◆✓
◆
✓
◆✓
◆
@U
µ0 !Q
3r2
µ0 !Q
r3
=
1
cos
✓
)
U
(r,
✓)
=
r
cos ✓ + g(✓).
@r
4⇡R
5R2
4⇡R
5R2
1
3 µ0
Equating the two expressions:
✓
◆✓
µ0 !Q
1
4⇡R
or
6r2
5R2
◆
r cos ✓ + f (r) =
✓
µ0 !Q
4⇡R3
◆
✓
µ0 !Q
4⇡R
◆✓
1
r2
5R2
◆
r cos ✓ + g(✓),
r3 cos ✓ + f (r) = g(✓).
But there is no way to write r3 cos ✓ as the sum of a function of ✓ and a function of r, so we’re stuck. The
reason is that you can’t have a scalar magnetic potential in a region where the current is nonzero.
Problem 5.55
Z
µ0
J
0
(a) r · B = 0, r ⇥ B = µ0 J, and r · A = 0, r ⇥ A = B ) A =
r d⌧ , so Z
4⇡
1
B
0
r · A = 0, r ⇥ A = B, and r · W = 0 (we’ll choose it so), r ⇥ W = A ) W =
r d⌧ .
4⇡
(b) W will be proportional to B and to two factors of r (since di↵erentiating twice must recover B), so I’ll
try something of the form W = ↵r(r · B) + r2 B, and see if I can pick the constants ↵ and in such a way
that r · W = 0 and r ⇥ W = A.
⇥
⇤
@x @y @z
r · W = ↵ [(r · B)(r · r) + r · r(r · B)] + r2 (r · B) + B · r(r2 ) . rr =
+
+
= 1 + 1 + 1 = 3;
@x @y @z
r(r · B) = r ⇥ (r ⇥ B) + B ⇥ (r ⇥ r) + (r · r)B + (B · r)r; but B is constant, so all derivatives of B vanish,
and r ⇥ r = 0 (Prob. 1.63),
so
✓
◆
@
@
@
r(r · B) = (B · r)r = Bx
+ By
+ Bz
(x x̂ + y ŷ + z ẑ) = Bx x̂ + By ŷ + Bz ẑ = B;
@x◆
@y
@z
✓
@
@
@
r(r2 ) = x̂
+ ŷ
+ ẑ
(x2 + y 2 + z 2 ) = 2x x̂ + 2y ŷ + 2z ẑ = 2r. So
@x
@y
@z
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129
CHAPTER 5. MAGNETOSTATICS
r · W = ↵ [3(r · B) + (r · B)] + [0 + 2(r · B)] = 2(r · B)(2↵ + ), which is zero if 2↵ + = 0.
⇥
⇤
r ⇥ W = ↵ [(r · B)(r ⇥ r) r ⇥ r(r · B)] + r2 (r ⇥ B) B ⇥ r(r2 ) = ↵ [0 (r ⇥ B)] + [0 2(B ⇥ r)]
1
= (r ⇥ B)(↵ 2 ) =
(r ⇥ B) (Prob. 5.25). So we want ↵ 2 = 1/2. Evidently ↵ 2( 2↵) = 5↵ = 1/2,
2
⇤
1 ⇥
or ↵ = 1/10; = 2↵ = 1/5. Conclusion: W =
r(r · B) 2r2 B . (But this is certainly not unique.)
10
R
R
H
(c)
R r ⇥ W = A ) (r ⇥ W) · da = A · da. Or W · dl =
A · da. Integrate around the amperian loop shown, taking
W to point parallel to the axis, and choosing W = 0 on the
axis:
◆
Z s✓
µ0 nI
µ0 nI s2 l
Wl =
ls̄ ds̄ =
(using Eq. 5.72 for A).
2
2 2
0
µ0 nIs2
ẑ (s < R).
4
◆
Z s✓
µ0 nIR2 l
µ0 nI R2
µ0 nIR2 l µ0 nIR2 l
For s > R, W l =
+
l ds̄ =
+
ln(s/R);
4
2
s̄
4
2
R
W=
W=
µ0 nIR2
[1 + 2 ln(s/R)] ẑ (s > R).
4
Problem 5.56
Apply the divergence theorem to the function [U ⇥ (r ⇥ V)], noting (from the product rule) that
r · [U ⇥ (r ⇥ V)] = (r ⇥ V) · (r ⇥ U) U · [r ⇥ (r ⇥ V)]:
Z
Z
I
r · [U ⇥ (r ⇥ V)] d⌧ = {(r ⇥ V) · (r ⇥ U) U · [r ⇥ (r ⇥ V)]} d⌧ = [U ⇥ (r ⇥ V)] · da.
As always, suppose we have two solutions, B1 (and A1 ) and B2 (and A2 ). Define B3 ⌘ B2 B1 (and
A3 ⌘ A2 A1 ), so that r ⇥ A3 = B3 and r ⇥ B3 = r ⇥ B1 r ⇥ B2 = µ0 J µ0 J = 0. Set U = V = A3
in
Z the above identity:
Z
Z
{(r ⇥ A3 ) · (r ⇥ A3 ) A3 · [r ⇥ (r ⇥ A3 )]} d⌧ = {(B3 ) · (B3 ) A3 · [r ⇥ B3 ]} d⌧ = (B3 )2 d⌧
I
I
=
[A3 ⇥ (r ⇥ A3 )] · da = (A3 ⇥ B3 ) · da. But either A is specified (in which case A3 = 0), or else B is
Z
H
specified (in which case B3 = 0), at the surface. In either case (A3 ⇥ B3 ) · da = 0. So (B3 )2 d⌧ = 0, and
hence B1 = B2 .
Problem 5.57
qed
µ0 m0
ˆ ThereFrom Eq. 5.88, Btot = B0 ẑ
(2 cos ✓ r̂ + sin ✓ ✓).
4⇡r3
⇣
µ0 m0
µ0 m0 ⌘
fore B · r̂ = B0 (ẑ · r̂)
2 cos ✓ = B0
cos ✓.
3
4⇡r
2⇡r3
µ0 m0
This is zero, for all ✓, when r = R, given by B0 =
, or
2⇡R3
✓
◆1/3
µ0 m0
R=
. Evidently no field lines cross this sphere.
2⇡B0
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130
CHAPTER 5. MAGNETOSTATICS
Problem 5.58
Q
Q!
Q!
Q
(a) I =
=
; a = ⇡R2 ; m =
⇡R2 ẑ = !R2 ẑ. L = RM v = M !R2 ; L = M !R2 ẑ.
(2⇡/!)
2⇡
2⇡
2
✓
◆
m
Q !R2
Q
Q
Q
=
=
. m=
L, and the gyromagnetic ratio is g =
.
2
L
2 M !R
2M
2M
2M
(b) Because g is independent of R, the same ratio applies to all “donuts”, and hence to the entire sphere
Q
(or any other figure of revolution): g =
.
2M
(c) m =
e ~
e~
(1.60 ⇥ 10 19 )(1.05 ⇥ 10
=
=
2m 2
4m
4(9.11 ⇥ 10 31 )
34
)
= 4.61 ⇥ 10
24
A m2 .
⇡
cos ✓ sin ✓
p
d✓.
R2 + (z 0 )2 2Rz 0 cos ✓
Problem 5.59
Z
Z
1
3
(a) Bave =
B
d⌧
=
(r ⇥ A) d⌧ =
3
(3/4)⇡R3
⇢Z
I
I4⇡R
3
3 µ0
J
A ⇥ da =
d⌧ 0 ⇥ da =
3 4⇡
r
4⇡R3
4⇡R
⇢I
Z
3µ0
1
0
J⇥
r da d⌧ . Note that J depends on
(4⇡)2 R3
the source point r0 , not on the field point r. To do the surface
0
integral, choose the (x, y, z) coordinates
p so that r lies on the
2
0
2
z axis (see diagram). Then r = R + (z )
2Rz 0 cos ✓,
2
while da = R sin ✓ d✓ d r̂. By symmetry, the x and y components must integrate to zero; since the z component of r̂ is
cos ✓, we have
I
1
r
da = ẑ
=
=
=
=
=
Z
p
cos ✓
R2 sin ✓ d✓ d = 2⇡R2 ẑ
Z
R2 + (z 0 )2 2Rz 0 cos ✓
0
Let u ⌘ cos ✓, so du = sin ✓ d✓.
Z 1
u
p
2⇡R2 ẑ
du
2
0
R + (z )2 2Rz 0 u
1
(
)
⇥
⇤
1
2 2(R2 + (z 0 )2 ) + 2Rz 0 u p 2
2
0 )2
0u
2⇡R ẑ
R
+
(z
2Rz
3(2Rz 0 )2
1
o
p
⇤
⇥
⇤p
2⇡R2 ẑ n⇥ 2
0 2
0
2
0 2
0
2 + (z 0 )2
0
2 + (z 0 )2 + 2Rz 0
R
+
(z
)
+
Rz
R
2Rz
R
+
(z
)
Rz
R
3(Rz 0 )2

⇥ 2
⇤
⇥ 2
⇤
2⇡
ẑ
R + (z 0 )2 + Rz 0 |R z 0 |
R + (z 0 )2 Rz 0 (R + z 0 )
0
2
3(z )
8
9
4⇡
4⇡ 0
>
>
>
z 0 ẑ =
r,
(r0 < R); >
>
>
< 3
=
3
>
>
>
>
4⇡ R3 0
4⇡R3
>
:
;
ẑ =
r , (r0 > R). >
0
2
0
3
3(z )
3 (r )
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131
CHAPTER 5. MAGNETOSTATICS
For now we want r0 < R, so Bave =
(Eq. 5.90), so Bave =
µ0 2m
.
4⇡ R3
3µ0 4⇡
(4⇡)2 R3 3
qed
Z
(J⇥r0 ) d⌧ 0 =
µ0
4⇡R3
Z
(J⇥r0 ) d⌧ 0 . Now m =
1
2
R
(r⇥J) d⌧
◆
Z ✓
Z
3µ0 4⇡ 3
r0
µ0
J ⇥ r̂
0
0
(b) This time r > R, so Bave =
R
J ⇥ 0 3 d⌧ =
r 2 d⌧ , where
(4⇡)2 R3 3
(r )
4⇡
goes from the source point to the center ( r = r0 ). Thus Bave = Bcen . qed
0
r
now
Problem 5.60
4⇡
(a) Problem 5.53 gives the dipole moment of a shell: m =
!R4 ẑ. Let R ! r, ! ⇢ dr, and integrate:
3
Z R
4⇡
4⇡ R5
Q
1
m=
!⇢ ẑ
r4 dr =
!⇢
ẑ. But ⇢ =
, so m = Q!R2 ẑ.
3
3
3
5
(4/3)⇡R
5
0
(b) Bave =
µ0 2m
µ0 2Q!
=
ẑ.
4⇡ R3
4⇡ 5R
µ0 m sin ✓ ˆ
µ0 Q!R2 sin ✓ ˆ
=
.
4⇡ r2
4⇡ 5
r2
(d) Use Eq. 5.69, with R ! r̄, ! ⇢ dr̄, and integrate:
(c) A ⇠
=
µ0 !⇢ sin ✓ ˆ
A=
3
r2
Z
R
0
r̄4 dr̄ =
µ0 ! 3Q sin ✓ R5 ˆ
µ0 Q!R2 sin ✓ ˆ
=
.
3 4⇡R3 r2 5
4⇡ 5
r2
This is identical to (c); evidently the field is pure dipole,
✓for points
◆ outside the
✓ sphere.
◆
µ0 !Q
3r2
6r2
(e) According to Prob. 5.30, the field is B =
1
cos
✓
r̂
1
sin ✓ ✓ˆ . The average
4⇡R
5R2
5R2
obviously points in the z direction, so take the z component of r̂ (cos ✓) and ✓ˆ ( sin ✓):
◆
✓
◆
Z ✓
µ0 !Q
1
3r2
6r2
2
Bave =
1
cos ✓ + 1
sin2 ✓ r2 sin ✓ dr d✓ d
4⇡R (4/3)⇡R3
5R2
5R2
◆
✓ 3
◆
Z ⇡ ✓ 3
3µ0 !Q
r
3 R5
R
6 R5
2
=
2⇡
cos ✓ +
sin2 ✓ sin ✓ d✓
(4⇡R2 )2
3
5 5R2
3
5 5R2
0
◆
Z ✓
Z ⇡
3µ0 !Q 3 ⇡ 16
7
3µ0 !Q 1
2
2
=
R
cos
✓
+
sin
✓
sin
✓
d✓
=
7 + 9 cos2 ✓ sin ✓ d✓
8⇡R4
75
75
8⇡R 75 0
0
⇡
µ0 !Q
µ0 !Q
µ0 !Q
=
7 cos ✓ 3 cos3 ✓
=
(20) =
(same as (b)). X
200⇡R
200⇡R
10⇡R
0
Problem 5.61
The issue (and the integral) is identical to the one in Prob. 3.48. The resolution (as before) is to regard
Eq. 5.89 as correct outside an infinitesimal sphere centered at the dipole. Inside this sphere the field is a
delta-function, A 3 (r), with A selected so as to make the average field consistent with Prob. 5.59:
Bave =
1
(4/3)⇡R3
Z
A 3 (r) d⌧ =
3
µ0 2m
2µ0 m
2µ0
A=
)A=
m 3 (r).
3
3
4⇡R
4⇡ R
3
3
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132
CHAPTER 5. MAGNETOSTATICS
Problem 5.62
For a dipole at the origin and a field point in the x z plane ( = 0), we have
µ0 m
ˆ = µ0 m [2 cos ✓(sin ✓ x̂ + cos ✓ ẑ) + sin ✓(cos ✓ x̂
(2 cos ✓ r̂ + sin ✓ ✓)
4⇡ r3
4⇡ r3
µ0 m
2
=
[3 sin ✓ cos ✓ x̂ + (2 cos ✓ sin2 ✓) ẑ].
4⇡ r3
B =
Here we have a stack of such dipoles, running from z =
L/2 to z = +L/2. Put the field point at s on the
x axis. The x̂ components cancel (because of symmetrically placed dipoles above and below z = 0), leaving B =
Z L/2
(3 cos2 ✓ 1)
µ0
2M ẑ
dz, where M is the dipole mo4⇡
r3
0
ment per unit length: m = I⇡R2 = ( vh)⇡R2 = !R⇡R2 h )
m
s
1
sin3 ✓
M =
= ⇡ !R3 . Now sin ✓ = , so 3 =
; z =
h
r
r
s3
s
s cot ✓ ) dz =
d✓. Therefore
sin2 ✓
Z ✓m
Z ✓m
µ0
sin3 ✓ s
µ0 !R3
3
2
B =
(⇡ !R ) ẑ
(3 cos ✓ 1) 3
d✓ =
ẑ
(3 cos2 ✓
2⇡
s sin2 ✓
2s2
⇡/2
⇡/2
=
µ0 !R3
ẑ
2s2
cos3 ✓ + cos ✓
✓m
⇡/2
=
µ0 !R3
cos ✓m 1
2s2
cos2 ✓m ẑ =
s
(L/2)
But sin ✓m = p
, and cos ✓m = p
, so B =
2
2
2
s + (L/2)
s + (L/2)2
sin ✓ ẑ)]
1) sin ✓ d✓
µ0 !R3
cos ✓m sin2 ✓m ẑ.
2s2
µ0 !R3 L
ẑ.
4[s2 + (L/2)2 ]3/2
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133
CHAPTER 6. MAGNETIC FIELDS IN MATTER
Chapter 6
Magnetic Fields in Matter
Problem 6.1
N = m2 ⇥B1 ; B1 =
Problem 6.3µ0 m1 m2
N=
(ŷ⇥ẑ) =
4⇡ r3
downward ( ẑ).
µ0 1
[3(m1 ·r̂)r̂
4⇡ r3
m1 ] ; r̂ = ŷ; m1 = m1 ẑ; 1m2 = m2 ŷ. B1 =
µ0 m1 m2
x̂. Here m1 = ⇡a2 I, m2 = b2 I. So N =
4⇡ r3
µ0 (abI)2
x̂.
4 r3
µ0 m1
ẑ.
4⇡ r3
Final orientation :
Problem 6.2
dF = I dl⇥B; dN = r⇥dF = I r⇥(dl⇥B). Now (Prob. 1.6): r⇥(dl⇥B) + dl⇥(B⇥r) + B⇥(r⇥dl) = 0.
But d [r⇥(r⇥B)] = dr⇥(r⇥B) + r⇥(dr⇥B) (since B is constant), and dr = dl, so dl⇥(B⇥r) = r⇥(dl⇥B)
1
d [r⇥(r⇥B)].
dN
H Hence 2r⇥(dl⇥B)
H = d [r⇥(r⇥B)] B⇥(r⇥dl).
H = 2 I {d [r⇥(r⇥B)] B⇥(r⇥dl)}.
1
) N = 2 I d [r⇥(r⇥B)] B⇥ (r⇥dl) . But the first term is zero ( d(· · · ) = 0), and the second integral is
2a (Eq. 1.107). So N = I(B⇥a) = m⇥B. qed
Problem 6.3
ẑ ✻
(a)
I
r̂
B
✕✸
θ✲
ŷ
R
❥
According to Eq. 6.2, F = 2⇡IRB cos ✓. But B =
µ0 [3(m1 ·r̂)r̂ m1 ]
, and B cos ✓ = B · ŷ, so B cos ✓ =
4⇡
r3
µ0 1
[3(m
·
r̂)(r̂
· ŷ) (m1 · ŷ)]. But m1 · ŷ = 0 and
3
1
4⇡ r
r̂ · ŷ = sin , while m1 · r̂ = m1 cos ✓. ) B cos ✓ =
µ0 1
4⇡ r 3 3m1 sin cos .
r
φ
m1 ✻
µ0 1
F = 2⇡IR 4⇡
r 3 3m1 sin cos . Now sin
p
R2 /r, so F = 3 µ20 m1 IR2
p
3µ0 m1 m2
r 2 R2
0
But IR2 ⇡ = m2 , so F = 3µ
, while for a dipole, R ⌧ r, so F =
.
2⇡ m1 m2
r5
2⇡ r4
⇥
⇤
µ0 1
µ0
d
d
1
(b) F = r(m2 ·B) = (m2 ·r)B = m2 dz
m1 ) = 2⇡
m1 m2 ẑ dz
3 ,
1 ·ẑ)ẑ
4⇡ z 3 (3(m
|
{z
}
| {zz }
2m1
3 z14
Problemor,
6.5
since z = r: F =
=
R
r
, cos
3µ0 m1 m2
ẑ.
2⇡ r4
=
r2
z
✻
! J = J0 ẑ
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m
❂
❂
x
✲
y
p
r 2 R2
.
r5
1
Problem 6.3
134
CHAPTER 6. MAGNETIC FIELDS IN MATTER
Problem 6.4
dF = I {(dy ŷ)⇥B(0, y, 0) + (dz ẑ)⇥B(0, ✏, z) (dy ŷ)⇥B(0, y, ✏) (dz ẑ)⇥B(0, 0, z)}
n
o
=I
(dy ŷ)⇥ [B(0, y, ✏) B(0, y, 0)] +(dz ẑ)⇥ [B(0, ✏, z) B(0, 0, z)]
|
{z
}
|
{z
}
@B
⇡ẑ✏ @B
⇡
✏
r̂
@z
@y
✻
B
⇢

✕
✸
R
@B
@B
R
θ✲ R
@B
ŷ dy @B
I ŷ⇥
) I✏2 ẑ⇥
. Note that
dz @B
⇡✏
@z 0,y,0 ⇡ ✏ @z 0,0,0 and
@y
@y ❥ @z
0,0,z
( x̂
0
F=m
@B
@y
0,0,0
.
r
x̂ φ ŷ
m 0✻ 1
⇢
ẑ )
@Bx
@By
@Bz
@Bx
0
= m ŷ
x̂
x̂
+ ẑ
1
@y
@y
@z
@z
@B
@Bx @By @Bz
@B
@B
y
z
x
@y
@y
@y
@z
@z
@z

✓
◆
@Bx
@Bx
@Bx
@By
@Bz
@Bx
= m x̂
+ ŷ
+ ẑ
using r·B = 0 to write
+
=
.
@x
@y
@z
@y
@z
@x
⇣
⌘
@Bx
@Bx
x
But m·B = mBx (since m = mx̂, here), so r(m·B) = mr(Bx ) = m @B
x̂
+
ŷ
+
ẑ
.
@x
@y
@z
Therefore F = r(m·B). qed
ŷ
0
ẑ
1
Problem 6.5
Problem 6.5
(a) B = µ0 J0 xŷ (Prob. 5.15).
m·B = 0, so Eq. 6.3 says F = 0.
z
✻
! J = J0 ẑ
(b) m·B = m0 µ0 J0 x, so F = m0 µ0 J0 x̂.
(c) Use product rule #4: r(p · E)
= p ⇥ (r ⇥ E) + E ⇥ (r ⇥ p) + (p · r)E + (E · r)p.
But p does not depend on (x, y, z), so the second
and fourth terms vanish, and r ⇥ E = 0, so the
first term is zero. Hence r(p · E) = (p · r)E. qed
❂
x
This argument does not apply to the magnetic analog,
since r ⇥ B 6= 0. In fact, r(m · B) = (m · r)B + µ0 (m ⇥ J).
@
@
(m·r)Ba = m0 @x
(B) = m0 µ0 J0 ŷ, (m·r)Bb = m0 @y
(µ0 J0 xŷ) = 0.
m
❂
✲
y
Problem 6.6
1
Aluminum, copper, copper chloride, and sodium all have an odd number of electrons, so we expect them to
be paramagnetic. The rest (having an even number) should be diamagnetic.
Problem
Problem
6.7 6.7
✻
Jb = r⇥M = 0; Kb = M⇥n̂ = M ˆ.
M ✲ n̂
ˆ
The field is that of a surface current Kb = M ,
but that’s just a solenoid, so the field
c
⃝2005
outside is zero,
and
inside B = µ0 Klaws
µ0they
M . currently
Moreover,
it points upward (in the drawing), so B = µ0 M.
b =as
exist. No portion of this material may be
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protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
135
CHAPTER
Problem
6.9 6. MAGNETIC FIELDS IN MATTER
Problem 6.8
Kb = M×n̂ = M φ̂.
▼
✿
1✲@
1 B2
(Essentially a long solenoid)
2
2
2
r⇥M = Jb =M (s ksK
)ẑ = (3ks
③ )ẑ = 3ksẑ, Kb = M⇥n̂ = ks ( ˆ⇥ŝ) = kR ẑ.
✎✎✎
s @s ✎ ✎ ✎
s
✌
✎
1
So the bound current flows up the cylinder, and returns down the surface. [Incidentally,
the total current should
R
RR
R
3
be zero. . . is it? Yes, for Jb da = 0 (3ks)(2⇡s ds) = 2⇡kR , while Kb dl = ( kR2 )(2⇡R) = 2⇡kR3 .] Since
these
currents
Problem
6.9have cylindrical symmetry, we can get the field by Ampère’s law:
Z s
B · 2⇡s = µ0 Ienc = µ0
Jb da = 2⇡kµ0 s3 ) B = µ0 ks2 ˆ = µ0 M.
0
Outside the cylinder Ienc = 0, so B = 0.
Problem 6.9
▼
✲
M
✎✎✎
✎✎✎
✿
③B
K
✌
Kb = M×n̂ = M φ̂.
(Essentially a long solenoid)
✎
✻✻
✶
B
'
K
!
❄❄
✛
(Essentially a physical dipole)
3
c
❖ ⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
M
✲ ✶ B
(Intermediate case)
"
K
[The
external
fields are the same as in the electrical
✎
✎
case;
the
internal
fields (inside the bar) are completely
✛
✎
di↵erent—in fact, opposite in direction.]
Problem 6.10
Kb = M , so the field inside
a complete ring would be µ0 M . The field of a square loop, at the center, is
p
given by Prob. 5.8: Bsq = 2 µ0 I/⇡R. Here I = M w, and R = a/2, so
Bsq =
p
p
2 µ0 M w
2 2 µ0 M w
=
;
⇡(a/2)
⇡a
net field in gap : B = µ0 M 1
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Problem
6.12
reproduced,
in any form or by any means, without permission in writing from the publisher.
!
p
2 2w
.
⇡a
z
✻
.
R✲
Jb
■
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
Kb
❃
}
l
136
1
CHAPTER 6. MAGNETIC
FIELDS IN MATTER
Problem
Problem6.12
6.11
As in Sec. 4.2.3, we want the average of B = Bout + Bin , where Bout is due to molecules outside a small
sphere around point P , and Bin is due to molecules inside the sphere. The average of Bout is same as field at
center (Prob. 5.59b), and for this it is OK to use Eq. 6.10, since the center is “far” from all the molecules in
question:
Z
µ0
M⇥ r̂
Aout =
r 2 d⌧
4⇡
.
outside
µ0
4⇡
2m
4
3
The average of Bin is
R3 —Eq. 5.93—where m = 3 ⇡R M. Thus the average Bin is 2µ0 M/3. But what is
left out of the integral Aout is the contribution of a uniformly magnetized sphere, to wit: 2µ0 M/3 (Eq. 6.16),
and this is precisely what Bin puts back in. So we’ll get the correct macroscopic field using Eq. 6.10. qed
Problem 6.12
z
✻
R✲
(a) M = ksẑ; Jb = r⇥M = k ˆ; Kb = M⇥n̂ = kR ˆ.
Jb
B is in the z direction (this is essentially a superposition of solenoids). So 1
■
l
loop shown
⇥R
⇤
HB = 0 outside. Use the amperian
Kb
B · dl = Bl = µ0 Ienc = µ0 Jb da + Kb l = µ0 [ kl(R s) + kRl] = µ0 kls.
❃
Problem 6.13
) B = µ0 ksẑ inside.
H
(b) By symmetry, H points in the z direction. That same amperian loop gives H·dl = Hl = µ0 Ifenc = 0, since
there is no free current here. So H = 0 , and hence B = µ0 M. Outside M = 0, so B = 0; inside M = ksẑ,
so B = µ0 ksẑ.
1
Problem 6.13
2
2
(a) The field
µ0 M, with the sphere removed.
Problem
6.13 of a magnetized sphere is 3 µ0 M (Eq. 6.16), so B = B0
3
In the cavity, H =
(b)
1
µ0 B,
so H =
1
µ0
B0
2
3 µ0 M
= H0 + M
2
3M
}
1
) H = H0 + M.
3
The field inside a long solenoid is µ0 K. Here K = M , so the field of the bound current on
the inside surface of the cavity is µ0 M , pointing down. Therefore
Kb ▼
B = B0
H=
µ0 M;
1
(B0
µ0
µ0 M) =
1
B0
µ0
M ) H = H0 .
(c)
✸
✰Kb
This time the bound currents are small, and far away from the center, so B = B0 ,
while H =
1
µ0 B0
= H0 + M ) H = H0 + M.
[Comment: In the wafer, B is the field in the medium; in the needle, H is the H in the medium; in the
sphere (intermediate case) both B and H are modified.]
c
⃝2005
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may beexist. No portion of this material may be
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under
laws
asmaterial
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137
1
CHAPTER
MAGNETIC FIELDS IN MATTER
Problem6.6.14
✲
Problem
✲
✲
M:
❯
◆
6.14
✲
✲
✲
❄
+
✲
✶
✯
✻
M.
✛
B
✛
H
✛
✛
✛
✍
1
µ0 B
✛
✻
✍
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
❥
✕
; B is the same as the field of a short solenoid; H =
✕
✯
✶
✲
+
❥
◆
❄
❯
✛
❯
◆
❄
❥
✶
✛
✛
✍
✻
(The continuity of W follows from the gradient theorem: W (b) W (a) =
if the two points are infinitesimally separated, this last integral ! 0.)
⇢
l
(i) ) Al Rl = RBl+1
) Bl = R2l+1 Al ,
P
P
Bl
(ii) ) (l + 1) Rl+2 Pl (cos ✓) + lAl Rl 2 Pl (cos ✓) = M cos ✓.
Combining these:
◆
❄
✛
✛
Rb
a
rW · dl =
c
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X all copyright laws as they currently exist. No portion of this material may be
protected under
1)Rorl by1 Aany
(cos ✓)without
= M cos
✓, so A
= 0 (lfrom
6= 1),
3A1 =
reproduced, in (2l
any+
form
permission
in lwriting
the and
publisher.
l Plmeans,
Thus Win (r, ✓) =
❥
✯
✕
c
⃝2005
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
c
⃝2005
Pearson
Inc.,
Upper
River,exist.
NJ. All
reserved.
This material
is
protected
under Education,
laws
as they
currently
No rights
portion
of this material
may be
protected
under
all form
they without
currentlypermission
exist. No portion
of from
this material
may be
reproduced,
in any
or bylaws
any as
means,
in writing
the publisher.
reproduced, in any form or by any means, without permission in writing from the publisher.
Rb
a
M ) A1 =
M
M
M
1
r cos ✓ =
z, and hence Hin = rWin =
ẑ =
M, so
3
3
3
3
✓
◆
1
2
B = µ0 (H + M) = µ0
M + M = µ0 M. X
3
3
✕
✯
✶
✲
*
✛
*
✲
✛
Problem 6.15
“Potentials”:
⇢
P
Win (r, ✓) =
A rl P (cos ✓), (r < R);
P Bl l l
Wout (r, ✓) =
P (cos ✓), (r > R).
r l+1 l
Boundary
Conditions:
⇢
(i) Win (R, ✓) = Wout (R, ✓),
?
out
in
(ii) @W
+ @W
= M ẑ · r̂ = M cos ✓.
@r
@r R = M
R
✻
✍
✛
H · dl;
M
.
3
❯
138
Problem 6.16
H
H·dl = Ifenc = I, so H =
I
2⇡s
CHAPTER 6. MAGNETIC FIELDS IN MATTER
ˆ. B = µ0 (1 +
1 @
s @s
Jb = r⇥M =
✓
s
mI
2⇡s
◆
m )H
ẑ = 0.
I ˆ
mI ˆ
. M = mH =
.
2⇡s
2⇡s
(
mI
at s = a;
2⇡a ẑ,
Kb = M⇥n̂ =
mI
2⇡b ẑ, at s = b.
= µ0 (1 +
m)
Total enclosed current, for an amperian loop between the cylinders:
I
µ0 (1 + m )I ˆ
mI
I+
2⇡a = (1 + m )I, so
B · dl = µ0 Ienc = µ0 (1 + m )I ) B =
.X
2⇡a
2⇡s
Problem 6.17
From Eq. 6.20:
H
H·dl = H(2⇡s) = Ifenc
H=
Jb =
m Jf
⇢
Is
2⇡a2 ,
I
2⇡s ,
(Eq. 6.33), and Jf =
Kb = M⇥n̂ =
m H⇥n̂
I
⇡a2 ,
) Kb =
Ib = Jb (⇡a2 ) + Kb (2⇡a) =
mI
(s < a)
(s > a)
(
I(s2 /a2 ), (s < a);
=
I
(s > a).
(
,
so Jb =
so B = µH =
mI
⇡a2
µ0 (1+ m )Is
,
2⇡a2
µ0 I
2⇡s ,
(s < a);
(s > a).
(same direction as I).
mI
(opposite direction to I).
2⇡a
m I = 0 (as it should be, of course).
Problem 6.18
By the method of Prob. 6.15:
For large r, we want B(r, ✓) ! B0 = B0 ẑ, so H =
1
µ0 B0 r cos ✓.
1
µ0 B
!
1
µ0 B0
ẑ, and hence W !
1
µ0 B0 z
=
“Potentials”:
P
⇢
Win (r, ✓) =
Al rl Pl (cos ✓),
(r < R);
P Bl
1
Wout (r, ✓) = µ0 B0 r cos ✓ +
P
(cos
✓),
(r > R).
l
l+1
r
Boundary
Conditions:
⇢
(i) Win (R, ✓) = Wout (R, ✓),
out
in
(ii) µ0 @W
+ µ @W
@r
@r R = 0.
R
(The latter follows from Eq. 6.26.)

X
X
1
Bl
(ii) ) µ0
B0 cos ✓ +
(l + 1) l+2 Pl (cos ✓) + µ
lAl Rl
µ0
R
1
Pl (cos ✓) = 0.
For l 6= 1, (i) ) Bl = R2l+1 Al , so [µ0 (l + 1) + µl]Al Rl 1 = 0, and hence Al = 0.
For l = 1, (i) ) A1 R = µ10 B0 R + B1 /R2 , and (ii) ) B0 + 2µ0 B1 /R3 + µA1 = 0, so A1 =
Win (r, ✓) =
3B0 /(2µ0 + µ).
3B0
r cos ✓ =
(2µ0 + µ)
3B0 z
3B0
3B0
. Hin = rWin =
ẑ =
.
(2µ0 + µ)
(2µ0 + µ)
(2µ0 + µ)
✓
◆
3µB0
1+ m
B = µH =
=
B0 .
(2µ0 + µ)
1 + m /3
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139
CHAPTER 6. MAGNETIC FIELDS IN MATTER
By the method of Prob. 4.23:
Step 1 : B0 magnetizes the sphere: M0 =
the sphere given by Eq. 6.16:
m H0
=
m
µ0 (1+
m)
2
2
2
m
µ0 M0 =
B0 = B0
3
31+ m
3
B1 =
B0 . This magnetization sets up a field within
(where  ⌘
m
1+
m
).
Step 2 : B1 magnetizes the sphere an additional amount M1 = µ0 B1 . This sets up an additional field in
the sphere:
✓ ◆2
2
2
2
B2 = µ0 M1 = B1 =
B0 , etc.
3
3
3
The total field is:
⇥
⇤
B = B0 + B1 + B2 + · · · = B0 + (2/3)B0 + (2/3)2 B0 + · · · = 1 + (2/3) + (2/3)2 + · · · B0 =
1
1
=
2/3
3
3
2 m /(1 +
Problem 6.19
2 2
m = e4mre B; M =
m
V
3+3
=
3+3 m
m)
e2 r 2
4me V B,
e2 r 2
m =
4me V
=
m
2
m
3(1 +
=
3+
m)
, so B =
m
✓
1+
1+
m
m /3
where V is the volume per electron. M =
◆
B0
.
(1 2/3)
B0 .
mH
(Eq. 6.29)
= µ0 (1+ m ) B (Eq. 6.30). So
µ0 . [Note: m ⌧ 1, so I won’t worry about the (1 + m )
term; for the same reason we need not distinguish B from Belse , as we
⇣ did
⌘in deriving the Clausius-Mossotti
µ0
3e2
10
equation in Prob. 4.41.] Let’s say V = 43 ⇡r3 . Then m = 4⇡
m for r.
4me r . I’ll use 1 Å= 10
⇣
⌘
19 2
3(1.6⇥10
)
Then m = (10 7 ) 4(9.1⇥10
= 2 ⇥ 10 5 , which is not bad—Table 6.1 says m = 1 ⇥ 10 5 .
31 )(10 10 )
However, I used only one electron per atom (copper has 29) and a very crude value for r. Since the orbital
radius is smaller for the inner electrons, they count for less ( m ⇠ r2 ). I have also neglected competing
paramagnetic e↵ects. But never mind. . . this is in the right ball park.
Problem 6.20
Place the object in a region of zero magnetic field, and heat it above the Curie point—or simply drop it on
a hard surface. If it’s delicate (a watch, say), place it between the poles of an electromagnet, and magnetize it
back and forth many times; each time you reverse the direction, reduce the field slightly.
Problem 6.21
m
(a) The magnetic force on the dipole is given by Eq. 6.3; to move the dipole in from infinity we must exert an
opposite force, so the work done is
Z r
Z r
U=
F · dl =
r(m · B) · dl = m · B(r) + m · B(1)
1
1
(I used the gradient theorem, Eq. 1.55). As long as the magnetic field goes to zero at infinity, then, U = m·B.
If the magnetic field does not go to zero at infinity, one must stipulate that the dipole starts out oriented
perpendicular to the field.
(b) Identical to Prob. 4.8, but starting with Eq. 5.89 instead of 3.104.
(c) U =
U=
µ0 1
4⇡ r 3 [3 cos ✓1 cos ✓2
µ0 m1 m2
(sin ✓1 sin ✓2
4⇡ r3
cos(✓2
✓1 )]m1 m2 . Or, using cos(✓2
✓1 ) = cos ✓1 cos ✓2 + sin ✓1 sin ✓2 ,
2 cos ✓1 cos ✓2 ) .
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140
CHAPTER 6. MAGNETIC FIELDS IN MATTER
Stable position occurs at minimum energy:
(
=
=
@U
@✓1
@U
@✓2
µ0 m1 m2
4⇡r 3 (cos ✓1 sin ✓2
µ0 m1 m2
4⇡r 3 (sin ✓1 cos ✓2
@U
@✓1
=
@U
@✓2
=0
+ 2 sin ✓1 cos ✓2 ) = 0 ) 2 sin ✓1 cos ✓2 =
+ 2 cos ✓1 sin ✓2 ) = 0 ) 2 sin ✓1 cos ✓2 =
8
<Either sin ✓1 = sin ✓2 = 0 :
Thus sin ✓1 cos ✓2 = sin ✓2 cos ✓1 = 0.
: or cos ✓1 = cos ✓2 = 0 :
1
! ! or
" "
or
3
cos ✓1 sin ✓2 ;
4 cos ✓1 sin ✓2 .
2
!
" #
4
Which of these is the stable minimum? Certainly not 2 or 3 —for these m2 is not parallel to B1 , whereas we
know m2 will line up along B1 . It remains to compare 1 (with ✓1 = ✓2 = 0) and 4 (with ✓1 = ⇡/2, ✓2 = ⇡/2):
m1 m2
m1 m2
U1 = µ04⇡r
( 2); U2 = µ04⇡r
( 1). U1 is the lower energy, hence the more stable configuration.
3
3
Conclusion: They line up parallel, along the line joining them:
(d) They’d line up the same way:
Problem 6.22
F=I
I
(because
dl ⇥ B = I
H
✓I
! !
! ! ! ! ! !
◆
dl ⇥ B0 + I
I
dl ⇥ [(r · r0 )B0 ]
I
✓I
◆
dl ⇥ [(r0 · r0 )B0 ] = I
I
dl ⇥ [(r · r0 )B0 ]
dl = 0). Now
Fi = I
(dl ⇥ B0 )i =
X
✏ijk
j,k,l
I
X
✏ijk dlj (B0 )k ,
j,k
rl dlj [(r0 )l (B0 )k ]
and (r · r0 ) =
(
Lemma 1 :
I
X
rl dlj =
rl (r0 )l , so
l
X
m
)
✏ljm am (proof below).
8
9
<
=
X
=I
✏ijk ✏ljm am (r0 )l (B0 )k
Lemma 2 :
✏ijk ✏ljm = il km
(proof
below).
im kl
:
;
j
j,k,l,m
X
X
=I
( il km
[ak (r0 )i (B0 )k ai (r0 )k (B0 )k ]
im kl ) am (r0 )l (B0 )k = I
X
k,l,m
= I [(r0 )i (a · B0 )
k
ai (r0 · B0 )] .
But r0 · B0 = 0 (Eq. 5.50), and m = Ia (Eq. 5.86), so F = r0 (m · B0 ) (the subscript just reminds us to take
the derivatives at the point where m is located). qed
Proof of Lemma
1:
H
P H
P
Eq. 1.108 says (c · r) dl = a ⇥ c = c ⇥ a. The jth component is p cp rp dlj =
✏jpm cp am . Pick
p,mP
H
P
cp = pl (i.e. 1 for the lth component, zero for the others). Then rl dlj =
m ✏jlm am =
m ✏ljm am . qed
Proof of Lemma 2:
✏ijk ✏ljm = 0 unless ijk and ljm are both permutations of 123. In particular, i must either be l or m, and k
must be the other, so
X
✏ijk ✏ljm = A il km + B im kl .
j
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141
CHAPTER 6. MAGNETIC FIELDS IN MATTER
To determine the constant A, pick i = l = 1, k = m = 3; the only contribution comes from j = 2:
✏123 ✏123 = 1 = A
11 33
+B
13 31
= A ) A = 1.
13 31
+B
11 33
=B)B=
To determine B, pick i = m = 1, k = l = 3:
✏123 ✏321 =
So
X
1=A
✏ijk ✏ljm =
qed
im kl .
il km
1.
j
Problem 6.23
(a) B1 =
µ0 2m
4⇡ z 3 ẑ
2
3µ0 m
2⇡z 4 ẑ. This
( md gẑ):
(Eq. 5.88, with ✓ = 0). So m2 ·B1 =
µ0 m2
2⇡ z 3 .
F = r(m·B) (Eq. 6.3) ) F =
@
@z
h
µ0 m2
2⇡ z 3
i
ẑ =
is the magnetic force upward (on the upper magnet); it balances the gravitational force downward
3µ0 m2
2⇡z 4
md g = 0 ) z =

1/4
3µ0 m2
2⇡md g
.
(b) The middle magnet is repelled upward by lower magnet and downward by upper magnet:
3µ0 m2
2⇡x4
3µ0 m2
2⇡y 4
md g = 0.
The top magnet is repelled upward by middle magnet, and attracted downward by lower magnet:
Subtracting:
3µ0 m2
2⇡
h
1
x4
Let ↵ ⌘ x/y; then 2 =
1
↵4
+
1
y4
1
y4
+
1
(↵+1)4 .
3µ0 m2
2⇡y 4
i
1
(x+y)4
3µ0 m2
2⇡(x + y)4
md g = 0.
md g +md g = 0, or
1
x4
2
y4
1
+ (x+y)
4 = 0, so: 2 =
1
(x/y)4
1
+ (x/y+1)
4.
Mathematica gives the numerical solution ↵ = x/y = 0.850115 . . .
Problem 6.24
(a) Forces on the upper charge:
1 q2
Fq =
ẑ,
4⇡✏0 z 2
At equilibrium,
Fm
✓
2µ0 m
= r(m · B) = r m
4⇡z 3
1 q2
3µ0 m2
=
2
4⇡✏0 z
2⇡z 4
) z2 =
6µ0 ✏0 m2
q2
)
◆
µ0 m2
=
2⇡
z=
✓
3
z4
◆
ẑ.
p m
6 ,
qc
p
where 1/ ✏0 µ0 = c, the speed of light.
(b) For electrons, q = 1.6 ⇥ 10 19 C (actually, it’s the magnitude of the charge we want in the expression
above), and m = 9.22 ⇥ 10 24 A m2 (the Bohr magneton—see Problem 5.58), so
z=
p
6
9.22 ⇥ 10 24
= 4.72 ⇥ 10
(1.6 ⇥ 10 19 )(3 ⇥ 108 )
13
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m.
142
CHAPTER 6. MAGNETIC FIELDS IN MATTER
(For comparison, the Bohr radius is 0.5 ⇥ 10 10 m, so the equilibrium separation is about 1% of the size of a
hydrogen atom.)
(c) Good question! Certainly the answer is no. Presumably this is an unstable equilibrium, so unless you
could find a way to maintain the orientation of the dipoles, and keep them on the z axis, the structure would
fall apart.
Problem 6.25
(a) The electric field inside a uniformly polarized sphere, E = 3✏10 P (Eq. 4.14) translates to H =
1
3 M. But B = µ0 (H+M). So the magnetic field inside a uniformly magnetized sphere is B = µ0 (
2
µ0 M (same as Eq. 6.16).
3
1
3µ0 (µ0 M)
1
3 M+M)
=
=
(b) The electric field inside a sphere of linear dielectric in an otherwise uniform electric field is E = 1+ 1e /3 E0
(Eq. 4.49). Now e translates to m , for then Eq. 4.30 (P = ✏0 e E) goes to µ0 M = µ0 m H, or M = m H
(Eq. 6.29). So Eq. 4.49 ) H = 1+ 1m /3 H0 . But B = µ0 (1 + m )H, and B0 = µ0 H0 (Eqs. 6.31 and 6.32),
so the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field is
✓
◆
B
1
B0
1+ m
=
, or B =
B0 (as in Prob. 6.18).
µ0 (1 + m )
(1 + m /3) µ0
1 + m /3
p
1
(c) The average electric field over a sphere, due to charges within, is Eave = 4⇡✏
3 . Let’s pretend the charges
0 R
are all due to the frozen-in polarization of some medium (whatever ⇢ might
be,
we
can
solve r·PR= ⇢ to find
R
1
1
the appropriate P). In this case there are no free charges, and p = P d⌧ , so Eave = 4⇡✏
P d⌧ , which
3
0 R
translates to
Z
1 1
1
Have =
µ0 M d⌧ =
m.
3
4⇡µ0 R
4⇡R3
µ0 2m
, in agreement
4⇡ R3
with Eq. 5.93. (We must assume for this argument that all the currents are bound, but again it doesn’t really
matter, since we can model any current configuration by an appropriate frozen-in magnetization. See G. H.
Goedecke, Am. J. Phys. 66, 1010 (1998).)
But B = µ0 (H + M), so Bave =
µ0 m
4⇡ R3
+ µ0 Mave , and Mave =
m
4
3
3 ⇡R
, so Bave =
Problem 6.26
n
o
R r̂
1
0
Eq. 2.15 : E = ⇢ 4⇡✏
d⌧
(for uniform charge density);
2
n 0 VR r
o
r̂ 2 d⌧ 0
1
Eq. 4.9 : V = P · 4⇡✏
(for uniform polarization);
0
nV r R
o
r̂
1
0
Eq. 6.11 : A = µ0 ✏0 M ⇥ 4⇡✏0 V r 2 d⌧ (for uniform magnetization).
8
⇣
⌘
< Ein = ⇢ 1 r
(Prob. 2.12),
⇣ 3✏0 3 ⌘
For a uniformly charged sphere (radius R):
: Eout = ⇢ 1 R2 r̂ (Ex. 2.3).
3✏0 r
(
Vin = 3✏10 (P · r),
3
So the scalar potential of a uniformly polarized sphere is:
Vout = 3✏10 Rr2 (P · r̂),
⇢
Ain = µ30 (M ⇥ r),
3
and the vector potential of a uniformly magnetized sphere is:
Aout = µ30 Rr2 (M ⇥ r̂),
(confirming the results of Ex. 4.2 and of Exs. 6.1 and 5.11).
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143
CHAPTER 6. MAGNETIC FIELDS IN MATTER
Problem 6.27
At the interface, the perpendicular component of B is continuous (Eq. 6.26), and the parallel component of
k
k
k
k
H is continuous (Eq. 6.25 with Kf = 0). So B1? = B2? , H1 = H2 . But B = µH (Eq. 6.31), so µ11 B1 = µ12 B2 .
k
k
Now tan ✓1 = B1 /B1? , and tan ✓2 = B2 /B2? , so
k
k
tan ✓2
B2 B1?
B
µ2
= ?
= 2k =
k
tan ✓1
µ1
B2 B1
B1
(the same form, though for di↵erent reasons, as Eq. 4.68).
Problem 6.28
In view of Eq. 6.33, there is a bound dipole at the center: mb = m m. So the net dipole moment at the
center is mcenter = m + mb = (1 + m )m = µµ0 m. This produces a field given by Eq. 5.89:
Bcenter =
dipole
µ 1
[3(m·r̂)r̂
4⇡ r3
m] .
This accounts for the first term in the field. The remainder must be due to the bound surface current (Kb ) at
r = R (since there can be no volume bound current, according to Eq. 6.33). Let us make an educated guess
(based either on the answer provided or on the analogous electrical Prob. 4.37) that the field due to the surface
bound current is (for interior points) of the form B surface = Am (i.e. a constant, proportional to m). In that
current
case the magnetization will be:
M=
mH
=
m
µ
B=
1
[3(m·r̂)r̂
4⇡ r3
m
m] +
m
µ
Am.
This will produce bound currents Jb = r⇥M = 0, as it should, for 0 < r < R (no need to calculate this
curl—the second term is constant, and the first is essentially the field of a dipole, which we know is curl-less,
except at r = 0), and
✓
◆
1
A
m
mA
Kb = M(R)⇥r̂ =
( m⇥r̂) +
(m⇥r̂) = m m
+
sin ✓ ˆ.
4⇡R3
µ
4⇡R3
µ
ˆ
But this
⇣ is exactly
⌘ the surface current produced by a spinning sphere: K = v = !R sin ✓ , with ( !R) \$
A
1
mm µ
4⇡R3 . So the field it produces (for points inside) is (Eq. 5.70):
◆
1
.
4⇡R3
current
⇣
⌘
⇣
1
Everything is consistent, therefore, provided A = 23 µ0 m A
,
or
A
1
3
µ
⇣ ⌘
⇣
⌘
⇣
⌘4⇡R
(µ
µ
)
(µ
µ)
µ
µ
2µ
0
1, so A 1 23 + 23 µ0 = 23 4⇡R30 , or A 1 + µ0 = 2 4⇡R
3 ; A =
m = µ0
B surface =
B=
µ
4⇡
⇢
2
2
µ0 ( ! R) = µ0
3
3
1
[3(m·r̂)r̂
r3
m] +
mm
✓
A
µ
2(µ0 µ)m
R3 (2µ0 + µ)
2µ0
3µ
m
⌘
=
µ 2(µ0 µ)
4⇡ R3 (2µ0 +µ) ,
2 µ0 m
3 4⇡R3 .
But
and hence
. qed
The exterior field is that of the central dipole plus that of the surface current, which, according to Prob. 5.37,
is also a perfect dipole field, of dipole moment
✓
◆
4
4
3
2⇡R3 µ 2(µ0 µ)m
µ(µ0 µ)m
m surface = ⇡R3 ( ! R) = ⇡R3
B surface =
=
.
3
3
3
2µ0 current
µ0 4⇡ R (2µ0 + µ)
µ0 (2µ0 + µ)
current
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144
CHAPTER 6. MAGNETIC FIELDS IN MATTER
So the total dipole moment is:
mtot =
µ
µ
(µ0 µ)
3µm
m+
m
=
,
µ0
µ0 (2µ0 + µ)
(2µ0 + µ)
and hence the field (for r > R) is
µ0
B=
4⇡
✓
3µ
2µ0 + µ
◆
1
[3(m·r̂)r̂
r3
m] .
Problem 6.29
The problem is that the field inside a cavity is not the same as the field in the material itself.
(a)
8 Ampére type. The 2field deep1inside the magnet is that of a long solenoid, B0 ⇡ µ0 M. From Prob. 6.13:
< Sphere : B = B0 3 µ0 M = 3 µ0 M;
Needle : B = B0 µ0 M = 0;
:
Wafer : B = µ0 M.
(b) Gilbert type. This is analogous to the electric case. The field at the center is approximately that midway
between
B0 ⇡ 0. From Prob. 4.16 (with E ! B, 1/✏0 ! µ0 , P ! M):
8 two distant pointµcharges,
< Sphere : B = B0 + 30 M = 13 µ0 M;
Needle : B = B0 = 0;
:
Wafer : B = B0 + µ0 M = µ0 M.
In the cavities, then, the fields are the same for the two models, and this will be no test at all. Yes. Fund it
with \$1 M from the Office of Alternative Medicine.
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145
CHAPTER 7. ELETRODYNAMICS
Chapter 7
Electrodynamics
Problem 7.1
(a) Let Q be the charge on the inner shell. Then E =
Ra
Ra
Q
1
1
1
E·dr = 4⇡✏
Q b r12 dr = 4⇡✏
a
b .
b
0
0
I=
(b) R =
Va
Vb
I
1
=
4⇡
Z
Z
J·da =
✓
1
a
1
b
◆
E·da =
1 Q
4⇡✏0 r 2 r̂
in the space between them, and (Va
Q
4⇡✏0 (Va Vb )
(Va
=
= 4⇡
✏0
✏0 (1/a 1/b)
(1/a
Vb ) =
Vb )
.
1/b)
.
(c) For large b (b
a), the second term is negligible, and R = 1/4⇡ a. Essentially all of the resistance is in
the region right around the inner sphere. Successive shells, as you go out, contribute less and less, because the
cross-sectional area (4⇡r2 ) gets larger and larger. For the two submerged spheres, R = 4⇡2 a = 2⇡1 a (one R as
the current leaves the first, one R as it converges on the second). Therefore I = V /R = 2⇡ aV.
Problem 7.2
(a) V = Q/C = IR. Because positive I means the charge on the capacitor is decreasing,
dQ
1
= I=
Q, so Q(t) = Q0 e t/RC . But Q0 = Q(0) = CV0 , so Q(t) = CV0 e t/RC .
dt
RC
dQ
1
V0 t/RC
Hence I(t) =
= CV0
e t/RC =
e
.
dt
RC
R
Z 1
Z 1
Z
V 2 1 2t/RC
(b) W = 12 CV02 . The energy delivered to the resistor is
P dt =
I 2 R dt = 0
e
dt =
R 0
0
0
✓
◆
1
V02
RC 2t/RC
1
e
= CV02 . X
R
2
2
0
dQ
1
dQ
(c) V0 = Q/C + IR. This time positive I means Q is increasing:
=I=
(CV0 Q) )
=
dt
RC
Q CV0
1
1
dt ) ln(Q CV0 ) =
t + constant ) Q(t) = CV0 + ke t/RC . But Q(0) = 0 ) k = CV0 , so
RC
RC
✓
◆
⇣
⌘
dQ
1
V0 t/RC
t/RC
t/RC
Q(t) = CV0 1 e
. I(t) =
= CV0
e
=
e
.
dt
RC
R
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146
(d) Energy from battery:
Z
1
CHAPTER 7. ELETRODYNAMICS
V0 I dt =
0
V02
R
Z
1
e
t/RC
dt =
0
V02 ⇣
RCe
R
t/RC
Since I(t) is the same as in (a), the energy delivered to the resistor is again
⌘
1
0
=
V02
RC = CV02 .
R
1
2
2 CV0 .
The final energy in
the capacitor is also 12 CV02 , so half the energy from the battery goes to the capacitor, and the other half
to the resistor.
Problem 7.3
R
(a) I = J · da, where the integral
R is taken over a surfaceR enclosing the positively charged conductor. But
J = E, and Gauss’s law says E · da = ✏10 Q, so I =
E · da = ✏0 Q. But Q = CV , and V = IR, so
✏0
I = ✏0 CIR, or R =
. qed
C
(b) Q = CV = CIR )
dQ
dt
=
I=
time constant is ⌧ = RC = ✏0 / .
1
RC Q
) Q(t) = Q0 e
t/RC
, or, since V = Q/C, V (t) = V0 e
t/RC
. The
Problem 7.4
I = J(s) 2⇡sL ) J(s) = I/2⇡sL. E = J/ = I/2⇡s L = I/2⇡kL.
Z a
I
b a
V =
E · dl =
(a b). So R =
.
2⇡kL
2⇡kL
b
Problem 7.5

E
E 2R
dP
1
2
2
I=
; P =I R=
;
=E
r+R
(r + R)2 dR
(r + R)2
2R
= 0 ) r + R = 2R ) R = r.
(r + R)3
Problem
H 7.6
H
E = E·dl = zero for all electrostatic fields. It looks as though E = E · dl = ( /✏0 )h, as would indeed
be the case if the field were really just /✏0 inside and zero outside. But in fact there is always a “fringing
field” at the edges (Fig. 4.31), and this is evidently just right to kill o↵ the contribution from the left end of
the loop. The current is zero.
Problem 7.7
Blv
. (Never mind the minus sign—it just tells you the
R
direction of flow: (v⇥B) is upward, in the bar, so downward through the resistor.)
(a) E =
d
dt
=
(b) F = IlB =
Bl dx
dt =
Blv; E = IR ) I =
B 2 l2 v
, to the left.
R
B 2 l2
dv
B 2 l2
dv
B 2 l2
=
v)
=
v ) v = v0 e mR t .
dt
R
dt
Rm
(d) The energy goes into heat in the resistor. The power delivered to resistor is I 2 R, so
(c) F = ma = m
dW
B 2 l2 v 2
B 2 l2 2
= I 2R =
R
=
v e
dt
R2
R 0
The total energy delivered to the resistor is W =
2↵t
↵mv02
, where ↵ ⌘
Z
0
1
e
2↵t
B 2 l2
;
mR
dt = ↵mv02
dW
= ↵mv02 e
dt
e
2↵t 1
2↵
0
2↵t
= ↵mv02
.
1
1
= mv02 . X
2↵
2
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147
CHAPTER 7. ELETRODYNAMICS
1
Problem 7.8
µ0 I ˆ
(a) The field of long wire is B =
, so
2⇡s
=
Z
µ0 I
B·da =
2⇡
s+a
Z
1
µ0 Ia
(a ds) =
ln
s
2⇡
s
✓
s+a
s
◆
.
✓
◆
✓
◆
d
µ0 Ia d
s+a
ds
µ0 Ia
1 ds 1 ds
µ0 Ia2 v
(b) E =
=
ln
, and
= v, so
=
.
dt
2⇡ dt
s
dt
2⇡
s + a dt
s dt
2⇡s(s + a)
The field points out of the page, so the force on a charge in the nearby side of the square is to the right. In
the far side it’s also to the right, but here the field is weaker, so the current flows counterclockwise.
Contents
(c) This time the flux is constant, so E = 0.
Problem 7.9
R
Since r·B = 0, Theorem 2(c) (Sect. 1.6.2) guarantees that B·da is the same for all surfaces with a given
boundary line.
Problem 7.10Problem 7.10
✲
Φ = B · a = Ba2 cos θ
✲
= B · a = Ba2 cos ✓
B
(view from above)
θ
✲
Here ✓ = !t, so
E = ddt = Ba2 ( sin !t)!;
❘a
E = B!a2 sin !t.
Problem 7.11
E = Blv = IR ) I =
downward:
mg
Bl
Rv
) upward magnetic force = IlB =
B 2 l2
dv dv
v=m ;
=g
R
dt dt
dv
= dt )
g ↵v
1
ln(g
↵
↵v, where ↵ ⌘
↵v) = t + const. ) g
↵v = g(1
e
↵t
);
B 2 l2
R v.
B 2 l2
. g
mR
↵v = Ae
This opposes the gravitational force
↵vt = 0 ) vt =
↵t
g
mgR
=
.
↵
B 2 l2
; at t = 0, v = 0, so A = g.
g
(1 e ↵t ) = vt (1 e ↵t ).
↵
) e ↵t = 1 0.9 = 0.1; ln(0.1) =
v=
At 90% of terminal velocity, v/vt = 0.9 = 1 e ↵t
↵t; ln 10 = ↵t;
v
t
t = ↵1 ln 10, or t90% =
ln 10.
g
Now the numbers: m = 4⌘Al, where ⌘ is the mass density of aluminum, A is the cross-sectional area, and
l is the length of a side. R = 4l/A , where is the conductivity of aluminum. So
8
9
⇢ = 2.8 ⇥ 10 8 ⌦ m >
>
>
>
<
=
4⌘Alg4l
16⌘g
16g⌘⇢
g = 9.8 m/s2
vt =
=
=
,
and
.
3
3
2
2
2
2
⌘ = 2.7 ⇥ 10 kg/m >
>
A B l
B
B
>
>
:
;
B = 1T
So vt =
(16)(9.8)(2.7⇥103 )(2.8⇥10
1
8
)
= 1.2 cm/s; t90% =
1.2⇥10
9.8
2
ln(10) = 2.8 ms.
If the loop were cut, it would fall freely, with acceleration g.
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by any Education,
means, without
in writing
fromAll
the
publisher.
cform orPearson
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River, NJ.
rights
reserved. This material is
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148
CHAPTER 7. ELETRODYNAMICS
Problem 7.12
=⇡
⇣ a ⌘2
2
Problem 7.14
B=
2
⇡a2
B0 cos(!t); E =
4
d
⇡a2
E
⇡a2 !
=
B0 ! sin(!t). I(t) =
=
B0 sin(!t).
dt
4
R
4R
Problem 7.13
=
Z
B dx dy = kt
2
Z
a
dx
0
Z
a
y 3 dy =
0
1 2 5
kt a . E =
4
d
=
dt
1
5
2 kta .
Problem 7.14
✿
Iind
pipe ✲
falling
magnet
❑
B✕
✻
✿
✿
I✿
✲
✗
ring ✲
✻
B❖
②
Iind
Suppose the current (I) in the magnet flows counterclockwise (viewed from
above), as shown, so its field, near the ends, points upward. A ring of
pipe below the magnet experiences an increasing upward flux, as the magnet
approaches, and hence (by Lenz’s law) a current (Iind ) will be induced in it
such as to produce a downward flux. Thus Iind must flow clockwise, which is
opposite to the current in the magnet. Since opposite currents repel, the force
on the magnet is upward. Meanwhile, a ring above the magnet experiences
a decreasing (upward) flux, so its induced current is parallel to I, and it
attracts the magnet upward. And the flux through rings next to the magnet
is constant, so no current is induced in them. Conclusion: the delay is due
to forces exerted on the magnet by induced eddy currents in the pipe.
Problem 7.15
In the quasistatic approximation, B =
⇢
µ0 nI ẑ, (s < a);
0,
(s > a).
Inside: for an “amperian loop” of radius s < a,
= B⇡s = µ0 nI⇡s ;
2
2
I
E · dl = E 2⇡s =
d
=
dt
µ0 n⇡s2
dI
;
dt
E=
µ0 ns dI ˆ
.
2 dt
Outside: for an “amperian loop” of radius s > a:
= B⇡a2 = µ0 nI⇡a2 ; E 2⇡s =
µ0 n⇡a2
dI
;
dt
E=
µ0 na2 dI ˆ
.
2s dt
3
Problem 7.16
(a) The magnetic field (in the quasistatic
is “circumferential”. This is analogous to the current
Problemapproximation)
7.16
in a solenoid, and hence the field is longitudinal.
✛
✲
l
❄
(b) Use the “amperian loop” shown.
Outside,
B = 0, so here E = 0R (like B outside
H
R a aµ0solenoid).
I
d
d
So E·dl = El = ddt = dt
B·da = dt
l ds0
s 2⇡s0
µ0 dI
a
dI
) E = 2⇡ dt ln s . But dt = I0 ! sin !t,
⇣a⌘
µ0 I0 !
so E =
sin(!t) ln
ẑ.
2⇡
s
I
✲
✻
a
s
✻
✻
✲
z
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149
CHAPTER 7. ELETRODYNAMICS
Problem 7.17
(a) The field inside the solenoid is B = µ0 nI. So
= ⇡a2 µ0 nI ) E =
⇡a2 µ0 n(dI/dt).
⇡a2 µ0 nk
.
R
B is to the right and increasing, so the field of the loop is to the left, so the current is counterclockwise, or
to the right, through the resistor.
In magnitude, then, E = ⇡a2 µ0 nk. Now E = Ir R, so Iresistor =
(b)
= 2⇡a2 µ0 nI; I =
Problem 7.18
=
Z
dQ
E
=
=
dt
R
1d
)
R dt
µ0 I ˆ
B·da; B =
;
2⇡s
E = Iloop R =
dQ =
Q=
µ0 Ia
=
2⇡
dQ
R=
dt
1
R
Z
s+a
s
d
=
dt
, in magnitude. So
Q=
2⇡a2 µ0 nI
.
R
ds0
µ0 Ia s + a
=
ln
;
s0
2⇡
s
µ0 a
dI
ln(1 + a/s) .
2⇡
dt
µ0 a
Iµ0 a
ln(1 + a/s) dI ) Q =
ln(1 + a/s).
2⇡R
2⇡R
The field of the wire, at the square loop, is out of the page, and decreasing, so the field of the induced
current must point out of page, within the loop, and hence the induced current flows counterclockwise.
Problem 7.19
In the quasistatic approximation, B =
⇢
µ0 N I
2⇡s
0,
(Eq. 5.60). The flux around the toroid is therefore
=
µ0 N I
2⇡
Z
a+w
a
ˆ, (inside toroid);
(outside toroid)
1
µ0 N Ih ⇣
w ⌘ µ0 N hw
h ds =
ln 1 +
⇡
I.
s
2⇡
a
2⇡a
d
µ0 N hw dI
µ0 N hwk
=
=
.
dt
2⇡a dt
2⇡a
The electric field is the same as the magnetic field of a circular current (Eq. 5.41):
B=
µ0 I
a2
ẑ,
2 (a2 + z 2 )3/2
with (Eq. 7.19)
I!
1 d
=
µ0 dt
N hwk
µ0
. So E =
2⇡a
2
✓
N hwk
2⇡a
◆
a2
ẑ =
(a2 + z 2 )3/2
Problem 7.20
@B/@t is nonzero along the left and right edges of the shaded rectangle:
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µ0 N hwka
ẑ.
4⇡ (a2 + z 2 )3/2
150
CHAPTER 7. ELETRODYNAMICS
E
The (inward) flux through the strip on the left is increasing; the (inward) flux through the strip on the right
is decreasing. This is analogous to two current sheets under Ampère’s law, with B ! E and µ0 Ienc ! d /dt
(Eq. 7.19). The one on the left is like a current flowing out (taking account of the minus sign), so its field is
counterclockwise and the one on the right is like a current flowing in, so its field is clockwise.
Problem 7.21
The answer is indeterminate, until some boundary conditions are supplied. We know that r ⇥ E =
(dB0 /dt) ẑ (and r · E = 0), but this is insufficient information to determine E. Ordinarily we would invoke
some symmetry of the configuration, or require that the field go to zero at infinity, to resolve the ambiguity,
0
but neither is available in this case. E = 12 dB
x ŷ) would do the job, in which case the force would be
dt (y x̂
zero, but we could add any constant vector to this, and make the force anything we like.
Problem 7.22
(a) From Eq. 5.41, the field (on the axis) is B =
is
=
µ0 I
b2
2 (b2 +z 2 )3/2 ẑ,
so the flux through the little loop (area ⇡a2 )
µ0 ⇡Ia2 b2
.
2(b2 + z 2 )3/2
µ0 m
2
ˆ
(b) The field (Eq. 5.88) is B = 4⇡
r 3 (2 cos ✓ r̂ + sin ✓ ✓), where m = I⇡a . Integrating over the spherical “cap”
(bounded by the big loop and centered at the little loop):
=
where r =
p
Z
B·da =
µ0 I⇡a2
4⇡ r3
Z
(2 cos ✓)(r2 sin ✓ d✓ d ) =
b2 + z 2 and sin ✓¯ = b/r. Evidently
(c) Dividing o↵ I (
1
= M12 I2 ,
2
=
µ0 I⇡a2 sin2 ✓
r
2
= M21 I1 ): M12 = M21 =
✓¯
0
µ0 Ia2
2⇡
2r
=
Z
✓¯
cos ✓ sin ✓ d✓
0
µ0 ⇡Ia2 b2
, the same as in (a)!!
2(b2 + z 2 )3/2
µ0 ⇡a2 b2
.
2(b2 + z 2 )3/2
Problem 7.23
E=
d
=
dt
M
dI
=
dt
M k.
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CHAPTER 7. ELETRODYNAMICS 4
151
Problem 7.21
E =−
dΦ
dI
= −M
= −M k.
dt
dt
I
❖
✻
a
❄
✻
a
❄
✻
a
❄
It’s hard to calculate M using a current in the little loop, so, exploiting the equality of the mutual inductances,
I’ll find the flux through the Rlittle loop when a current I flows in the big loop: = M I. The field of one long
2a 1
0I
0I
0 Ia
wire is B = µ2⇡s
) 1 = µ2⇡
a ds = µ2⇡
ln 2, so the total flux is
a s
=2
1
=
µ0 Ia ln 2
µ0 a ln 2
µ0 ka ln 2
) M=
) E=
,
⇡
⇡
⇡
in magnitude.
Direction: The net flux (through the big loop), due to I in the little loop, is into the page. (Why? Field
lines point in, for the inside of the little loop, and out everywhere outside the little loop. The big loop encloses
all of the former, and only part of the latter, so net flux is inward.) This flux is increasing, so the induced
current in the big loop is such that its field points out of the page: it flows counterclockwise.
Problem 7.24
B = µ0 nI ) 1 = µ0 nI⇡R2 (flux through a single turn). In a length l there are nl such turns, so the
total flux is = µ0 n2 ⇡R2 Il. The self-inductance is given by = LI, so the self-inductance per unit length is
L = µ0 n2 ⇡R2 .
Problem 7.25
The field of one wire is B1 =
µ0 I
2⇡ s ,
so
= 2·
µ0 I
2⇡
·l
dR ✏
✏
ds
s
=
µ0 Il
⇡
ln
d ✏
✏
. The ✏ in the numerator is
negligible (compared to d), but in the denominator we cannot let ✏ ! 0, else the flux is infinite.
µ0 l
L=
ln(d/✏). Evidently the size of the wire itself is critical in determining L.
⇡
Problem 7.26
(a) In the quasistatic approximation B =
µ0 ˆ
. So
2⇡s
1
This is the flux through one turn; the total flux is N times
E=
d
µ0 N h
(4⇡ ⇥ 10 )(10 )(10
=
ln(b/a)I0 ! sin(!t) =
dt
2⇡
2⇡
7
= 2.61 ⇥ 10
4
= 5.22 ⇥ 10
7
3
=
µ0 I
2⇡
1:
2
=
)
Z
b
1
h ds =
a s
µ0 N h
2⇡
µ0 Ih
ln(b/a).
2⇡
ln(b/a)I0 cos(!t). So
ln(2)(0.5)(2⇡ 60) sin(!t)
sin(!t) (in volts), where ! = 2⇡ 60 = 377/s. Ir =
E
2.61 ⇥ 10
=
R
500
4
sin(!t)
sin(!t) (amperes).
7
6
dIr
)(10 2 )
N 2h
(b) Eb = L
; where (Eq. 7.28) L = µ02⇡
ln(b/a) = (4⇡⇥10 )(10
ln(2) = 1.39 ⇥ 10
2⇡
dt
Therefore Eb = (1.39 ⇥ 10 3 )(5.22 ⇥ 10 7 !) cos(!t) = 2.74 ⇥ 10 7 cos(!t) (volts).
Ratio of amplitudes:
2.74 ⇥ 10
2.61 ⇥ 10
7
4
= 1.05 ⇥ 10
Problem 7.27
With I positive clockwise, E =
2
d Q
dt2
=
1
LC Q
=
! Q, where ! =
2
3
=
3
(henries).
µ0 N 2 h!
ln(b/a).
2⇡R
L dI
dt = Q/C, where Q is the charge on the capacitor; I =
p1 .
LC
dQ
dt ,
so
The general solution is Q(t) = A cos !t + B sin !t. At t = 0,
c
⃝2005
Pearson
Education,
Inc.,
Upper
c 2012 Pearson Education, Inc., Upper Saddle River,
NJ.
All rights
reserved.
This
material
protected
laws as they
exist. No portion of this material may be
protected under all copyright laws as they currently
exist.under
No portion
of this material
maycurrently
be
reproduced,
in in
any
form or
by the
any publisher.
means, without permission in writing from the publisher.
reproduced, in any form or by any means, without
permission
writing
from
1
152
CHAPTER 7. ELETRODYNAMICS
Q = CV , so A = CV ; I(t) = dQ
A! sin !t + B! sin !t. At t = 0, I = 0, so B = 0, and
dt =
r
✓
◆
C
t
I(t) = CV ! sin !t = V
sin p
.
L
LC
Contents
2
Q
d Q
dQ
1
If you put in a resistor, the oscillation is “damped”. This time L dI
dt = C + IR, so L dt2 + R dt + C Q = 0.
For an analysis of this case, see Purcell’s Electricity and Magnetism (Ch. 8) or any book on oscillations and
waves.
Problem 7.28
1
(a) W = 12 LI 2 . L = µ0 n2 ⇡R2 l (Prob. 7.24) W = µ0 n2 ⇡R2 lI 2 .
Problem 7.26
2
H
1
ˆ
(b) W = 2 (A·I)dl. A = (µ0 nI/2)R , at the surface (Eq. 5.72 or 5.73). So W1 = 12 µ02nI RI · 2⇡R, for one
turn. There areR nl such turns in length l, so W = 12 µ0 n2 ⇡R2 lI 2 . X
R
(c) W = 2µ1 0 B 2 d⌧. B = µ0 nI, inside, and zero outside; d⌧ = ⇡R2 l, so W = 2µ1 0 µ20 n2 I 2 ⇡R2 l =
1
2
2 2
5
.X
2 µ0 n ⇡R lI ⇥R
⇤
H
R
(d) W = 2µ1 0 B 2 d⌧
(A⇥B)·da . This time B 2 d⌧ = µ20 n2 I 2 ⇡(R2 a2 )l. Meanwhile,
A⇥B = 07.26
outside (at s = b). Inside, A = µ02nI a ˆ (at s = a), while B = µ0 nI ẑ.
Problem
A⇥B = 12 µ20 n2 I 2 a( ˆ⇥ẑ)
ŝ
points inward (“out” of the volume)
| {z }
✕
☛
ŝ
✲
ẑ
H
R
(A⇥B) · da = ( 12 µ20 n2 I 2 aŝ) · [a d dz( ŝ)] = 12 µ20 n2 I 2 a2 2⇡l.
✲ ẑ
⇥
⇤
W = 2µ1 0 µ20 n2 I 2 ⇡(R2 a2 )l + µ20 n2 I 2 ⇡a2 l = 12 µ0 n2 I 2 R2 ⇡l. X
φ̂
❖
Problem 7.29
✓ ◆
Z
Z
µ0 nI
1
1 µ20 n2 I 2
1
µ0 n2 I 2
b
1
B=
; W =
B 2 d⌧ =
hs
d
ds
=
h2⇡
ln
=
µ0 n2 I 2 h ln(b/a).
2⇡s
2µ0
2µ0 4⇡ 2
s2
8⇡ 2
a
4⇡
µ0 2
L=
n h ln (b/a) (same as Eq. 7.28).
2⇡
→
Problem
7.30
I
✛
✲
l
µ0 Is
B·dl = B(2⇡s) = µ0 Ienc = µ0✕I(s2 /R2 ) ) B =
.
2
→I2⇡R
w
Z
2 2 Z R
2
1
1 ❄
µ0 I
µ0 I l s4 R Bµ0 l 2
1
☛ 2
W =
B 2 d⌧ =
s
(2⇡s)l
ds
=
=
I = LI 2 .
h
2
4
4
2µ0
2µ0 4⇡ R 0
4⇡R 4 0
16⇡
2
✛
µ0
✻
So L =
l, and L = L/l = µ0 /8⇡, independent ofI R!
8⇡
Problem 7.31
dI
dI
R
E0
(a) Initial current: I0 = E0 /R. So L
= IR )
=
I ) I = I0 e Rt/L , or I(t) = e
dt
dt
L
R
E02 2Rt/L
dW
2
2
2Rt/L
(b) P = I R = (E0 /R) e
R=
e
=
.
R
dt
✓
◆
Z
1
E 2 1 2Rt/L
E2
L
E2
1
2
W = 0
e
dt = 0
e 2Rt/L
= 0 (0 + L/2R) = L (E0 /R) .
R 0
R
2R
R
2
0
2
(c) W0 = 12 LI02 = 12 L (E0 /R) . X
Problem 7.32
µ0 1
(a) Pearson
B1 = 4⇡
a1 ], River,
since NJ.
m1All
= rights
I1 a1 .reserved.
The flux
through loop 2 is then
1 · r̂Upper
r 3 I1 [3(a
c
⃝2005
Education,
Inc.,
This material is
µ0 1 laws as they currently exist. No portion of this material may
µ0 be
)(a2 · r̂permission
) a1 ·ain
= M Ifrom
M publisher.
=
[3(a1 · r̂ )(a2 · r̂ )
reproduced,
any
form or by3 I
any
means,
the
2 = Bin
1 ·a
2 =
1 [3(a
1 · r̂ without
2 ] writing
1.
4⇡ r
4⇡ r 3
Rt/L
.
a1 ·a2 ].
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
153
CHAPTER 7. ELETRODYNAMICS
dW
2
2
(b) E1 = M dI
E1 I1 = M I1 dI
dt ,
dt 1 =
dt . (This is the work done per unit time against the mutual emf in
loop 1—hence the minus sign.) So (since I1 is constant) W1 = M I1 I2 , where I2 is the final current in loop 2:
µ0
W =
[3(m1 · r̂ )(m2 · r̂ ) m1 ·m2 ].
4⇡ r 3
Notice that this is opposite in sign to Eq. 6.35. In Prob. 6.21 we assumed that the magnitudes of the dipole
moments were fixed, and we did not worry about the energy necessary to sustain the currents themselves—only
the energy required to move them into position and rotate them into their final orientations. But in this
problem we are including it all, and it is a curious fact that this merely changes the sign of the answer. For
commentary on this subtle issue see R. H. Young, Am. J. Phys. 66, 1043 (1998), and the references cited
there.
Problem 7.33
(a) The (solenoid) magnetic field is
(
µ0 K ẑ = µ0 !R ẑ (s < R),
B=
0
(s > R).
From Example 7.7, the electric field is
8
>
s dB
sR
>
>
µ0 !˙ ˆ
< 2 dt ˆ =
2
E=
>
R2 dB ˆ
R3
>
>
=
µ0 !˙ ˆ
:
2s dt
2s
At the surface (s = R) E =
1
2
2 µ0 R
N=
(s < R),
(s > R).
!˙ ˆ, so the torque on a length ` of the cylinder is
✓
◆
1
2
R( 2⇡R`)
µ0 R !˙ ẑ = ⇡µ0 2 R4 !`
˙ ẑ,
2
and the work done per unit length is
W
=
`
But d = ! dt, and the integral becomes
Z !f
1
! d! = !f2 ;
2
0
⇡µ0
)
Z
d!
d .
dt
W
=
`
µ0 ⇡
2
2
R4
!f R 2
2
.
(This is the work done by the field; the work you must do does not include the minus sign.)
(b) Because B = µ0 K ẑ = µ0 !f R ẑ is uniform inside the solenoid (and zero outside), W =
W
1
µ0 ⇡
=
(µ0 !f R)2 ⇡R2 =
`
2µ0
2
Problem 7.34
The displacement current density (Sect. 7.3.2) is Jd = ✏0 @E
@t =
I
B·dl = B · 2⇡s = µ0 Idenc = µ0
I
A
1 2 2
B ⇡R `.
2µ0
!f R 2
2
=
ẑ. Drawing an “amperian loop” at
I
⇡a2
.
I
s2
µ0 Is2
2
·
⇡s
=
µ
I
)
B
=
;
0
⇡a2
a2
2⇡sa2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
B=
µ0 Is ˆ
.
2⇡a2
154
Problem 7.35
(t)
(a) E =
ẑ;
✏0
CHAPTER 7. ELETRODYNAMICS
Q(t)
It
It
=
;
ẑ.
⇡a2
⇡a2 ⇡✏0 a2
I
dE 2
s2
s2
µ0 I ˆ
(b) Idenc = Jd ⇡s2 = ✏0
⇡s = I 2 .
B · dl = µ0 Idenc ) B 2⇡s = µ0 I 2 ) B =
s .
dt
a
a
2⇡a2
(c) A surface current flows radially outward over the left plate; let I(s) be the total current crossing a circle
of radius s. The charge density (at time t) is
(t) =
(t) =
[I
I(s)]t
.
⇡s2
Since we are told this is independent of s, it must be that I
so a2 = I, or = I/a2 . Therefore I(s) = I(1 s2 /a2 ).
B 2⇡s = µ0 Ienc = µ0 [I
I(s)] = µ0 I
I(s) = s2 , for some constant . But I(a) = 0,
s2
µ0 I ˆ
) B=
s . X
a2
2⇡a2
Problem 7.36
µ0 I0 ! 2
µ0 ✏0 2
(a) Jd = ✏0
cos(!t) ln (a/s) ẑ. But I0 cos(!t) = I. So Jd =
! I ln(a/s) ẑ.
2⇡
2⇡
Z
Z
Z
a
µ0 ✏0 ! 2 I a
(b) Id = Jd · da =
ln(a/s)(2⇡s ds) = µ0 ✏0 ! 2 I
(s ln a s ln s)ds
2⇡
0
0
h
ia
h 2
i
2
µ0 ✏0 ! 2 Ia2
s2
s2
a
a2
a2
2
= µ0 ✏0 ! 2 I (ln a) s2
ln
s
+
=
µ
✏
!
I
ln
a
ln
a
+
=
.
0
0
2
4
2
2
4
4
0
(c)
Id
µ0 ✏0 ! 2 a2
=
.
I
4
Since µ0 ✏0 = 1/c2 , Id /I = (!a/2c)2 . If a = 10
8
m/s
10
2c
11
! = 10a
= 3⇥10
5⇥10 3 m , or ! = 0.6 ⇥ 10 /s = 6 ⇥ 10 /s; ⌫ =
microwave region, way above radio frequencies.)
!
2⇡
3
m, and
Id
I
=
1
100 ,
so that
!a
2c
=
1
10 ,
⇡ 1010 Hz, or 104 megahertz. (This is the
Problem 7.37
Physically, this is the field of a point charge q at the origin, out to an expanding spherical shell of radius vt;
outside this shell the field is zero. Evidently the shell carries the opposite charge, q. Mathematically, using
product rule #5 and Eq. 1.99:
✓
◆
1 q
1 q
q 3
1 q
@
r · E = ✓(vt r)r ·
r̂ +
r̂ · r[✓(vt r)] =
(r)✓(vt r) +
(r̂ · r̂) ✓(vt r).
2
2
2
4⇡✏0 r
4⇡✏0 r
✏0
4⇡✏0 r
@r
But
3
(r)✓(vt
r) =
3
(r)✓(t), and
@
@r ✓(vt
r) =
(vt
⇢ = ✏0 r · E = q 3 (r)✓(t)
r) (Prob. 1.46), so
q
(vt
4⇡r2
r).
(For t < 0 the field and the charge density are zero everywhere.)
Clearly r · B = 0, and r ⇥ E = 0 (since E has only an r component, and it is independent of ✓ and ).
There remains only the Ampére/Maxwell law, r ⇥ B = 0 = µ0 J + µ0 ✏0 @E/@t. Evidently
⇢
@E
q
@
q
J = ✏0
= ✏0
[✓(vt r)] r̂ =
v (vt r) r̂.
2
@t
4⇡✏0 r @t
4⇡r2
protected under all copyright laws as they currently exist. No portion of this material may be
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155
CHAPTER 7. ELETRODYNAMICS
(The stationary charge at the origin does not contribute to J, of course; for the expanding shell we have J = ⇢v,
as expected—Eq. 5.26.)
Problem 7.38
µ0 qm
From r·B = µ0 ⇢m it follows that the field of a point monopole is B = 4⇡
r 2 r̂ . The force law has the
1
2
form F / qm B c2 v⇥E (see Prob. 5.22—the c is needed on dimensional grounds). The proportionality
✓
◆
1
constant must be 1 to reproduce “Coulomb’s law” for point charges at rest. So F = qm B
v⇥E
.
c2
Problem 7.39
Integrate the “generalized Faraday law” (Eq. 7.44iii), r ⇥ E = µ0 Jm @B
@t , over the surface of the loop:
Z
I
Z
Z
d
d
(r ⇥ E) · da = E · dl = E = µ0 Jm · da
B · da = µ0 Imenc
.
dt
dt
dI
dI
µ0
1d
µ0
1
But E = L , so
=
Im +
, or I =
Qm +
, where Qm is the total magnetic charge
dt
dt
L enc L dt
L
L
passing through the surface, and
is the change in flux through the surface. If we use the flat surface, then
Qm = qm and
= 0 (when the monopole is far away,
= 0; the flux builds up to µ0 qm /2 just before
it passes through the loop; then it abruptly drops to µ0 qm /2, and rises back up to zero as the monopole
disappears into the distance). If we use a huge balloon-shaped surface, so that qm remains inside it on the far
side, then Qm = 0, but rises monotonically from 0 to µ0 qm . In either case,
I=
µ0 qm
.
L
[The analysis is slightly di↵erent for a superconducting loop, but the conclusion is the same.]
Problem 7.40

V
1
V
@D
@
@ V0 cos(2⇡⌫t)
✏V0
E=
) Jc = E = E =
. Jd =
= (✏E) = ✏
=
[ 2⇡⌫ sin(2⇡⌫t)].
d
⇢
⇢d
@t
@t
@t
d
d
The ratio of the amplitudes is therefore:
⇥
Jc
V0
d
1
=
=
= 2⇡(4 ⇥ 108 )(81)(8.85 ⇥ 10
Jd
⇢d 2⇡⌫✏V0
2⇡⌫✏⇢
12
Problem 7.41
Begin with7.39
a di↵erent problem: two parallel
Problem
wires carrying charges + and
as shown.
Field of one wire: E =
2⇡✏0 s ŝ;
potential: V =
⇤
)(0.23)
9
1
= 2.41.
z✻
2⇡✏0
b
ln(s/a).
Potential of combination: V = 2⇡✏0 ln(s /s+ ),
n
o
(y+b)2 +z 2
or V (y, z) = 4⇡✏0 ln (y
b)2 +z 2 .
−λ ✰
x
b
✲y
+λ
Find the locus of points of fixed V (i.e. equipotential surfaces):
e4⇡✏0 V / ⌘ µ =
y (µ
2
(y
(y + b)2 + z 2
=) µ(y 2
(y b)2 + z 2
1) + b (µ
2
b )2 + z 2 + b2
1) + z (µ
b2
2
1)
2
= 0 =) (y
2yb + b2 + z 2 ) = y 2 + 2yb + b2 + z 2 ;
2yb(µ + 1) = 0 =) y + z + b
b
2
) + z 2 = b2 (
2
2
2
1).
2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
2yb = 0
✓
µ+1
⌘
µ 1
◆
;
156
CHAPTER 7. ELETRODYNAMICS
q 2
p
p
2b µ
(µ2 2µ+1)
2
This is a circle, with center at y0 = b = b µµ+11 and radius = b
1 = b (µ +2µ+1)
=
2
(µ 1)
µ 1 .
This suggests an image solution to the problem at hand. We want y0 = d, radius = a, and V = V0 . These
determine the parameters b, µ, and of the image solution:
b µµ+11
d
y0
µ+1
d
=
= 2bpµ = p . Call ⌘ ↵.
a
2 µ
a
µ 1
4↵ µ = (µ + 1) = µ2 + 2µ + 1 =) µ2 + (2 4↵2 )µ + 1 = 0;
p
p
p
4↵2 2 ± 4(1 2↵2 )2 4
µ=
= 2↵2 1 ± 1 4↵2 + 4↵4 1 = 2↵2 1 ± 2↵ ↵2 1;
2
4⇡✏0 V0
4⇡✏0 V0
p
= ln µ =) =
. That’s the line charge in the image problem.
2
ln 2↵
1 ± 2↵ ↵2 1
Z
Z
1
I = J·da =
E·da =
Qenc =
l.
✏0
✏0
2
2
I
4⇡ V0
p
=
=
. Which sign do we want? Suppose
l
✏0
ln 2↵2 1 ± 2↵ ↵2 1
the cylinders are far apart, d
a, so that ↵
1.

p
1
1
( ) = 2↵2 1 ± 2↵2 1 1/↵2 = 2↵2 1 ± 2↵2 1
+ ···
2
2↵
8↵4
(
4↵2 2 1/2↵2 + · · · ⇡ 4↵2 (+ sign),
1
2
= 2↵ (1 ± 1) (1 ± 1) ⌥ 2 ± · · · =
4↵
1/4↵2
( sign).
The current per unit length is i =
The current must surely decrease with increasing ↵, so evidently the + sign is correct:
i=
Problem 7.42
From Prob. 3.24,
4⇡ V0
p
1 + 2↵ ↵2
ln 2↵2
8
1
X
>
>
>
V
(s,
)
=
sk bk sin(k ),
>
in
>
>
<
k=1
>
1
>
X
>
>
>
V
(s,
)
=
s
>
out
:
k
1
, where ↵ =
d
.
a
(s < a);
dk sin(k ), (s > a).
k=1
(We don’t need the cosine terms, because V is clearly an odd function of .) At s = a, Vin = Vout = V0 /2⇡.
Let’s start with Vin , and use Fourier’s trick to determine bk :
Z ⇡
Z
1
1
X
X
V0 ⇡
V0
ak bk sin(k ) =
)
ak bk
sin(k ) sin(k 0 ) d =
sin(k 0 ) d . But
2⇡
2⇡
⇡
⇡
k=1
k=1
Z ⇡
0
sin(k ) sin(k ) d = ⇡ kk0 , and
⇡

Z ⇡
⇡
0
1
2⇡
2⇡
0
0
0
sin(k ) d =
sin(k
)
cos(k
)
=
cos(k 0 ) =
( 1)k . So
0
2
0
0
0
(k )
k
k
k
⇡

✓
◆k ⇡
1
V
2⇡
V
1
V 0 X 1 ⇣ s ⌘k
0
0
⇡ak bk =
( 1)k , or bk =
, and hence Vin (s, ) =
sin(k ).
2⇡
k
⇡k
a
⇡
k
a
k=1
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157
CHAPTER 7. ELETRODYNAMICS
1
1
X
V0 X 1 ⇣ a ⌘k
1
sin(k ). Both sums are of the form S ⌘
( x)k sin(k ) (with
⇡
k
s
k
k=1
k=1
x = s/a for r < a and x = a/s for r > a). This series can be summed explicitly, using Euler’s formula
1
1
X
X
1
1
k
(ei✓ = cos ✓ + i sin ✓): S = Im
( x)k eik = Im
xei
.
k
k
k=1
k=1
1
X
⇥
⇤
1 2 1 3 1 4
1
But ln(1 + w) = w
w + w
w ··· =
( w)k , so S = Im ln 1 + xei
.
2
3
4
k
Similarly, Vout (s, ) =
k=1
Now ln Rei✓ = ln R + i✓, so S =
Im 1 + xei
tan ✓ =
=
Re (1 + xei )
✓, where
1
2i
1
2
⇥
8
V0
>
>
Vin (s, ) =
tan
>
>
<
⇡
Conclusion:
>
>
V0
>
>
tan
: Vout (s, ) =
⇡
=
@Vin
=
@s
=
@Vin
@s
1
1
✓
✓
s sin
a + s cos
a sin
s + a cos
◆
◆
i
i
)]
⇤
=
x ei
e i
i [2 + x (ei + e
i
)]
=
x sin
.
1 + x cos
, (s < a);
, (s > a).
@Vin
.
@s s=a
s=a
8
9
>
>
>
>

V0 <
1
( a sin ) =
V0
a sin

⇣
⌘2 (s + a cos )2 > = ⇡ (s + a cos )2 + (a sin )2
⇡ >
a sin
>
>
: 1 + s+a
;
cos
✓
◆
V0
a sin
;
2
⇡ s + 2as cos + a2
8
9
>
>
>
>

V0 <
1
[(a + s cos ) sin
s sin cos ] = V0
a sin

=
⇣
⌘
2
2
>
>
⇡ >
(a + s cos )
⇡ (a + s cos )2 + (s sin )2
s sin
>
: 1 + a+s
;
cos
✓
◆
V0
a sin
.
⇡ s2 + 2as cos + a2
(b) From Eq. 2.36, ( ) =
@Vout
=
@s
⇢
1 + xei
1 + xe
i
[(1 + xe ) + (1 + xe
=
s=a
@Vout
@s
✏0
s=a
@Vout
@s
V0
=
2⇡a
✓
sin
1 + cos
◆
, so ( ) =
✏0 V0
sin
=
⇡a (1 + cos )
Problem 7.43
✏0 V0
tan( /2).
⇡a
✓
◆
✓
◆
✓
◆
1 @
@(zf )
@ 2 (zf )
z d
df
d
df
df
(a) r V =
s
+
=
s
=0)
s
=0)s
= A (a constant) )
s @s
@s
@z 2
s ds
ds
ds
ds
ds
ds
A
= df ) f = A ln(s/s0 ) (s0 another constant). But (ii) ) f (b) = 0, so ln(b/s0 ) = 0, so s0 = b, and
s
I⇢
1
I⇢z ln(s/b)
V (s, z) = Az ln(s/b). But (i) ) Az ln(a/b) = (I⇢z)/(⇡a2 ), so A =
; V (s, z) =
.
⇡a2 ln(a/b)
⇡a2 ln(a/b)
2
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158
(b) E =
rV =
CHAPTER 7. ELETRODYNAMICS
⇣z
⇣s⌘ ⌘
@V
I⇢z
1
I⇢ ln(s/b)
I⇢
ẑ =
ŝ
+
ẑ
=
ŝ
+
ln
ẑ .
@z
⇡a2 s ln(a/b)
⇡a2 ln(a/b)
⇡a2 ln(a/b) s
b

⇣z ⌘
⇤
I⇢
✏0 I⇢z
Es (a ) = ✏0
0 =
.
2
⇡a ln(a/b) a
⇡a3 ln(a/b)
@V
ŝ
@s
⇥
(c) (z) = ✏0 Es (a+ )
Problem 7.44
@B
(a) Faraday’s law says r⇥E = @B
@t , so E = 0 )H @t = 0 ) B(r) is independent of t.
(b) Faraday’s law in integral form (Eq. 7.19) says E · dl = d /dt. In the wire itself E = 0, so through
the loop is constant.
(c) Ampère-Maxwell) r⇥B = µ0 J + µ0 ✏0 @E
@t , so E = 0, B = 0 ) J = 0, and hence any current must be
at the surface.
(d) From Eq. 5.70, a rotating shell produces a uniform magnetic field (inside): B = 23 µ0 !aẑ. So to cancel
3 B0
3B0
such a field, we need !a =
. Now K = v = !a sin ✓ ˆ, so K =
sin ✓ ˆ.
2 µ0
2µ0
Problem 7.45
(a) To make the field parallel to the plane, we need image monopoles of the same sign (compare Figs. 2.13
and 2.14), so the image dipole points down (-z).
(b) From Prob. 6.3 (with r ! 2z):
F =
3µ0 m2
.
2⇡ (2z)4
3µ0 m2
1
= Mg ) h =
4
2⇡ (2h)
2
Problem 7.43
(c) Using Eq. 5.89, and referring to the figure:
B=
=
=
=
=
✓
3µ0 m2
2⇡M g
◆1/4
.
11
m✻
✻
θ
r1
h
µ0 1
{[3(m ẑ · r̂1 ) r̂1 mẑ] + [3( m ẑ · r̂2 ) r̂2 + mẑ]}
4⇡ (r1 )3
r̂2
3µ0 m
[(ẑ · r̂1 ) r̂1 (ẑ · r̂2 ) r̂2 ] . But ẑ · r̂1 = ẑ · r̂2 = cos ✓.
r ✲
❯ ✕✲ r̂
4⇡(r1 )3
✕❯
3µ0 m
r̂1
cos
✓(r̂
+
r̂
).
But
r̂
+
r̂
=
2
sin
✓
r̂.
1
2
1
2
4⇡(r1 )3
r2
h
p
3µ0 m
r
h
2 + h2 .
sin
✓
cos
✓
r̂.
But
sin
✓
=
,
cos
✓
=
,
and
r
=
r
1
θ
2⇡(r1 )3
r1
r1
3µ0 mh
r
−m ❄
❄
r̂.
2
2
5/2
2⇡ (r + h )
Now B = µ0 (K ⇥ ẑ) ) ẑ ⇥ B = µ0 ẑ ⇥ (K ⇥ ẑ) = µ0 [K ẑ(K · ẑ)] = µ0 K. (I used the BAC-CAB rule,
and noted that K · ẑ = 0, because the surface current is in the x y plane.)
K=
1
(ẑ ⇥ B) =
µ0
3mh
r
(ẑ ⇥ r̂) =
2
2⇡ (r + h2 )5/2
3mh
r
ˆ. qed
2
2⇡ (r + h2 )5/2
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12
CHAPTER 7. ELETRODYNAMICS
159
Problem
Problem7.44
7.46
Say the angle between the dipole (m1 ) and the z axis is ✓ (see diagram).
−
B(z) =
µ0
1
[3(m2 · ẑ) ẑ
4⇡ (h + z)3
m2 ]
✲x
h
for points on the z axis (Eq. 5.89). The torque on m1 is (Eq. 6.1)
−
µ0
[3(m2 · ẑ)(m1 ⇥ ẑ)
4⇡(2h)3
But m1 = m(sin ✓ x̂ + cos ✓ẑ), m2 = m(sin ✓ x̂
m1 ⇥ m2 = 2m2 sin ✓ cos ✓ŷ.
N=
✻
θ +
✒
m
1
The field of the image dipole (m2 ) is
N = m1 ⇥ B =
z
cos ✓ẑ), so m2 · ẑ =
⇥ 2
µ0
3m sin ✓ cos ✓ ŷ
3
4⇡(2h)
m cos ✓, m1 ⇥ ẑ =
m2
❘+
θ
(m1 ⇥ m2 )] .
m sin ✓ ŷ, and
⇤
µ0 m2
2m2 sin ✓ cos ✓ ŷ) =
sin ✓ cos ✓ ŷ.
4⇡(2h)3
Evidently the torque is zero for ✓ = 0, ⇡/2, or ⇡. But 0 and ⇡ are clearly unstable, since the nearby
ends of the dipoles (minus, in the figure) dominate, and they repel. The stable configuration is ✓ = ⇡/2:
parallel to the surface (contrast Prob. 4.6).
µ0 m
In this orientation, B(z) = 4⇡(h+z)
3 x̂, and the force on m1 is (Eq. 6.3):
F=r

µ0 m2
4⇡(h + z)3
=
z=h
3µ0 m2
ẑ
4⇡(h + z)4
=
z=h
3µ0 m2
ẑ.
4⇡(2h)4
At equilibrium this force upward balances the weight M g:
3µ0 m2
1
= Mg ) h =
4⇡(2h)4
2
✓
3µ0 m2
4⇡M g
◆1/4
.
1
Incidentally, this is (1/2)1/4 = 0.84 times the height it would adopt in the orientation perpendicular to the
plane (Prob. 7.45b).
Problem 7.47
R
ˆ
f = v⇥B; v = !a sin ✓ ˆ; f = !aB0 sin ✓( ˆ⇥ẑ). E = f ·dl, and dl = a d✓ ✓.
R
⇡/2
ˆ ˆ⇥ẑ) = ẑ·(✓⇥
ˆ ˆ) = ẑ·r̂ = cos ✓.
So E = !a2 B0 0 sin ✓( ˆ⇥ẑ)·✓ˆ d✓. But ✓·(
Z ⇡/2
2
⇥ sin ✓ ⇤ ⇡/2
1
E = !a2 B0
sin ✓ cos ✓ d✓ = !a2 B0
= !a2 B0 (same as the rotating disk in Ex. 7.4).
0
2
2
0
z✻
Problem 7.48
Ry p
F = IBl; = 2B a a2 x2 dx (a = radius of circle).
p
p
l/2
✲
E = ddt = 2B a2 y 2 dy
= 2Bv a2 y 2 = IR.
dt
x
p
p
2Bv
4B 2 v
2
2
2
2
2
2
I= R
a
y ; l/2 = a
y . So F = R (a
y ) = mg.
mgR
✛✲
vcircle =
;
y
4B 2 (a2 y 2 )
Z a
Z
a
a
dy
4B 2
4B 2
1 3
4B 2 4 3
16 B 2 a3
tcircle =
=
(a2 y 2 )dy =
(a2 y
y )
=
( a )=
.
v
mgR a
mgR
3
mgR 3
3 mgR
+a
a
Contents
}
⃝2005
Pearson
Inc.,
River, NJ.
c 2012 Pearson Education, Inc., Upper
River, Education,
NJ. All rights
reserved.
This material
underexist.
lawsofas
they
currently
protected under all copyright laws as protected
they currently
No portion
this
material
mayexist.
be No portion of this material may be
reproduced,
any form or
any means,
without
permission in writing from the publisher.
reproduced, in any form or by any means,
withoutinpermission
in by
writing
from the
publisher.
160
CHAPTER 7. ELETRODYNAMICS
Problem 7.49
(a) From Prob. 5.52a,
A(r, t) =
@
@t (r
[Check: r ⇥ E =
⇥ A) =
@B
@t ,
1
4⇡
Z
B(r0 , t) ⇥
r
r̂
2
d⌧ 0 , so E =
@A
.
@t
2
2
1 Q
1
4⇡R
(b) The Coulomb field is zero inside and 4⇡✏
r̂ = ✏0Rr2 r̂ outside. The Faraday field is
2 r̂ = 4⇡✏
r2
0 r
0
@A
@t , where A is given (in the quasistatic approximation) by Eq. 5.69, with ! a function of time. Letting
!˙ ⌘ d!/dt,
8 µ R!˙
0
>
r sin ✓ ˆ
(r < R),
>
>
<
3
E(r, ✓, , t) =
>
>
µ0 R4 !˙ sin ✓ ˆ
> R2
:
r̂
(r > R).
✏0 r2
3
r2
Problem 7.50
dB
dv
dB
= m
= ma = F = qE, or E = R
. But
dt
dt
dt
✓
◆
I
d
d
1 d
dB
1
1
E·dl =
, so E 2⇡R =
, so
=R
, or B =
+ constant. If at time t = 0
2
dt
dt
2⇡R dt
dt ✓
2
⇡R
◆
1
1
the field is o↵, then the constant is zero, and B(R) =
(in magnitude). Evidently the field at R
2 ⇡R2
must be half the average field over the cross-section of the orbit. qed
qBR = mv (Eq. 5.3). If R is to stay fixed, then qR
Problem 7.51
In the quasistatic approximation the magnetic field of the wire is B = (µ0 I/2⇡s) ˆ, or, in Cartesian
coordinates,
µ0 I
µ0 I ⇣ y
x ⌘ µ0 I ( y x̂ + x ŷ)
B=
( sin x̂ + cos ŷ) =
x̂ + ŷ =
,
2⇡s
2⇡s
s
s
2⇡
x2 + y 2
where x and y are measured from the wire. To convert to the stationary coordinates in the diagram, y ! y
B=
vt:
µ0 I [ (y vt) x̂ + x ŷ]
.
2⇡
x2 + (y vt)2
r⇥E=
@B
=
@t
µ0 I
2⇡
⇢
x2
[ (y vt) x̂ + x ŷ]
( 2v)(y
[x2 + (y vt)2 ]2
v x̂
+ (y vt)2
vt) .
At t = 0, then,
r⇥E =
µ0 Iv
2⇡
⇢
x̂
[ y 2 x̂ + xy ŷ]
+2
2
2
x +y
[x2 + y 2 ]2
=
µ0 Iv
2⇡
⇢
(x2
y 2 ) x̂ + 2xy ŷ]
[x2 + y 2 ]2
=
µ0 Iv ⇣
cos ŝ + sin
2⇡s2
⌘
ˆ .
Our problem is to find a vector function of s and (it obviously doesn’t depend on z) whose divergence is
zero, whose curl is as given above, that goes to zero at large s:
E(s, ) = Es (s, ) ŝ + E (s, ) ˆ + Ez (s, ) ẑ,
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161
CHAPTER 7. ELETRODYNAMICS
with
1 @
1 @E
(sEs ) +
= 0,
s @s
s @
1 @Ez
µ0 Iv
cos
(r ⇥ E)s =
=
s @
2⇡s2
@Ez
µ0 Iv
(r ⇥ E) =
=
sin ,
2⇡s2
@s
1 @
@Es
(r ⇥ E)z =
(sE )
= 0.
s @s
@
r·E =
The first and last of these are satisfied if Es = E = 0, the middle two are satisfied by Ez =
Evidently E =
µ0 Iv
sin .
2⇡s
µ0 Iv
sin ẑ. The electric and magnetic fields ride along with the wire.
2⇡s
Problem 7.52
2
1 qQ
1
2
Initially, mv
r = 4⇡✏0 r 2 ) T = 2 mv =
new orbit, of radius r1 and velocity v1 :
1 1 qQ
2 4⇡✏0 r .
After the magnetic field is on, the electron circles in a
mv12
1 qQ
1
1 1 qQ 1
=
+ qv1 B ) T1 = mv12 =
+ qv1 r1 B.
2
r1
4⇡✏0 r1
2
2 4⇡✏0 r1
2
1
⇠
= r 1 1 + dr
= r 1 1 dr
r
r , while v1 = v + dv, B = dB. To first order, then,
✓
◆
1 1 qQ
dr
1
qvr
1 1 qQ
T1 =
1
+ q(vr) dB, and hence dT = T1 T =
dB
dr.
2 4⇡✏0 r
r
2
2
2 4⇡✏0 r2
But r1 = r + dr, so (r1 )
1
qr dB
qr
dv
Now, the induced electric field is E = 2r dB
dt (Ex. 7.7), so m dt = qE = 2 dt , or m dv = 2 dB. The increase in
qvr
1
2
kinetic energy is therefore dT = d( 2 mv ) = mv dv = 2 dB. Comparing the two expressions, I conclude that
dr = 0. qed
Problem 7.53
d
↵
E=
= ↵. So the current in R1 and R2 is I =
; by Lenz’s law, it flows counterclockwise. Now
dt
R1 + R 2
↵R1
the voltage across R1 (which voltmeter #1 measures) is V1 = IR1 =
(Vb is the higher potential),
R 1 + R2
and V2 =
IR2 =
↵R2
(Vb is lower ).
R 1 + R2
Problem 7.54
d
dB
⇡↵r2
(a) E =
= ⇡r2
= ↵⇡r2 = IR ) (in magnitude) I =
. If B is out
dt
dt
R
law says the current is clockwise.
I
↵s ˆ
↵s
↵
E · dl = E2⇡s = ⇡s2 ↵ ) E =
=
( sin x̂ + cos ŷ) = (s sin x̂ s cos
2
2
2
Z
Z
p
↵
Along the line from P to Q, dl = dx x̂, and y = r/ 2, so V =
E · dl =
y dx =
2
↵r2
P is at the higher voltage, and the meter reads
.
2
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of the page, Lenz’s
↵
(y x̂ x ŷ).
2
↵ r p
p (r 2). Thus
2 2
ŷ) =
162
CHAPTER 7. ELETRODYNAMICS
That’s the simplest way to do it. But you might instead regard the 3/4-circle+chord as a circuit, and use
Kirchho↵’s rule (the total emf around a closed loop is zero): V + IR1 ↵A1 = 0, where R1 = 3/4R is the
resistance of the curved portion, and A1 = (3/4)⇡r2 + r2 /2 = (r2 /4)(3⇡ + 2) is the area (3/4 of the circle, plus
the triangle). Then
r2
⇡r2 ↵ 3
↵r2
↵r2
V = ↵ (3⇡ + 2)
R=
(3⇡ + 2 3⇡) =
. X
4
R 4
4
2
Or we could do the same thing, using the small loop at the top: V + IR2 ↵A2 = 0, where R2 = (1/4)R
and A2 = (1/4)⇡r2 r2 /2 = (r2 /4)(⇡ 2).
V =
⇡↵r2 R
R 4
↵
r2
(⇡
4
2) =
↵r2
(⇡
4
⇡ + 2) =
↵r2
. X
2
Problem 7.55
E = vBh =
dI
dv d2 v
hB dI
L ; F = IhB = m ;
=
=
dt
dt dt2
m dt
Problem 7.56
A point on the upper loop: r2 = (a cos
r
2
= (r2
r1 )2 = (a cos
= a2 cos2
2
= a2 + b2 + z 2
b cos
2
2ab cos
2
cos
2ab(cos
⇥
= (a2 + b2 + z 2 ) 1
2 , a sin
2
2
1)
1
2
1
✓
◆
hB
L
v,
d2 v
=
dt2
hB
! 2 v, with ! = p
.
mL
a point on the lower loop: r1 = (b cos
+ (a sin
+ b2 cos2
1
cos
2 cos(
2 , z);
hB
m
b sin
2
+ a2 sin2
+ sin 2 sin
⇤ ab ⇥
1
1) =
2
1)
1 , 0).
+ z2
2ab sin
2
1 , b sin
1
sin
2
+ b2 sin2
1)
= a2 + b2 + z 2 2ab cos(
⇤
2 cos( 2
1) .
2
1
+ z2
1)
dl1 = b d 1 ˆ1 = b d 1 [ sin 1 x̂ + cos 1 ŷ]; dl2 = a d 2 ˆ2 = a d 2 [ sin 2 x̂ + cos 2 ŷ], so
dl1 ·dl2 = ab d 1 d 2 [sin 1 sin 2 + cos 1 cos 2 ] = ab cos( 2
1) d 1 d 2.
II
ZZ
µ0
dl1 ·dl2
µ0 ab
cos( 2
1)
p
p
M=
=
d 2 d 1.
r
4⇡
4⇡ ab/
1 2 cos( 2
1)
Both integrals run from 0 to 2⇡. Do the
2⇡
Z
1
2
integral first, letting u ⌘
cos u
p
du =
1 2 cos u
1
Z2⇡
0
p
2
1:
cos u
du
1 2 cos u
(since the integral runs over a complete cycle of cos u, we may as well change the limits to 0 ! 2⇡). Then the
1 integral is just 2⇡, and
Z 2⇡
Z 2⇡
µ0 p
cos u
µ0 p
cos u
p
p
M=
ab 2⇡
du =
ab
du.
4⇡
2
1 2 cos u
1 2 cos u
0
0
(a) If a is small, then
p
1
1
⇠
=1+
2 cos u
⌧ 1, so (using the binomial theorem)
cos u, and
Z
0
2⇡
p
cos u
du ⇠
=
1 2 cos u
Z
0
2⇡
cos u du +
Z
2⇡
cos2 u du = 0 + ⇡,
0
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163
CHAPTER 7. ELETRODYNAMICS
p
and hence M = (µ0 ⇡/2) ab
3.
µ0 ⇡a2 b2
(same as in Prob. 7.22).
2(b2 + z 2 )3/2
⇠
= ab/(b2 + z 2 ), so M ⇠
=
Moreover,
(b) More generally,
(1 + ✏)
1/2
1
3
✏ + ✏2
2
8
=1
5 3
✏ + · · · =) p
16
1
1
=1+
2 cos u
cos u +
3
2
2
cos2 u +
5
2
3
cos3 u + · · · ,
so
M=
µ0 p
ab
2
µ0 p
=
ab
2
⇢Z
2⇡
Z
cos u du +
0

3
0 + (⇡) +
2
2⇡
cos2 u du +
0
2
5
(0) +
2
3
3
2
2
Z
2⇡
cos3 u du +
0
3
µ0 ⇡ p
( ⇡) + · · · =
ab
4
2
Problem 7.57
Let be the flux of B through a single loop of either coil, so that
E1 =
N1
d
,
dt
E2 =
N2
= I1 L1 + M I2 = N1 ;
In case I1 = 0, we have
(b)
E1 =
d 1
dt
=
1
M 2 dI
dt
M RI2
Z
0
15
1+
8
= N1
2⇡
2
and
cos4 u du + · · ·
+()
2
4
+ ···
◆
qed
.
= N2 . Then
= I2 L2 + M I1 = N2 ,
or
is the flux through one turn):
= I1
L1
M
L2
M
+ I2
= I2
+ I1
.
N1
N1
N2
N2
L2
L1
L2
M
M
N2 ; if I2 = 0, we have N1 = N2 . Dividing: L1 = M , or L1 L2 =
dI1
2
2
+ M dI
E2 = ddt2 = L2 dI
I2 R. qed
dt = V1 cos(!t);
dt + M dt =
dI1
dI2
2
equation by L2 : L1 L2 dt + L2 dt M = L2 V1 cos !t. Plug in L2 dI
dt =
M
N1
1
L1 dI
dt
(c) Multiply the first
2
1
✓
3
d
E2
N2
, so
=
. qed
dt
E1
N1
Problem 7.58
(a) Suppose current I1 flows in coil 1, and I2 in coil 2. Then (if
1
3
5
2
=
M 2.
qed
I2 R
1
M dI
dt .
L2 V1
L2 V1
1
cos !t. L1 dI
dt + M M R ! sin !t = V1 cos !t.
MR
◆
✓
◆
L2
V1 1
L2
! sin !t ) I1 (t) =
sin !t +
cos !t .
R
L1 !
R
1
M 2 dI
dt = L2 V1 cos !t ) I2 (t) =
dI1
V1
=
dt
L1
✓
cos !t
cos !t R
L2
N2
N2
=
=
. The ratio of the amplitudes is
. qed
V1 cos !t
M
N1
N1
✓
◆
✓
◆
V1
1
L2
(V1 )2 1
L2
2
(e) Pin = Vin I1 = (V1 cos !t)
sin !t +
cos !t =
sin !t cos !t +
cos !t .
L1
!
R
L1
!
R
(d)
Vout
I2 R
=
=
Vin
V1 cos !t
L2 V1
MR
(L2 V1 )2
cos2 !t. Average of cos2 !t is 1/2; average of sin !t cos !t is zero.
M 2R
✓
◆


1
L2
1
(L2 )2
1
(L2 )2
(V1 )2 L2
2
So hPin i = (V1 )2
; hPout i = (V1 )2
=
(V
)
; hPin i = hPout i =
.
1
2
2
L1 R
2
M R
2
L1 L2 R
2L1 R
Pout = Vout I2 = (I2 )2 R =
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164
CHAPTER 7. ELETRODYNAMICS
Problem 7.59
(a) The charge flowing into dz in time dt is
dq = I(z) dt
I(z + dz) dt =
dI
dz dt = (t + dt) dz
dt
(t) dz =
d
d
dt dz )
=
dt
dt
dI
. X
dz
Since the left side is a function only of t, and the right side is a function only of z, they must both be constant;
call it k:
d
dI
= k ) (t) = kt + C1 ;
= k ) I(z) = kz + C2 .
dt
dz
If (0) = 0 and I(0) = 0 the constants C1 and C2 must both be zero: (t) = kt, I(z) = kz. X
kt
µ0 I ˆ
µ0 kz ˆ
(b) In the quasistatic approximation, E =
ŝ =
ŝ; B =
=
.
2⇡✏0 s
2⇡✏0 s
2⇡s
2⇡s
✓
◆
1 @
kt
r·E =
s
=0 X
s @s
2⇡✏0 s
✓
◆
1 @
µ0 kz
r·B =
=0 X
s@
2⇡s
@E ˆ 1 @E
@B
r⇥E =
ẑ = 0 =
X
@z
s@
@t
@B ˆ 1 @(sB)
µ0 k
@E
r⇥B =
+
ẑ =
ẑ = µ0 ✏0
X
@z
s @s
2⇡s
@t
For a gaussian cylinder of radius s about the z axis, at height z and with width dz:
I
I
kt
kt
dz
Qenc
E · da =
(2⇡s) dz =
dz =
=
. X
B · da = 0. X
2⇡✏0 s
✏0
✏0
✏0
For a circular amperian loop of radius s about the z axis, at height z:
I
Z
I
Z
d
µ0 kz
d
E · dl = 0 =
B · da, X
B · dl =
(2⇡s) = µ0 I = µ0 Ienc + µ0 ✏0
E · da. X
dt
2⇡s
dt
R
R
(Note that B · da and E · da are zero through this loop.)
Problem 7.60
(a) The continuity equation says @⇢
r·J. Here the right side is independent of t, so we can integrate:
@t =
⇢(t) = ( r·J)t+ constant. The “constant” may be a function of r—it’s only constant with respect to t. So,
putting in the r dependence explicitly, and noting that r·J = ⇢(r,
˙ 0), ⇢(r, t) = ⇢(r,
˙ 0)t + ⇢(r, 0). qed
R ⇢ r̂
R J⇥ r̂
µ0
1
(b) Suppose E = 4⇡✏0 r 2 d⌧ and B = 4⇡
r 2 d⌧ . We want to show that r·B = 0, r⇥B =
1
@B
µ0 J + µ0 ✏0 @E
;
r·E
=
⇢,
and
r⇥E
=
,
provided
that J is independent of t.
@t
✏0
@t ✓
◆
R
⇢ r̂
1
We know from Ch. 2 that Coulomb’s law E = 4⇡✏
d⌧
satisfies r·E = ✏10 ⇢ and r⇥E = 0. Since
0
r2
B is constant (in time), the r·E and r⇥E equations are satisfied. From Chapter 5 (specifically, Eqs. 5.475.50) we know that the Biot-Savart law satisfies r·B = 0. It remains only to check r⇥B. The argument in
Sect. 5.3.2 carries through until the equation following Eq. 5.54, where I invoked r0 ·J = 0. In its place we
now put r0 ·J = ⇢:
˙
Z
r̂
µ0
r⇥B = µ0 J
(J·r) 2 d⌧
(Eqs. 5.51-5.53)
r
4⇡
| {z }
( J·r0 ) rr̂ 2
(Eq. 5.54)
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165 1
CHAPTER 7. ELETRODYNAMICS
Integration by parts yields two terms, one of which becomes a surface integral, and goes to zero. The other is
r 0
r̂
r 3 r ·J = r 2 (
⇢).
˙ So:
µ0
4⇡
r⇥B = µ0 J
Z
r̂ 14
@
( ⇢)d⌧
˙
= µ0 J + µ0 ✏0
2
r
@t
Contents
Problem
Problem
7.61 7.56
1 ( )dz
(a) dEz =
sin ✓
4⇡✏0 r 2
p
z
2
2
sin ✓ =
r ;Z r = z + s

vt
z dz
1
p
Ez =
=
2 + s2 )3/2
4⇡✏0 (z
4⇡✏0
z 2 + s2 vt
Problem
7.26
(
)
1
1
p
p
Ez =
.
4⇡✏0
(vt ✏)2 + s2
(vt)2 + s2
⇢
1
4⇡✏0
Z
⇢ r̂
r
3
= µ0 J + µ0 ✏0
d⌧
@E
. qed
@t
y✻
dE
θ✼
r
✏
s
\$
vt − ϵ ! "#
−z
✲z
vt
(b)
E
=
=
4⇡✏0
2✏0
(c) Id = ✏0
Z
0
a
(
hp
(vt
p
(vt
d E
=
dt
2
1
✏)2 + s2
✏)2 + a2
(
p
p
1
(vt)2 + s2
(vt)2 + a2
v(vt ✏)
p
(vt ✏)2 + a2
)
(✏
2⇡s ds =
i
vt) + (vt) .
v(vt)
p
2✏0
(vt)2 + a2
+ 2v
)
hp
(vt
✏)2 + s2
ia
(vt)2 + s2
ŝ 0
✕
✲ ẑ
p
✲ ẑ
❖
φ̂
.
w
☛
❄
h
✻
✕
→I
→
As ✏ ! 0, vt < ✏ also ! 0, so Id ! 2 (2v) = v = I. With an infinitesimal gap we attribute the magnetic field
to displacement current, instead of real current, but we get the same answer. qed
Problem 7.62
✛
✲
l
B
✛
I
1
1 Q
Q
✏0 wl
✏0 w
(a) Parallel-plate capacitor: E =
; V = Eh =
h)C=
=
) C=
.
✏0
✏0 wl
V
h
h
(b) B = µ0 K = µ0
I
;
w
= Bhl =
µ0 I
µ0 h
µ0 h
hl = LI ) L =
l) L=
.
w
w
w
(c) CL = µ0 ✏0 = (4⇡ ⇥ 10 7 )(8.85 ⇥ 10 12 ) = 1.112 ⇥ 10 17 s2 /m2 .
p
p
(Propagation speed 1/ LC = 1/ µ0 ✏0 = 2.999 ⇥ 108 m/s = c.)
(d) D = , E = D/✏ = /✏, so just replace ✏0 by ✏;
H = K, B = µH = µK, so just replace µ0 by µ.
LC = ✏µ;
p
v = 1/ ✏µ.
c 2012
Education,
Inc.,Inc.,
Upper
River,
NJ. NJ.
All rights
reserved.
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material
is is
c Pearson
⃝2005
Pearson
Education,
Upper
River,
All rights
reserved.
material
protected
under
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as they
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No portion
of this
material
maymay
be be
protected
under
as they
currently
exist.
No portion
of this
material
reproduced,
in any
formform
or by
means,
without
permission
in writing
fromfrom
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reproduced,
in any
orany
by any
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c
⃝2005
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166
CHAPTER 7. ELETRODYNAMICS
Problem 7.63
(a) J = (E + v⇥B); J finite, = 1 ) E + (v⇥B) = 0. Take the curl: r⇥E + r⇥(v⇥B) = 0. But
@B
@B
. So
= r⇥(v⇥B). qed
@t
@t
H
(b) r·B = 0 ) B·da = 0 for any closed surface. Apply this at time (t + dt) to the surface consisting of
S, S 0 , and R:
Z
Z
Z
B(t + dt)·da +
B(t + dt)·da
B(t + dt)·da = 0
S0
R
S
(the sign change in the third term comes from switching outward da to inward da).
Z
Z
Z
Z
⇥
⇤
d =
B(t + dt)·da
B(t)·da =
B(t + dt) B(t) ·da
B(t + dt)·da
{z
}
S0
S
S |
R
@B
(for infinitesimal dt)
@t dt
d
=
⇢Z
S
@B
·da dt
@t
Z
R
⇥
⇤
B(t + dt)· (dl⇥v) dt (Figure 7.13).
Since the second term is already first order in dt, we can replace B(t + dt) by B(t) (the distinction would be
second order):
⇢Z ✓
◆
Z
I
Z
@B
@B
·da dt B·(dl⇥v) = dt
·da
r⇥(v⇥B)·da .
d = dt
{z }
@t
S @t
C|
S
S
(v⇥B)·dl
d
=
dt
Z 
S
@B
@t
r⇥(v⇥B) ·da = 0. qed
Problem 7.64
(a)
1
⇢e cos ↵ + cµ0 ⇢m sin ↵
✏0
1
1
1
1
(⇢e cos ↵ + cµ0 ✏0 ⇢m sin ↵) = (⇢e cos ↵ + ⇢m sin ↵) = ⇢0e . X
✏0
✏0
c
✏0
1
1
(r · B) cos ↵
(r · E) sin ↵ = µ0 ⇢m cos ↵
⇢e sin ↵
c
c✏0
1
µ0 (⇢m cos ↵
⇢e sin ↵) = µ0 (⇢m cos ↵ c⇢e sin ↵) = µ0 ⇢0m . X
cµ0 ✏0
✓
◆
✓
◆
@B
@E
(r ⇥ E) cos ↵ + c(r ⇥ B) sin ↵ =
µ0 Jm
cos ↵ + c µ0 Je + µ0 ✏0
sin ↵
@t
@t
✓
◆
@
1
@B0
µ0 (Jm cos ↵ cJe sin ↵)
B cos ↵
E sin ↵ = µ0 J0m
.X
@t
c
@t
✓
◆
✓
◆
1
@E
1
@B
(r ⇥ B) cos ↵
(r ⇥ E) sin ↵ = µ0 Je + µ0 ✏0
cos ↵
µ0 Jm
sin ↵
c
@t
c
@t
1
@
@E0
µ0 (Je cos ↵ + Jm sin ↵) + µ0 ✏0 (E cos ↵ + cB sin ↵) = µ0 J0e + µ0 ✏0
.X
c
@t
@t
r · E0 = (r · E) cos ↵ + c(r · B) sin ↵ =
=
r · B0 =
=
r ⇥ E0 =
=
r ⇥ B0 =
=
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CHAPTER 7. ELETRODYNAMICS
167
(b)
1
0
F0 = qe0 (E0 + v ⇥ B0 ) + qm
(B0
v ⇥ E0 )
2
c
✓
◆
✓
◆
1
1
= qe cos ↵ + qm sin ↵ (E cos ↵ + cB sin ↵) + v ⇥ B cos ↵
E sin ↵
c
c
✓
◆
1
1
+ (qm cos ↵ cqe sin ↵) B cos ↵
E sin ↵
v ⇥ (E cos ↵ + cB sin ↵)
c
c2
h
= qe E cos2 ↵ + cB sin ↵ cos ↵ cB sin ↵ cos ↵ + E sin2 ↵
✓
◆i
1
1
+v ⇥ B cos2 ↵
E sin ↵ cos ↵ + E sin ↵ cos ↵ + B sin2 ↵
c
c
◆
h✓ 1
1
+qm
E sin ↵ cos ↵ + B sin2 ↵ + B cos2 ↵
E sin ↵ cos ↵
c
c
✓
◆i
1
1
1
1
2
2
+v ⇥
B sin ↵ cos ↵
E
sin
↵
E
cos
↵
B
sin
↵
cos
↵
c
c2
c2
c
✓
◆
1
= qe (E + v ⇥ B) + qm B
v ⇥ E = F. qed
c2
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168
CHAPTER 8. CONSERVATION LAWS
Chapter 8
Conservation Laws
Problem 8.1
Example 7.13.
9
1 >
E=
ŝ >
>
2⇡✏0 s =
1
I 1
S=
(E ⇥ B) =
ẑ;
2 ✏ s2
>
µ
4⇡
0
0
µ0 I 1 ˆ >
>
;
B=
2⇡ s
Z
Zb
Zb
I
1
I
P = S · da = S2⇡s ds =
ds =
ln(b/a).
2⇡✏0
s
2⇡✏0
a
But V =
Zb
a
Problem 7.62.
E=
E · dl =
✏0
a
2⇡✏0
ẑ
Zb
a
1
ds =
ln(b/a), so P = IV.
s
2⇡✏0
9
>
>
=
S=
1
I
(E ⇥ B) =
ŷ;
µ0
✏0 w
>
µ0 I >
x̂ ;
w
Z
Z
Ih
P = S · da = Swh =
, but V = E · dl = h, so P = IV.
✏0
✏0
B = µ0 K x̂ =
Problem 8.2
(a) E =
✏0
ẑ;
=
Q
It
; Q(t) = It ) E(t) =
ẑ.
2
⇡a
⇡✏0 a2
@E 2
I⇡s2
µ0 Is ˆ
⇡s = µ0 ✏0
) B(s, t) =
.
@t
⇡✏0 a2
2⇡a2
"
#
✓
◆
✓
◆2
✓
◆2
⇤
1
1 2
1
It
1 µ0 Is
µ0 I 2 ⇥
2
(b) uem =
✏0 E +
B
=
✏0
+
=
(ct)2 + (s/2)2 .
2
µ0
2
⇡✏0 a2
µ0 2⇡a2
2⇡ 2 a4
✓
◆✓
◆
1
1
It
µ0 Is
I 2t
S=
(E ⇥ B) =
( ŝ) =
s ŝ.
2
2
µ0
µ0 ⇡✏0 a
2⇡a
2⇡ 2 ✏0 a4
B 2⇡s = µ0 ✏0
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169
CHAPTER 8. CONSERVATION LAWS
@uem
µ0 I 2 2
I 2t
I 2t
I 2t
@uem
=
2c t = 2 4 ;
r·S=
r · (s ŝ) = 2 4 =
.X
2
4
2
4
@t
2⇡ a
⇡ ✏0 a
2⇡ ✏0 a
⇡ ✏0 a
@t

Z
Z b
b
2
µ0 I 2
µ0 wI 2
1 s4
2s
(c) Uem = uem w2⇡s ds = 2⇡w 2 4
[(ct)2 + (s/2)2 ]s ds =
(ct)
+
2⇡ a 0
⇡a4
2
4 4 0

Z
µ0 wI 2 b2
b2
I 2t
I 2 wtb2
2
=
(ct)
+
.
Over
a
surface
at
b:
P
=
S
·
da
=
[b
ŝ
·
(2⇡bw
ŝ)]
=
.
in
2⇡a4
8
2⇡ 2 ✏0 a4
⇡✏0 a4
dUem
µ0 wI 2 b2 2
I 2 wtb2
=
2c t =
= Pin . X (Set b = a for total.)
4
dt
2⇡a
⇡✏0 a4
Problem 8.3
The force is clearly in the z direction, so we need
1
\$
( T · da)z = Tzx dax + Tzy day + Tzz daz =
µ0

1
1 2
=
Bz (B · da)
B daz .
µ0
2
✓
Bz Bx dax + Bz By day + Bz Bz daz
1 2
B daz
2
◆
2
µ0 m
ˆ (outside), where m = 4 ⇡R3 ( !R). (From
µ0 R! ẑ (inside) and B =
(2 cos ✓ r̂ + sin ✓ ✓)
3
4⇡r3
3
Eq. 5.70, Prob. 5.37, and Eq. 5.88.) We want a surface that encloses the entire upper hemisphere—say a
hemispherical cap just outside r = R plus the equatorial circular disk.
Now B =
Hemisphere:
Bz =
da =
B2 =
\$
( T · da)z =
=
=
i
⇥
⇤
µ0 m h
ˆ z = µ0 m 2 cos2 ✓ sin2 ✓ = µ0 m 3 cos2 ✓ 1 .
2
cos
✓
(r̂)
+
sin
✓
(
✓)
z
3
3
4⇡R
4⇡R
4⇡R3
µ0 m
2
2
R sin ✓ d✓ d r̂; B · da =
(2 cos ✓)R sin ✓ d✓ d ; daz = R2 sin ✓ d✓ d cos ✓;
4⇡R3
⇣ µ m ⌘2
⇣ µ m ⌘2
0
0
4 cos2 ✓ + sin2 ✓ =
3 cos2 ✓ + 1 .
3
3
4⇡R
4⇡R

1 ⇣ µ0 m ⌘2
1
3 cos2 ✓ 1 2 cos ✓R2 sin ✓ d✓ d
3 cos2 ✓ + 1 R2 sin ✓ cos ✓ d✓ d
3
µ0 4⇡R
2
✓
◆2 
!R
1 2
µ0
R sin ✓ cos ✓ d✓ d
12 cos2 ✓ 4 3 cos2 ✓ 1
3
2
✓
◆2
!R2
µ0
9 cos2 ✓ 5 sin ✓ cos ✓ d✓ d .
2
3
µ0
(Fhemi )z =
2
✓
= µ0 ⇡
✓
!R2
3
!R
3
◆2
2
2⇡
◆2 ✓
Z⇡/2
9 cos ✓
0
0+
5 cos ✓ sin ✓ d✓ = µ0 ⇡
3
9
4
5
2
◆
=
µ0 ⇡
4
✓
wR
3
2
◆2
✓
.
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
!R2
3
◆2 
9
5
cos4 ✓ + cos2 ✓
4
2
⇡/2
0
170
CHAPTER 8. CONSERVATION LAWS
Disk:
2
µ0 R!; da = r dr d ˆ = r dr d ẑ;
3
✓
◆2
2
2
2
B · da =
µ0 R!r dr d ; B =
µ0 R! ; daz = r dr d .
3
3
✓
◆2 
✓
◆2
1 2
1
1
2
\$
µ0 R!
r dr d + r dr d =
µ0 R! r dr d .
( T · da)z =
µ0 3
2
2µ0 3
✓
◆2 ZR
✓
◆
2
!R
!R2
(Fdisk )z = 2µ0
2⇡ r dr = 2⇡µ0
.
3
3
Bz =
0
Total:
F=
⇡µ0
✓
!R2
3
◆2 ✓
2+
1
4
◆
ẑ =
⇡µ0
✓
!R2
2
◆2
ẑ (agrees with Prob. 5.44).
Alternatively, we could use a surface consisting of the entire equatorial plane, closing it with a hemispherical
surface “at infinity,” where (since the field is zero out there) the contribution is zero. We have already done
the integral over the disk; it remains to do rest of the integral over the plane, from R to 1. On the plane,
µ0 m ˆ
µ0 m
✓ = 0, and (for r > R) B =
✓=
ẑ, so
Hello
4⇡r3
4⇡r3

1
1 2
1 2
1 ⇣ µ0 m ⌘2
\$
( T · da)z =
Bz (Bz daz )
Bz daz ) =
Bz daz =
( r dr d ).
µ0
2
2µ0
2µ0 4⇡r3
The contribution from the rest of the plane is therefore
✓
◆2 
Z 1
1 ⇣ µ0 m ⌘2
1
µ0 4 4
1
(Frest )z =
dr =
2⇡
⇡R !
5
4
2µ0 4⇡
r
16⇡
3
4r
R
✓
◆2
µ0 ⇡
!R2
=
.
4
3
1
=
µ0 ⇡
R
✓
R4 !
3
◆2 
1
4R4
This is the same as (Fhemi )z , so—when added to (Fdisk )z —it will yield the same total force as before.
Problem 8.4
(a)
\$
( T · da)z = Tzx dax + Tzy day + Tzz daz .
z
✻
r
But for the x y plane dax = day = 0, and daz =
a
r dr d (I’ll calculate
the upper charge).
✓ the force on ◆
r
✼θ
1
\$
E 2 ( r dr d ).
( T · da)z = ✏0 Ez Ez
✇
a
2
1
q
r
Now E =
2 2 cos ✓ r̂, and cos ✓ =
, so Ez =
x❂
r
r
4⇡✏
✓ 0 ◆2
2
q
r
0, E 2 =
3 . Therefore
2
2⇡✏0
(r + a2 ) ✓
◆2 Z1
Z1
1
q
r3 dr
q2 1
u du
(letting u ⌘ r2 )
Fz = ✏0
2⇡
3 = 4⇡✏ 2
2
2
2
2⇡✏0
(u
+ a2 )3
0
(r + a )
0
0
"
#1

q2 1
1
a2
q2 1
1
a2
q2
1
=
+
=
0+ 2
=
. X
3
2
4
4⇡✏0 2
(u + a ) 2 (u + a2 )
4⇡✏0 2
a
2a
4⇡✏0 (2a)2
✲y
0
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171
CHAPTER 8. CONSERVATION LAWS
1
q
a
2 2 sin ✓ ẑ, and sin ✓ =
r , so
4⇡✏0 r
◆2
✓
qa
1
\$
2
2
E = Ez =
3 , and hence ( T · da)z =
2
2
2⇡✏0
(r + a )
(b) In this case E =
Fz =
✏0
2
✓
qa
2⇡✏0
◆2
2⇡
Z1
0
r dr
3
(r2 + a2 )
=
q 2 a2
4⇡✏0
"
1
1
2
4 (r + a2 )2
✏0
2
#1
✓
qa
2⇡✏0
◆2
r dr d
Therefore
3.
(r2 + a2 )

q 2 a2
1
0+ 4 =
4⇡✏0
4a
=
0
q2
1
. X
4⇡✏0 (2a)2
Problem 8.5
(a) E =
ẑ, B = µ0 v x̂, g = ✏0 (E ⇥ B) = µ0 2 v ŷ, p = (dA)g = dAµ0 2 v ŷ.
✏0
(b) (i) There is a magnetic force, due to the (average) magnetic field at the upper plate:
F = q(u ⇥ B) = A[( u ẑ) ⇥ ( 12 µ0 v x̂)] = 12 µ0 2 Avu ŷ,
Z
Z
2
1
I1 = F dt = 2 µ0 Av ŷ u dt = 12 dµ0 2 Av ŷ.
[The velocity of the patch (of area A) is actually v + u = v ŷ u ẑ, but the y component produces a magnetic
force in the z direction (a repulsion of the plates) which reduces their (electrical) attraction but does not deliver
(horizontal) momentum to the plates.]
(ii) Meanwhile, in the space immediately above the upper plate the magnetic field drops abruptly to
zero (as the plate moves past), inducing an electric field by Faraday’s law. The magnetic field in the vicinity
of the top plate (at d(t) = d0 ut) can be written, using Problem 1.46(b),
B(z, t) =
µ0 v ✓(d
z) x̂,
)
@B
= µ0 vu (d
@t
z) x̂.
In the analogy at the beginning of Section 7.2.2, the Faraday-induced electric field is just like the magnetostatic
field of a surface current K =
vu x̂. Referring to Eq. 5.58, then,
(
1
2 µ0 vu ŷ, for z < d,
Eind =
+ 12 µ0 vu ŷ, for z > d.
This induced electric field exerts a force on area A of the bottom plate, F = (
an impulse
Z
I2 = 12 µ0 2 Av ŷ u dt = 12 µ0 2 Avd ŷ.
A)(
1
2 µ0
vu ŷ), and delivers
(I dropped the subscript on d0 , reverting to the original notation: d is the initial separation of the plates.)
The total impulse is thus I = I1 + I2 = dAµ0 2 v ŷ, matching the momentum initially stored in the fields,
from part (a). [I thank Michael Ligare for untangling this surprisingly subtle problem. Incidentally, there is
also “hidden momentum” in the original configuration. It is not relevant here; it is (relativistic) mechanical
momentum (see Example 12.13), and is delivered to the plates as they come together, so it does not a↵ect the
overall conservation of momentum.]
Problem 8.6
(a) gem = ✏0 (E ⇥ B) = ✏0 EB ŷ; pem = ✏0 EBAd ŷ.
Z 1
Z 1
Z 1
Z
(b) I =
F dt =
I(l ⇥ B) dt =
IBd(ẑ ⇥ x̂) dt = (Bd ŷ)
0
0
0
0
1
✓
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reproduced, in any form or by any means, without permission in writing from the publisher.
dQ
dt
◆
dt
172
= (Bdŷ)[Q(1)
CHAPTER 8. CONSERVATION LAWS
Q(0)] = BQd ŷ.
Problem 8.7
(a) Ex = Ey = 0, Ez =
But the original field was E = /✏0 = Q/✏0 A, so Q = ✏0 EA, and hence
/✏0 . Therefore
Txy = Txz = Tyz = · · · = 0;
\$
T =
(b) F =
I
✏0 2
E =
2
Txx = Tyy =
2
2✏0
;
✓
Tzz = ✏0 Ez2
1 2
E
2
◆
=
2
✏0 2
E =
.
2
2✏0
1
1 0 0
@ 0
1 0 A.
2✏0
0 0 +1
2
0
\$
T · da (S = 0, since B = 0); integrate over the x y plane: da =
dx dy ẑ (negative because
outward with respect to a surface enclosing the upper plate). Therefore
Fz =
Z
Tzz daz =
2
2✏0
A, and the force per unit area is f =
F
=
A
2
2✏0
ẑ.
2
(c) Tzz =
/2✏0 is the momentum in the z direction crossing a surface perpendicular to z, per unit
area, per unit time.
(d) The recoil force is the momentum delivered per unit time, so the force per unit area on the top plate is
2
f=
2✏0
ẑ
(same as (b)).
Problem 8.8
B = µ0 nI ẑ (for s < R; outside the solenoid B = 0). The force on a segment ds of spoke is
dF = I 0 dl ⇥ B = I 0 µ0 nI ds(ŝ ⇥ ẑ) =
I 0 µ0 nI ds ˆ.
The torque on the spoke is
N=
Z
r ⇥ dF = I µ0 nI
0
ZR
a
1
s ds( ŝ ⇥ ˆ) = I 0 µ0 nI R2
2
Z
1
µ0 nI(R2
2
a2 ( ẑ).
Therefore the angular momentum of the cylinders is L = N dt =
a ) ẑ
so
1
L=
µ0 nIQ(R2 a2 ) ẑ (in agreement with Eq. 8.34).
2
Problem 8.9
Q 1
(a) Between the shells, E =
r̂,
4⇡✏0 r2
L=
Z
(r ⇥ g) d⌧ =
QB0
4⇡
Z
g = ✏0 (E ⇥ B) =
2
Z
I 0 dt. But
R
I 0 dt = Q,
QB0
(r̂ ⇥ ẑ).
4⇡r2
1
QB0
[r ⇥ (r̂ ⇥ ẑ)]r2 sin ✓ dr d✓ d =
r2
4⇡
Z
r[r̂ ⇥ (r̂ ⇥ ẑ)] sin ✓ dr d✓ d .
protected under all copyright laws as they currently exist. No portion of this material may be
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173
CHAPTER 8. CONSERVATION LAWS
Now r̂ ⇥ (r̂ ⇥ ẑ) = r̂(r̂ · ẑ) ẑ(r̂ · r̂) = r̂ cos ✓ = ẑ, but L has to be along the z direction, so we pick o↵ the z
component of r̂: [r̂ ⇥ (r̂ ⇥ ẑ)]z = cos2 ✓ 1 = sin2 ✓.
Lz =
QB0
4⇡
Z
r sin ✓ dr d✓ d =
3
QB0
2⇡
4⇡
Z
⇡
sin ✓ d✓
3
0
Z
b
QB0
2
r dr =
a
1
QB0 (b2
3
L=
✓ ◆✓ 2
◆
4
b
a2
=
3
2
1
QB0 (b2 a2 ).
3
a2 ) ẑ.
(b) The changing magnetic field induces an electric field (Example 7.7). Assuming symmetry about the z
s dB ˆ
axis, E =
. The force on a patch of area da is dF = E da, and the torque on this patch is dN = s⇥dF.
2 dt
The net torque on the sphere at radius a is (using s = a sin ✓ ŝ and da = a2 sin ✓ d✓ d ):
✓
◆✓
◆Z
Z
Q
1 dB
Qa2 dB
Qa2 dB
2
Na =
ẑ
s
da
=
ẑ
2⇡
sin3 ✓ d✓ =
ẑ.
2
4⇡a
2 dt
8⇡ dt
3 dt
Qb2 dB
Q dB 2
Similarly, Nb =
ẑ, so the total torque is N =
(b
3 dt
3 dt
to the spheres is
Z
Z 0
Q 2
dB
2
L = N dt = (b
a ) ẑ
dt =
3
B0 dt
Problem 8.10
(a)
(where m =
8
0,
>
<
a2 ) ẑ, and the angular momentum delivered
Q 2
(b
3
X
a2 )B0 ẑ.
9
(r < R) >
=
82
9
(r < R) >
>
< 3 µ0 M ẑ,
=
; B=
(Ex. 6.1)
E=
h
i
>
>
: 1 Q r̂, (r > R) >
;
: µ0 m 2 cos ✓ r̂ + sin ✓ ✓ˆ , (r > R) >
;
4⇡ r3
4⇡✏0 r2
4 3
µ0 Qm
ˆ sin ✓, and (r̂ ⇥ ✓)
ˆ = ˆ, so
⇡R M ); g = ✏0 (E ⇥ B) =
(r̂ ⇥ ✓)
3
(4⇡)2 r5
`=r⇥g =
But (r̂ ⇥ ˆ) =
µ0 mQ
L=
ẑ
(4⇡)2
µ0 mQ
sin ✓(r̂ ⇥ ˆ).
(4⇡)2 r4
ˆ and only the z component will survive integration, so (since (✓)
ˆz=
✓,
Z
sin2 ✓ 2
r sin ✓ dr d✓ d
r4
.
Z2⇡
0
µ0 mQ
L=
ẑ (2⇡)
(4⇡)2
d = 2⇡;
Z⇡
4
sin ✓ d✓ = ;
3
3
0
✓ ◆✓ ◆
4
1
=
3
R
Z1
R
2
µ0 M QR2 ẑ.
9
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
sin ✓):
1
dr =
r2
✓
1
r
◆
1
R
=
1
.
R
a
✼θ
✇
r
a
✲x
y❂
−q
174
CHAPTER 8. CONSERVATION LAWS
z
✻
(b) Apply Faraday’s law to the ring shown:
✓
◆
I
d
2 dM
E · dl = E(2⇡r sin ✓) =
= ⇡(r sin ✓)2
µ0
dt
3
dt
) E=
θR
µ0 dM
(r sin ✓) ˆ.
3 dt
✓
µ0 dM
The force on a patch of surface (da) is dF = E da =
(r sin ✓) da ˆ
z
3 dt
✻
µ0 dM 2
✶
The torque on the′ patch is✸dN = r ⇥ dF =
r sin ✓ da (r̂ ⇥ ˆ). But
r
3 dt
only the
qmz component (✓ˆz = sin ✓):
Z
r
µ0 dM
d
N=
ẑ r2 sin2 ✓ r2 sin ✓ d✓ d .
θ
3 dt
✲x
qe
✓ ◆
Z⇡
Z2⇡
4
µ0 dM
4
Here r = R;
sin3 ✓ d✓ = ;
d = 2⇡, so N =
ẑ R4
(2⇡) =
3
3 dt
3
0
◆
Q
.
4⇡R2
ˆ and we want
(r̂ ⇥ ˆ) = ✓,
=
2µ0
dM
QR2
ẑ.
9
dt
0
L=
Z
2µ0
QR2 ẑ
9
N dt =
Z0
dM =
2µ0
M QR2 ẑ
9
(same as (a)).
M
(c) Let the charge on the sphere at time t be q(t); the charge density is
=
(“south of”) the ring in the figure is
qs =
2⇡R
2
Z⇡
sin ✓0 d✓0 =
q(t)
. The charge below
4⇡R2
q
q
⇡
( cos ✓0 )|✓ = (1 + cos ✓).
2
2
0
So the total current crossing the ring (flowing “north”) is I(t) =
1 dq
(1 + cos ✓), and hence
2 dt
I
ˆ = 1 dq (1 + cos ✓) ✓.
ˆ The force on a patch of area da is dF = (K ⇥ B) da.
( ✓)
2⇡R sin ✓
4⇡R dt
sin ✓

4
c
⃝2005
Pearson Education,
Thisµmaterial
is
⇡R3River,
2 Inc., Upper
0M
ˆ this1material
ˆ
protected under
portion
may
µ0 Mlaws
ẑ +as they3 currently
(2exist.
cos ✓Nor̂ +
sin ✓of✓)
=
[2 ẑbe+ 2 cos ✓ r̂ + sin ✓ ✓];
3
3
4⇡
R
2
6
reproduced, in any form or by any means, without permission in writing from the publisher.
K(t) =
K⇥B=
1 dq µ0 M (1 + cos ✓) ˆ
[2(✓ ⇥ ẑ) + 2 cos ✓ (✓ˆ ⇥ r̂)].
| {z }
4⇡R dt 6
sin ✓
ˆ
dN = R r̂ ⇥ dF =
=
µ0 M
12⇡
✓
dq
dt
◆
µ0 M
24⇡
✓
dq
dt
◆
(1 + cos ✓)
2[ r̂ ⇥ (✓ˆ ⇥ ẑ)
|
{z
}
sin ✓
ˆ · ẑ) ẑ(r̂ · ✓)
ˆ
✓(r̂
cos ✓ (r̂ ⇥ ˆ)]R2 sin ✓ d✓ d
| {z }
2
ˆ d✓ d = µ0 M R
(1 + cos ✓)R2 [cos ✓ ✓ˆ + cos ✓ ✓]
6⇡
ˆ
✓
✓
dq
dt
◆
ˆ
(1 + cos ✓) cos ✓ d✓ d ✓.
protected under all copyright laws as they currently exist. No portion of this material may be
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175
CHAPTER 8. CONSERVATION LAWS
ˆz=
The x and y components integrate to zero; (✓)
sin ✓, so (using
Z2⇡
d = 2⇡):
0
Nz =
=
µ0 M R2
6⇡
µ0 M R2
3
✓
✓
dq
dt
◆
(2⇡)
Z⇡
(1 + cos ✓) cos ✓ sin ✓ d✓ =
µ0 M R2
3
0
◆✓ ◆
dq
2
=
dt
3
2µ0
dq
M R2 .
9
dt
✓
dq
dt
◆✓
sin2 ✓
2
cos3 ✓
3
◆
⇡
0
2µ0
dq
M R2 ẑ.
9
dt
N=
Therefore
L=
Z
N dt =
2µ0
M R2 ẑ
9
Z0
dq =
2µ0
M R2 Q ẑ (same as (a)).
9
Q
(I used the average field at the discontinuity—which is the correct thing to do—but in this case you’d get the
same answer using either the inside field or the outside field.)
µ0 Ib
b2
ẑ. (I put the
2 (b2 + z 2 )3/2
origin at the center of the upper loop, with the z axis pointing up, so z = h at the location of the lower loop.)
Treat the lower loop as a magnetic dipole, with moment m = Ia ⇡a2 ẑ. The force on m is given by Eq. 6.3:
✓
◆
µ0 Ib
b2
µ0 ⇡a2 b2 Ia Ib @ 2
3/2
F = r(m · B) = r Ia ⇡a2
=
b + z2
ẑ
2
2
3/2
2 (b + z )
2
@z
✓
◆
µ0 ⇡a2 b2 Ia Ib
3
3⇡
a2 b2 h
5/2
=
b2 + z 2
(2z) ẑ =
µ0 Ia Ib 2
ẑ. X
2
2
2
(b + h2 )5/2
Problem 8.11 The magnetic field of the upper loop is given by Eq. 5.41: B =
Problem 8.12 Following the method of Section 7.2.4 (leading up to Eq. 7.30), the power delivered to the two
loops is
dW
= Ea Ia Eb Ib
dt
(where these are the changing values as the currents are turned on, not the final values). Now
Ea =
so
dW
=
dt
✓
La Ia
La
dIa
dIb
+ M Ia
dt
dt
◆
dIa
dt
M
dIb
,
dt
Eb =
Lb
dIb
dt
M
dIa
,
dt
✓
◆
✓
◆
dIb
dIa
d 1
1
+ Lb Ib
+ M Ib
=
La Ia2 + Lb Ib2 + M Ia Ib .
dt
dt
dt 2
2
The total work done, then, to increase the currents from zero to their final values, is
W =
Problem 8.13
d
(a) E =
;
dt
1
1
La Ia2 + Lb Ib2 + M Ia Ib . qed
2
2
= ⇡a2 B; B = µ0 nIs ; E = Ir R. So Ir =
1
dIs
µ0 ⇡a2 n
.
R
dt
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reproduced, in any form or by any means, without permission in writing from the publisher.
176
(b)
I
E · dl =
S=
CHAPTER 8. CONSERVATION LAWS
d
) E(2⇡a) =
dt
1
1
(E ⇥ B) =
µ0
µ0
✓
1
dIs ˆ
µ0 Ir
b2
µ0 an
. B=
ẑ (Eq. 5.41).
2
dt
2 (b2 + z 2 )3/2
!
µ0 Ir
b2
1
dIs
ab2 n
( ˆ ⇥ ẑ) =
µ0 Ir
ŝ.
3/2
2 (b2 + z 2 )
4
dt (b2 + z 2 )3/2
µ0 ⇡a2 n
µ0 an dIs
2 dt
◆
dIs
)E=
dt
Power:
Z
Z1
Z1
1
dIs
1
2 2
P =
S · da =
(S)(2⇡a) dz =
⇡µ0 a b nIr
dz
3/2
2
dt
(b2 + z 2 )
1
1
✓
◆
1
z
1
1
2
The integral is p
= 2
= 2.
2
2
2
2
b
b
b
1
b z +b
✓
◆
dI
s
=
⇡µ0 a2 n
Ir = (RIr )Ir = Ir 2 R. qed
dt
Problem 8.14 The fields are zero for s < a and s > b; between the cylinders,
E=
1
ŝ,
2⇡✏0 s
B=
µ0 I ˆ
,
2⇡ s
where I = v.
(a) From Eq. 8.5:
1
u=
2
✓
1 2
✏0 E +
B
µ0
2
◆
" ✓
#
◆2 2
2
1
1
1 ⇣ µ0 ⌘2 2 v 2
1
=
✏0
+
=
(1 + ✏0 µ0 v 2 ) 2 .
2
2
2
2
2⇡✏0
s
µ0 2⇡
s
8⇡ ✏0
s
Writing ✏0 µ0 = 1/c2 , and integrating over the volume between the cylinders:
2
W
=
`
8⇡ 2 ✏0
✓
v2
1+ 2
c
◆Z
a
(b) From Eq. 8.29:
g = ✏0 (E ⇥ B) = ✏0
p
µ0 2 v
=
ẑ
`
4⇡ 2
(c) From Eq. 8.10:
S=
b
Z
a
b
2
1
2⇡s
ds
=
s2
4⇡✏0
✓
2⇡✏0 s
1 q
r̂,
4⇡✏0 r2
B=
µ0 v
2⇡s
◆
v2
1+ 2
c
ẑ =
◆
ln(b/a).
µ0 2 v
ẑ.
4⇡ 2 s2
1
µ0 2 v
2⇡s ds =
ln(b/a) ẑ.
2
s
2⇡
1
1
(E ⇥ B) =
g = c2 g.
µ0
✏0 µ0
Problem 8.15
(a) The fields are E =
◆✓
✓
dW
=
dt
Z
S · da =
µ0
2 2
c v
ln(b/a) ẑ.
2⇡
µ0 N I ˆ
(for points inside the toroid—see Eq. 5.60).
2⇡ s
g = ✏0 (E ⇥ B) = ✏0
q
µ0 N I
µ0 qN I
ẑ =
ẑ.
4⇡✏0 a2 2⇡a
8⇡ 2 a3
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177
CHAPTER 8. CONSERVATION LAWS
(Note that inside the toroid r ⇡ a ŝ and s ⇡ a.)
p=
✓
◆
µ0 qN I
µ0 qN Iwh
ẑ (2⇡awh) =
ẑ.
2
3
8⇡ a
4⇡ a2
(b) The changing magnetic field induces an electric field, as given by Problem 7.19 (with z = 0):
µ0
dI
E=
N hw ẑ. The impulse delivered to q is therefore
4⇡a2
dt
Z
Z
Z
µ0
dI
µ0 q
I = F dt = qE dt = q
N
hw
ẑ
dt =
N hwI ẑ. X
4⇡a2
dt
4⇡a2
Problem 8.16
According to Eqs. 3.104, 4.14, 5.89, and 6.16, the fields are
8
9
8
9
1
2
>
>
>
>
>
>
P,
(r
<
R),
>
>
>
>
µ
M,
(r
<
R),
0
< 3✏0
=
<3
=
E=
B=
>
>
>
>
>
>
>
1 1
>
>
: µ0 m [3(m · r̂) r̂ m] , (r > R), >
;
:
;
[3(p
·
r̂)
r̂
p],
(r
>
R),
3
4⇡ r
4⇡✏0 r3
R
where p = (4/3)⇡R3 P, and m = (4/3)⇡R3 M. Now p = ✏0 (E ⇥ B) d⌧ , and there are two contributions, one
from inside the sphere and one from outside.
Inside:
◆ ✓
◆
Z ✓
Z
1
2
2
2
4
8
pin = ✏0
µ0 M d⌧ =
µ0 (P ⇥ M) d⌧ =
µ0 (P ⇥ M) ⇡R3 =
µ0 ⇡R3 (M ⇥ P).
P ⇥
3✏0
3
9
9
3
27
Outside:
pout = ✏0
1 µ0
4⇡✏0 4⇡
Z
1
{[3(p · r̂) r̂
r6
p] ⇥ [3(m · r̂) r̂
m]} d⌧.
Now r̂⇥(p⇥m) = p(r̂·m) m(r̂·p), so r̂⇥[r̂⇥(p⇥m)] = (r̂·m)(r̂⇥p) (r̂·p)(r̂⇥m), whereas using the BACCAB rule directly gives r̂⇥[r̂⇥(p⇥m)] = r̂[r̂·(p⇥m)] (p⇥m)(r̂·r̂). So {[3(p · r̂) r̂ p] ⇥ [3(m · r̂) r̂ m]} =
3(p·r̂)(r̂⇥m)+3(m·r̂)(r̂⇥p)+(p⇥m) = 3 {r̂[r̂ · (p ⇥ m)] (p ⇥ m)}+(p⇥m) = 2(p⇥m)+3r̂[r̂·(p⇥m)].
Z
µ0
1
pout =
{ 2(p ⇥ m) + 3 r̂[r̂ · (p ⇥ m)]} r2 sin ✓ dr d✓ d .
2
16⇡
r6
To evaluate the integral, set the z axis along (p ⇥ m); then r̂ · (p ⇥ m) = |p ⇥ m| cos ✓. Meanwhile, r̂ =
sin ✓ cos x̂ + sin ✓ sin ŷ + cos ✓ ẑ. But sin and cos integrate to zero, so the x̂ and ŷ terms drop out, leaving
✓Z 1
◆⇢
Z
Z
µ0
1
pout =
dr
2(p
⇥
m)
sin
✓
d✓
d
+
3|p
⇥
m|
ẑ
cos2 ✓ sin ✓ d✓ d
16⇡ 2
r4
0
✓
◆1
µ0
1
4⇡
µ0
=
2(p ⇥ m)4⇡ + 3(p ⇥ m)
=
(p ⇥ m)
16⇡ 2
3r3 R
3
12⇡R3
✓
◆ ✓
◆
µ0
4 3
4 3
4µ0 3
=
⇡R
P
⇥
⇡R
M
=
R (M ⇥ P).
12⇡R3 3
3
27
✓
◆
8
4
4
ptot =
+
µ0 R3 (M ⇥ P) = µ0 R3 (M ⇥ P).
27 27
9
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178
CHAPTER 8. CONSERVATION LAWS
Problem 8.17
(a) From Eq. 5.70 and Prob. 5.37,
8
2
e
>
< r < R : E = 0, B = µ0 R! ẑ, with =
;
3
4⇡R2
1 e
µ0 m
>
ˆ with m = 4 ⇡ !R4 .
:r > R : E =
r̂, B =
(2 cos ✓ r̂ + sin ✓ ✓),
4⇡✏0 r2
4⇡ r3
3
The energy stored in the electric field is (Ex. 2.9):
WE =
1 e2
.
8⇡✏0 R
The energy density of the internal magnetic field is:
✓
◆2
1 2
1
2
e
µ0 ! 2 e2
µ0 ! 2 e2 4 3
µ0 e2 ! 2 R
uB =
B =
µ0 R!
=
,
so
W
=
⇡R
=
.
B
in
2µ0
2µ0 3
4⇡R2
72⇡ 2 R2
72⇡ 2 R2 3
54⇡
The energy density in the external magnetic field is:
uB =
WBout
1 µ20 m2
e2 ! 2 R4 µ0 1
4 cos2 ✓ + sin2 ✓ =
3 cos2 ✓ + 1 , so
2
6
2µ0 16⇡ r
18(16⇡ 2 ) r6
µ0 e2 ! 2 R4
=
(18)(16)⇡ 2
Z1
1 2
r dr
r6
Z⇡
3 cos ✓ + 1 sin ✓ d✓
2
0
R
WB = WBin + Wbout =
Z2⇡
d =
µ0 e2 ! 2 R4
(18)(16)⇡ 2
0
✓
1
3R3
◆
(4)(2⇡) =
µ0 e2 ! 2 R
.
108⇡
µ0 e ! R
µ0 e ! R
1 e2
µ0 e2 ! 2 R
(2 + 1) =
; W = WE + WB =
+
.
108⇡
36⇡
8⇡✏0 R
36⇡
2
2
2
(b) Same as Prob. 8.10(a), with Q ! e and m !
2
1
µ0 e2 !R
e!R2 : L =
ẑ.
3
18⇡
µ0 e2
~
9⇡~
(9)(⇡)(1.05 ⇥ 10 34 )
!R = ) !R =
=
= 9.23 ⇥ 1010 m/s.
2
18⇡
2
µ0 e
(4⇡ ⇥ 10 7 )(1.60 ⇥ 10 19 )2
"
"
✓
◆2 #
✓
◆2 #
✓
◆2
1 e2
2 !R
2 !R
2 9.23 ⇥ 1010
2
1+
= mc ; 1 +
=1+
= 2.10 ⇥ 104 ;
8⇡✏0 R
9
c
9
c
9
3 ⇥ 108
(c)
(2.10 ⇥ 104 )(1.6 ⇥ 10 19 )2
9.23 ⇥ 10 10
11
=
2.95
⇥
10
m;
!
=
8⇡(8.85 ⇥ 10 12 )(9.11 ⇥ 10 31 )(3 ⇥ 108 )2
2.95 ⇥ 10 11
Since !R, the speed of a point on the equator, is 300 times the speed of light, this “classical” model is clearly
unrealistic.
Problem 8.18
Maxwell’s equations with magnetic charge (Eq. 7.44):
R=
(i) r · E =
1
⇢e ,
✏0
(ii) r · B = µ0 ⇢m ,
(iii) r ⇥ E =
µ0 Jm
@B
,
@t
(iv) r ⇥ B = µ0 Je + µ0 ✏0
@E
.
@t
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179
CHAPTER 8. CONSERVATION LAWS
The Lorentz force law becomes (Eq. 8.44)
✓
F = qe (E + v ⇥ B) + qm B
Following the argument in Section 8.1.2:

✓
F · dl = qe (E + v ⇥ B) + qm B
dW
=
dt
Z
◆
1
v
⇥
E
.
c2
◆
1
v ⇥ E · v dt = (qe E + qm B) · v dt,
c2
(E · Je + B · Jm ) d⌧
(which generalizes Eq. 8.6). Use (iii) and (iv) to eliminate Je and Jm :
(E · Je + B · Jm ) =
but r · (E ⇥ B) = B · (r ⇥ E)
1
E · (r ⇥ B)
µ0
✏0 E ·
@E
@t
1
B · (r ⇥ E)
µ0
E · (r ⇥ B), so
1
r · (E ⇥ B)
µ0
(E · Je + B · Jm ) =
1
@B
B·
,
µ0
@t
1 @
2 @t
✓
1 2
✏0 E +
B
µ0
1
µ0
I
(E ⇥ B) · da,
2
◆
(generalizing Eq. 8.8). Thus
dW
=
dt
d
dt
Z
1
2
✓
✏0 E 2 +
1 2
B
µ0
◆
d⌧
which is identical to Eq. 8.9. Evidently Poynting’s theorem is unchanged(!), and
u=
1
2
✓
✏0 E 2 +
◆
1 2
B ,
µ0
S=
1
(E ⇥ B),
µ0
the same as before.
To construct the stress tensor we begin with the generalization of Eq. 8.13:
✓
◆
Z 
1
F=
(E + v ⇥ B) ⇢e + B
v ⇥ E ⇢m d⌧.
c2
The force per unit volume (Eq. 8.14)
✓
f = (⇢e E + Je ⇥ B) + ⇢m B
✓
1
= ✏0 (r · E)E +
r⇥B
µ0
becomes
◆
1
Jm ⇥ E
c2
◆
✓
◆
@E
1
1
1 @B
✏0
⇥B+
(r · B)B µ0 ✏0
r⇥E
⇥E
@t
µ0
µ0
µ0 @t
1
@
= ✏0 [(r · E)E E ⇥ (r ⇥ E)] +
[(r · B)B B ⇥ (r ⇥ B)] ✏0 (E ⇥ B)
µ0
@t


1
1
1
@
= ✏0 (r · E)E
r(E 2 ) + (E · r)E +
(r · B)B
r(B 2 ) + (B · r)B
✏0 (E ⇥ B).
2
µ0
2
@t
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z
✻
+q
a
r
a
✼θ
✇
y❂
CHAPTER 8. CONSERVATION
LAWS
−q
180
This is identical to Eq. 8.16, so the stress tensor is the same as before:
✓
Tij ⌘ ✏0 Ei Ej
1
2
ij E
2
◆
1
+
µ0
✓
1
2
Bi Bj
ij B
2
◆
.
Likewise, Eq. 8.20 is still valid. In fact, this argument is more straightforward when you include magnetic
charge, since you don’t need artificially to insert the (r · B)B term (after Eq. 8.15).
The electromagnetic momentum density (Eq. 8.29) also stays the same, since the argument in Section 8.2.3
is formulated entirely in terms of the fields:
g = ✏0 (E ⇥ B).
Problem 8.19
z
qe r
E=
;
4⇡✏0 r3
µ0 qm r0
µ0 qm
(r d ẑ)
B=
=
.
3
2
2
0
4⇡ r
4⇡ (r + d
2rd cos ✓)3/2
✻
✶
✸
r′
qm
d
r
θ
✲x
qe
Momentum density (Eq. 8.32):
g = ✏0 (E ⇥ B) =
µ0 qe qm
( d)(r ⇥ ẑ)
.
2
3
2
(4⇡) r (r + d2 2rd cos ✓)3/2
Angular momentum density (Eq. 8.33):
` = (r ⇥ g) =
µ0 qe qm d
r ⇥ (r ⇥ ẑ)
. But r ⇥ (r ⇥ ẑ) = r(r · ẑ)
(4⇡)2 r3 (r2 + d2 2rd cos ✓)3/2
r2 ẑ = r2 cos ✓ r̂
r2 ẑ.
The x and y components will integrate to zero; using (r̂)z = cos ✓, we have:
Z
µ0 qe qm d
r2 (cos2 ✓ 1)
L=
ẑ
r2 sin ✓ dr d✓ d . Let u ⌘ cos ✓ :
2
3/2
3
2
2
(4⇡)
r (r + d
2rd cos ✓)
Z1 Z1
r 1 u2
µ0 qe qm d
=
ẑ
(2⇡)
du dr.
3/2
(4⇡)2
(r2 + d2 2rdu)
1 0
Do the r integral first:
Z1
0
r dr
(r2 + d2
3/2
2rdu)
=
d(1
(ru d)
p
2
u ) r 2 + d2
1
2rdu
0
=
u
d (1
u2 )
+
c
⃝2005
Pearson Education, Inc., Upper Saddle River, NJ. A
protected under all copyright laws as they currently exist. N
d
u +or1 by any means,
1 without permissio
reproduced, in any form
=
=
.
2
2
d (1 u ) d
d (1 u )
d(1 u)
Then
µ0 qe qm d 1
L=
ẑ
8⇡
d
Z1
1
(1 u2 )
µ0 qe qm
du =
ẑ
(1 u)
8⇡
Z1
1
(1 + u)du =
µ0 qe qm
ẑ
8⇡
✓
u+
u2
2
◆
1
=
1
µ0 qe qm
ẑ.
4⇡
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181
CHAPTER 8. CONSERVATION LAWS
Problem 8.20
Z
Z
Z
p = ✏0 (E ⇥ B) d⌧ = ✏0 (rV ) ⇥ B d⌧ = ✏0
[r ⇥ (V B)
V
V
V
I
Z
Z
1
= ✏0
V B ⇥ da + ✏0 µ0
V J d⌧ = 2
V J d⌧.
c V
S
V
V r ⇥ B] d⌧
(I used Problem 1.61(b) in the penultimate step. Here V is all of space, and S is its surface at infinity, where
B = 0, so the surface integral vanishes.) Using V (r) ⇡ V (0) + (rV ) · r = V (0) E(0) · r,
Z
Z
1
1
p = 2 V (0) J d⌧
[E(0) · r]J d⌧.
c
c2
R
R
R
For a current loop, J d⌧ ! I dl = I dl = 0, and (Eq. 1.108):
Z
Z
Z
[E(0) · r]J d⌧ ! [E(0) · r]I dl = I [E(0) · r] dl = Ia ⇥ E(0) = m ⇥ E.
So
p=
1
(m ⇥ E). X
c2
Problem 8.21
(a) The rotating shell at radius b produces a solenoidal magnetic field:
B = µ0 K ẑ, where K =
b !b b,
and
b
Q
. So B =
2⇡bl
=
µ0 !b Q
ẑ (a < s < b).
2⇡l
(Note that if angular velocity is defined with respect to the z axis, then !b is a negative number.) The shell at
a also produces a magnetic field (µ0 !a Q/2⇡l) ẑ, in the region s < a, so the total field inside the inner shell is
B=
µ0 Q
(!a
2⇡l
!b ) ẑ, (s < a).
Meanwhile, the electric field is
1
Q
ŝ =
ŝ, (a < s < b).
2⇡✏0 s
2⇡✏0 ls
✓
◆✓
◆
Q
µ0 !b Q
µ0 !b Q2 ˆ
µ0 !b Q2
g = ✏0 (E ⇥ B) = ✏0
(ŝ ⇥ ẑ) =
;
`
=
r
⇥
g
=
(r ⇥ ˆ).
2⇡✏0 ls
2⇡l
4⇡ 2 l2 s
4⇡ 2 l2 s
E=
Now r ⇥ ˆ = (s ŝ + z ẑ) ⇥ ˆ = s ẑ
L=
µ0 !b Q2
ẑ
4⇡ 2 l2
z ŝ, and the ŝ term integrates to zero, so
Z
d⌧ =
µ0 !b Q2
⇡(b2
4⇡ 2 l2
a2 )l ẑ =
µ0 !b Q2 (b2
4⇡l
a2 )
ẑ.
(b) The extra electric field induced by the changing magnetic field due to the rotating shells is given by
d
1 d ˆ
E 2⇡s =
)E=
, and in the region a < s < b
dt
2⇡s dt
✓
◆
µ0 Q
µ0 Q!b
µ0 Q
1 µ0 Q
d!a
d!b ˆ
=
(!a !b ) ⇡a2
⇡ s2 a2 =
!a a2 !b s2 ; E(s) =
a2
s2
.
2⇡l
2⇡l
2l
2⇡s 2l
dt
dt
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182
CHAPTER 8. CONSERVATION LAWS
In particular,
µ0 Qa
4⇡l
E(a) =
✓
d!a
dt
d!b
dt
◆
ˆ,
and E(b) =
µ0 Q
4⇡lb
✓
2 d!a
a
dt
b
2 d!b
dt
◆
ˆ.
The torque on a shell is N = r ⇥ qE = qsE ẑ, so
✓
◆✓
◆
Z 1
µ0 Qa
d!a
d!b
µ0 Q2 a2
Na = Qa
ẑ; La =
Na dt =
(!a !b ) ẑ.
4⇡l
dt
dt
4⇡l
0
✓
◆✓
◆
Z 1
µ0 Q
d!a
d!b
µ0 Q2 2
Nb = Qb
a2
b2
ẑ; Lb =
Nb dt =
a !a b2 !b ẑ.
4⇡lb
dt
dt
4⇡l
0
Ltot = La + Lb =
µ0 Q2 2
a !a
4⇡l
a2 !a + a2 !b ẑ =
b2 ! b
µ0 Q2 !b 2
(b
4⇡l
a2 ) ẑ.
Thus the reduction in the final mechanical angular momentum (b) is equal to the residual angular momentum
in the fields (a). X
Problem 8.22
B = µ0 nI ẑ, (s < R);
E=
q
4⇡✏0
r
r 3 , where r
g = ✏0 (E ⇥ B) = ✏0 (µ0 nI)
✓
q
4⇡✏0
◆
= (x
1
r
3
a, y, z).
( r ⇥ ẑ) =
µ0 qnI
[y x̂
4⇡ r 3
(x
a) ŷ].
Linear Momentum.
Z
Z
y x̂ (x a) ŷ
dx dy dz. The x̂ term is odd in y; it integrates to zero.
[(x a)2 + y 2 + z 2 ]3/2
µ0 qnI
(x a)
=
ŷ
dx dy dz. Do the z integral first :
4⇡
[(x a)2 + y 2 + z 2 ]3/2
1
z
2
p
=
.
[(x a)2 + y 2 ]
[(x a)2 + y 2 ] (x a)2 + y 2 + z 2 1
Z
µ0 qnI
(x a)
=
ŷ
dx dy. Switch to polar coordinates :
2⇡
[(x a)2 + y 2 ]
p=
g d⌧ =
µ0 qnI
4⇡
Z
x = s cos , y = s sin , dx dy ) s ds d ; [(x a)2 + y 2 ] = s2 + a2 2sa cos .
Z
µ0 qnI
(s cos
a)
=
ŷ
s ds d
2⇡
(s2 + a2 2sa cos )
✓
◆
Z 2⇡
Z 2⇡
cos d
2⇡
A
d
2⇡
Now
=
1 p
;
=p
.
2
2
2
(A
+
B
cos
)
B
(A
+
B
cos
)
A
B
A
B2
0
0
p
Here A2 B 2 = (s2 + a2 )2 4s2 a2 = s4 + 2s2 a2 + a4 4s2 a2 = (s2 a2 )2 ; A2 B 2 = a2
✓ 2
◆
Z 
Z R
µ0 qnI
a + s2
2a2
µ0 qnI
µ0 qnIR2
=
ŷ
1
+
s
ds
=
ŷ
s
ds
=
ŷ.
2a
a2 s2
(a2 s2 )
a
2a
0
s2 .
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183
CHAPTER 8. CONSERVATION LAWS
Angular Momentum.
µ0 qnI
µ0 qnI
r ⇥ [y x̂ (x a) ŷ] =
z(x a) x̂ + zy ŷ [x(x a) + y 2 ] ẑ .
4⇡ r 3
4⇡ r 3
The x̂ and ŷ terms are odd in z, and integrate to zero, so
Z
µ0 qnI
x2 + y 2 xa
L=
ẑ
dx dy dz. The z integral is the same as before.
4⇡
[(x a)2 + y 2 + z 2 ]3/2
Z
Z
µ0 qnI
x2 + y 2 xa
µ0 qnI
s a cos
=
ẑ
dx
dy
=
ẑ
s2 ds d
2
2
2
2⇡
[(x a) + y ]
2⇡
(s + a2 2sa cos )
✓
◆
Z 
Z R 2
s2
a2 + s2
s
s2
+
1
s
ds
=
µ
qnI
ẑ
s ds = zero.
= µ0 qnI ẑ
0
2
2
2
2
2
a
s
a
s
s2
0 a
` = r⇥g =
Problem 8.23
(a) If
Z we’re only interested in the work done on free charges and currents, Eq. 8.6 becomes
dW
@D
@D
= (E · Jf ) d⌧ . But Jf = r ⇥ H
(Eq. 7.56), so E · Jf = E · (r ⇥ H) E ·
. From product
dt
@t
@t
V
@B
rule #6, r · (E ⇥ H) = H · (r ⇥ E) E · (r ⇥ H), while r ⇥ E =
, so
@t
@B
@B
@D
E · (r ⇥ H) = H ·
r · (E ⇥ H). Therefore E · Jf = H ·
E·
r · (E ⇥ H), and hence
@t
@t
@t
◆
Z ✓
I
dW
@D
@B
=
E·
+H·
d⌧
(E ⇥ H) · da.
dt
@t
@t
V
S
This is Poynting’s theorem for the fields in matter. Evidently the Poynting vector, representing the power per
unit area transported by the fields, is S = E ⇥ H, and the rate of change of the electromagnetic energy density
@uem
@D
@B
is
=E·
+H·
.
@t
@t
@t
1
For linear media, D = ✏E and H = B, with ✏ and µ constant (in time); then
µ
@uem
@E
1
@B
1 @
1 @
1 @
= ✏E ·
+ B·
= ✏ (E · E) +
(B · B) =
(E · D + B · H) ,
@t
@t
µ
@t
2 @t
2µ @t
2 @t
so uem = 12 (E · D + B · H). qed
(b) If we’re only interested in the force on free charges and currents, Eq. 8.13
f = ⇢f E + Jf ⇥ B.
✓ becomes
◆
@D
@D
But ⇢f = r · D, and Jf = r ⇥ H
, so f = E(r · D) + (r ⇥ H) ⇥ B
⇥ B. Now
@t
✓
◆@t
@
@D
@B
@B
@D
@
(D ⇥ B) =
⇥B+D⇥
, and
= r ⇥ E, so
⇥B =
(D ⇥ B) + D ⇥ (r ⇥ E), and
@t
@t
@t
@t
@t
@t
@
hence f = E(r · D) D ⇥ (r ⇥ E) B ⇥ (r ⇥ H)
(D ⇥ B). As before, we can with impunity add the
@t
term H(r · B), so
f = {[E(r · D)
D ⇥ (r ⇥ E)] + [H(r · B)
B ⇥ (r ⇥ H)]}
@
(D ⇥ B).
@t
The term in curly brackets can be written as the divergence of a stress tensor (as in Eq. 8.19), and the last
term is (minus) the rate of change of the momentum density, g = D ⇥ B.
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184
CHAPTER 8. CONSERVATION LAWS
Problem 8.24
(a) Initially, the disk will rise like a helicopter. The force on one charge (velocity v = !R ˆ + vz ẑ) is
Fi = q(v ⇥ B) = qk
ŝ ˆ ẑ
⇣
0 !R vz = qk 2!Rz ŝ
R 0 2z
⌘
Rvz ˆ + !R2 ẑ .
The net force on all the charges is
n
X
d2 z
=
dt2
d2 z
F=
Fi = nqkR ! ẑ = M 2 ẑ;
dt
i=1
2
✓
nqkR2
M
◆
!.

The net torque on the disk is
N=
n
X
i=1
(ri ⇥ Fi ) = n(R ŝ) ⇥ ( qkRvz ˆ) =
nqkR2 vz ẑ = I
d!
ẑ
dt
where I is the moment of inertia of the disk. So
✓
d!
=
dt
nqkR2
I
◆
dz
.
dt

Di↵erentiate , and combine with :
d2 z
=
dt2
✓
I
nqkR2
◆
d2 !
=
dt2
✓
nqkR2
M
◆
!
)
d2 !
= ↵!,
dt2
where
nqkR2
↵⌘ p
.
MI
The solution (with initial angular velocity !0 and initial angular acceleration 0) is
!(t) = !0 cos ↵t.
Meanwhile,
dz
=
dt
r
◆
✓
◆
I
d!
I
I
=
!0 ↵ sin ↵t = !0
sin ↵t.
nqkR2 dt
nqkR2
M
r Z t
r
I
!0
I
z(t) = !0
sin ↵t dt =
(1 cos ↵t) .
M 0
↵ M
✓
[The problem is a little cleaner if you make the disk massless, and assign a mass m to each of the charges.
Then M ! nm and I ! nmR2 , so ↵ = qkR/m and z(t) ! (!0 R/↵)(1 cos ↵t).]
(b) The disk rises and falls harmonically, as its rotation slows down and speeds up. The total energy is
E=
1
1
1
I
1
1
1
M vz2 + I! 2 = M !02
sin2 ↵t + I!02 cos2 ↵t = I!02 (sin2 ↵t + cos2 ↵t) = I!02 ,
2
2
2
M
2
2
2
which is constant (and equal to the initial energy). [Of course, if you didn’t notice that the rotation rate is
changing, you might think the magnetic force is doing work, as the disk oscillates up and down.]
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185
CHAPTER 9. ELECTROMAGNETIC WAVES
Chapter 9
Electromagnetic Waves
Problem 9.1
h
i
2
@f1
@ 2 f1
b(z vt)2
2
b(z vt)2
= 2Ab(z vt)e b(z vt) ;
=
2Ab
e
2b(z
vt)
e
;
@z
@z 2
h
i
2
2
2
@f1
@ 2 f1
@ 2 f1
= 2Abv(z vt)e b(z vt) ;
= 2Abv ve b(z vt) + 2bv(z vt)2 e b(z vt) = v 2 2 . X
2
@t
@t
@z
@f2
@ 2 f2
= Ab cos[b(z vt)];
= Ab2 sin[b(z vt)];
@z
@z 2
@f2
@ 2 f2
@ 2 f2
= Abv cos[b(z vt)];
= Ab2 v 2 sin[b(z vt)] = v 2 2 . X
2
@t
@t
@z
2Ab(z vt)
@ 2 f3
2Ab
8Ab2 (z vt)2
@f3
=
;
=
+
;
@z
[b(z vt)2 + 1]2 @z 2
[b(z vt)2 + 1]2
[b(z vt)2 + 1]3
2
@f3
@ 2 f3
2Abv 2
8Ab2 v 2 (z vt)2
2Abv(z vt)
2 @ f3
=
;
=
+
=
v
.X
@t
[b(z vt)2 + 1]2 @t2
[b(z vt)2 + 1]2
[b(z vt)2 + 1]3
@z 2
h
i
2
@f4
@ 2 f4
2
b(bz 2 +vt)
2 2
b(bz 2 +vt)
= 2Ab2 ze b(bz +vt) ;
=
2Ab
e
2b
z
e
;
@z
@z 2
2
2
2
2
@f4
@ f4
@ f4
= Abve b(bz +vt) ;
= Ab2 v 2 e b(bz +vt) 6= v 2 2 .
2
@t
@t
@z
2
@f5
@f5
3 @ f5
2
= Ab cos(bz) cos(bvt) ;
= Ab sin(bz) cos(bvt)3 ;
= 3Ab3 v 3 t2 sin(bz) sin(bvt)3 ;
@z
@z 2
@t
2
@ 2 f5
3 3
3
6 6 4
3
2 @ f5
=
6Ab
v
t
sin(bz)
sin(bvt)
9Ab
v
t
sin(bz)
cos(bvt)
=
6
v
.
@t2
@z 2
Problem 9.2
@f
@2f
= Ak cos(kz) cos(kvt);
= Ak 2 sin(kz) cos(kvt);
@z
@z 2
2
@f
@2f
2 2
2@ f
= Akv sin(kz) sin(kvt);
=
Ak
v
sin(kz)
cos(kvt)
=
v
.X
@t
@t2
@z 2
Use the trig identity sin ↵ cos = 12 [sin(↵ + ) + sin(↵
)] to write
f=
A
{sin[k(z + vt)] + sin[k(z
2
vt)]} ,
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186
CHAPTER 9. ELECTROMAGNETIC WAVES
which is of the form 9.6, with g = (A/2) sin[k(z
vt)] and h = (A/2) sin[k(z + vt)].
Problem 9.3
(A3 )2 = A3 ei
3
A3 e
i
3
= A1 ei
1
+ A2 ei
2
A1 e
= (A1 )2 + (A2 )2 + A1 A2 ei 1 e i 2 + e i 1 ei
p
A3 =
(A1 )2 + (A2 )2 + 2A1 A2 cos( 1
2 ).
A3 ei
3
3
= A3 (cos
3
+ i sin
= (A1 cos
1
+ A2 cos
= tan
1
✓
A1 sin
A1 cos
3)
= A1 (cos
2)
1
+ i sin
+ i(A1 sin
+ A2 sin
1 + A2 cos
1
2
2
◆
1
1)
2
i
1
+ A2 e
2
= (A1 )2 + (A2 )2 + A1 A2 2 cos(
+ A2 (cos
+ A2 sin
i
2 ).
+ i sin 2 )
A3 sin
tan 3 =
A3 cos
2 );
1
2
3
3
=
A1 sin
A1 cos
+ A2 sin
1 + A2 cos
1
2
;
2
.
Problem 9.4
@2f
1 @2f
= 2 2 . Look for solutions of the form f (z, t) = Z(z)T (t). Plug
2
@z
v @t
d2 Z
1 d2 T
1 d2 Z
1 d2 T
this in: T 2 = 2 Z 2 . Divide by ZT :
=
. The left side depends only on z, and the
dz
v
dt
Z dz 2
v 2 T dt2
right
only on t, so both must be constant. Call 9
the constant k 2 .
8 side
2
d Z
>
>
2
>
) Z(z) = Aeikz + Be ikz , >
>
>
< dz 2 = k Z
=
> 2
>
>
>
>
: d T = (kv)2 T ) T (t) = Ceikvt + De ikvt . >
;
dt2
(Note that k must be real, else Z and T blow up; with no loss of generality we can assume k is positive.)
f (z, t) = Aeikz + Be ikz Ceikvt + De ikvt = A1 ei(kz+kvt) + A2 ei(kz kvt) + A3 ei( kz+kvt) + A4 ei( kz kvt) .
The general linear combination of separable solutions is therefore
Z 1h
i
f (z, t) =
A1 (k)ei(kz+!t) + A2 (k)ei(kz !t) + A3 (k)ei( kz+!t) + A4 (k)ei( kz !t) dk,
The wave equation (Eq. 9.2) says
0
where ! ⌘ kv. But we can combine the third term with the first, by allowing k to run negative (! = |k|v
remains positive); likewise the second and the fourth:
Z 1h
i
f (z, t) =
A1 (k)ei(kz+!t) + A2 (k)ei(kz !t) dk.
1
Because (in the end) we shall only want the the real part of f , it suffices to keep only one of these terms (since
k goes negative, both terms include waves traveling in both directions); the second is traditional (though either
would do). Specifically,
Z 1
Re(f ) =
[Re(A1 ) cos(kz + !t) Im(A1 ) sin(kz + !t) + Re(A2 ) cos(kz !t) Im(A2 ) sin(kz !t)] dk.
1
The first term, cos(kz + !t) = cos( kz !t), combines with the third, cos(kz !t), since the negative k is
picked up in the other half of the range of integration, and the second, sin(kz +!t) = sin( kz !t), combines
with the fourth for the same reason. So the general solution, for our purposes, can be written in the form
Z 1
f˜(z, t) =
Ã(k)ei(kz !t) dk qed (the tilde0 s remind us that we want the real part).
1
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187
CHAPTER 9. ELECTROMAGNETIC WAVES
Problem 9.5
@gI
1 @gI @hR
1 @hR @gT
1 @gT
=
;
=
;
=
.
@z
v1 @t @z
v1 @t @z
v2 @t
1 @gI ( v1 t)
1 @hR (v1 t)
1 @gT ( v2 t)
v1
Equation 9.27 )
+
=
) gI ( v1 t) hR (v1 t) =
gT ( v2 t) + 
v1
@t
v1
@t
v2
@t
v2
(where  is a constant).
✓
◆
✓
◆
v1
2v2
Adding these equations, we get 2gI ( v1 t) = 1 +
gT ( v2 t)+, or gT ( v2 t) =
gI ( v1 t)+0
v2
v1 + v2
v2
(where 0 ⌘ 
). Now gI (z, t), gT (z, t), and hR (z, t) are each functions of a single variable u (in the
v1 + v2
first case u = z v1 t, in the second u = z v2 t, and in the third u = z + v1 t). Thus
✓
◆
2v2
gT (u) =
gI (v1 u/v2 ) + 0 .
v1 + v2
✓
◆
✓
◆
v1
v1
Multiplying the first equation by v1 /v2 and subtracting, 1
gI ( v1 t)
1+
hR (v1 t) =  )
v2
v2
✓
◆
✓
◆
✓
◆
v2 v1
v2
v2 v1
hR (v1 t) =
gI ( v1 t) 
, or hR (u) =
gI ( u) + 0 .
v1 + v2
v1 + v2
v1 + v2
Equation 9.26 ) gI ( v1 t) + hR (v1 t) = gT ( v2 t). Now
[The
8notation is tricky, so here’s an example: for a sinusoidal wave,
< gI = AI cos(k1 z !t) = AI cos[k1 (z v1 t)] ) gI (u) = AI cos(k1 u).
gT = AT cos(k2 z !t) = AT cos[k2 (z v2 t)] ) gT (u) = AT cos(k2 u).
:
hR = AR cos( k1 z !t) = AR cos[ k1 (z + v1 t)] ) hR (u) = AR cos( k1 u).
AT
2v2
AR
v2 v1
v1
Here 0 = 0, and the boundary conditions say
=
,
=
(same as Eq. 9.32), and 2 k1 = k2
AI
v1 + v2 AI
v1 + v2
v
(consistent with Eq. 9.24).]
Problem 9.6
(a) T sin ✓+
T sin ✓ = ma ) T
✓
@f
@z
0+
@f
@z
0
◆
=m
@2f
@t2
.
0
◆
✓
im! 2
ÃR )] = m( ! 2 ÃT ), or k1 (ÃI ÃR ) = k2
ÃT .
T
✓
◆
✓
◆
m! 2
2k1
Multiply first equation by k1 and add: 2k1 ÃI = k1 + k2 i
ÃT , or ÃT =
ÃI .
T
k1 + k2 im! 2 /T
✓
◆
2k1 (k1 + k2 im! 2 /T )
k1 k2 + im! 2 /T
ÃR = ÃT ÃI =
ÃI =
ÃI .
2
k1 + k2 im! /T
k1 + k2 im! 2 /T
✓
◆
p
2
If the second string is massless, so v2 = T /µ2 = 1, then k2 /k1 = 0, and we have ÃT =
ÃI ,
1 i
✓
◆
✓
◆
1+i
m! 2
m(k1 v1 )2
mk1 T
k1
1+i
ÃR =
ÃI , where ⌘
=
=
, or
= m . Now
= Aei , with
1 i
k1 T
k1 T
T µ1
µ1
1 i
✓
◆✓
◆
2
1+i
1 i
(1 + i )2
1 + 2i
A2 =
= 1 ) A = 1, and ei =
=
)
1 i
1+i
(1 i )(1 + i )
1+ 2
✓
◆
2
2
i R
i
i I
1
tan =
.
Thus
A
e
=
e
A
e
)
A
=
A
,
=
+
tan
.
R
I
R
I
R
I
2
2
1
1
✓
◆
✓
◆✓
◆
2
2
2
4
2
i
2
Similarly,
= Ae ) A =
=
)A= p
.
1 i
1 i
1+i
1+ 2
1+ 2
(b) ÃI + ÃR = ÃT ; T [ik2 ÃT
ik1 (ÃI
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188
Aei =
(1
2(1 + i )
2(1 + i )
=
) tan
i )(1 + i )
(1 + 2 )
2
AT = p
CHAPTER 9. ELECTROMAGNETIC WAVES
1+
2
AI ;
Problem 9.7
@2f
(a) F = T 2
@z
T
=
I
+ tan
1
.
= . So AT ei
T
2
=p
1+
2
ei AI ei I ;
@2f
@2f
@2f
@f
, or T 2 = µ 2 +
.
2
@t
@z
@t
@t
d2 F̃
(b) Let f˜(z, t) = F̃ (z)e i!t ; then T e i!t 2 = µ( ! 2 )F̃ e i!t + ( i!)F̃ e i!t )
dz
d2 F̃
d2 F̃
!
2
T 2 = !(µ! + i )F̃ ,
= k̃ F̃ , where k̃ 2 ⌘ (µ! + i ). Solution : F̃ (z) = Ãeik̃z + B̃e ik̃z .
2
dz
dz
T
!
Resolve k̃ into its real and imaginary parts: k̃ = k + i ) k̃ 2 = k 2 2 + 2ik = (µ! + i ).
T
⇣ ! ⌘2 1
!
!
µ! 2
2k =
)=
; k 2 2 = k 2
=
; or k 4 k 2 (µ! 2 /T ) (! /2T )2 = 0 )
2
T
2kT
2T
k
T
i µ! 2 h
i
p
p
1h
k2 =
(µ! 2 /T ) ± (µ! 2 /T )2 + 4(! /2T )2 =
1 ± 1 + ( /µ!)2 . But k is real, so k 2 is positive, so
2
2T
r q
h
i 1/2
p
p
µ
!
we need the plus sign: k = !
1 + 1 + ( /µ!)2 .  =
=p
1 + 1 + ( /µ!)2
.
2T
2kT
2T µ
Plugging this in, F̃ = Aei(k+i)z + Be i(k+i)z = Ae z eikz + Bez e ikz . But the B term gives an exponentially increasing function, which we don’t want (I assume the waves are propagating in the +z direction),
so B = 0, and the solution is f˜(z, t) = Ãe z ei(kz !t) . (The actual displacement of the string is the real part
z
@f
@t
z=µ z
of this, of course.)
(c) The wave is attenuated by the factor e z , which becomes 1/e when
p
q
p
1
2T µ
z= =
1 + 1 + ( /µ!)2 ; this is the characteristic penetration depth.

✓
◆
k1 k i
(d) This is the same as before, except that k2 ! k + i. From Eq. 9.29, ÃR =
ÃI ;
k1 + k + i
s
✓
◆2 ✓
◆✓
◆
AR
k1 k i
k1 k + i
(k1 k)2 + 2
(k1 k)2 + 2
=
=
.
A
=
AI
R
AI
k1 + k + i
k1 + k i
(k1 + k)2 + 2
(k1 + k)2 + 2
p
(where k1 = !/v1 = ! µ1 /T , while k and  are defined in part b). Meanwhile
✓
◆
✓
◆
k1 k i
(k1 k i)(k1 + k + i)
(k1 )2 k 2 2 2ik1
2k1 
1
=
=
) R = tan
.
k1 + k + i
(k1 + k)2 + 2
(k1 + k)2 + 2
(k1 )2 k 2 2
Problem 9.8
(a) fv (z, t) = A cos(kz !t) x̂; fh (z, t) = A cos(kz !t + 90 ) ŷ =
A sin(kz !t) ŷ. Since fv2 +fh2 = A2 , the vector sum f = fv +fh lies
on a circle of radius A. At time t = 0, f = A cos(kz) x̂ A sin(kz) ŷ.
At time t = ⇡/2!, f = A cos(kz 90 ) x̂ A sin(kz 90 ) ŷ =
A sin(kz) x̂ + A cos(kz) ŷ. Evidently it circles counterclockwise .
To make a wave circling the other way, use h = 90 .
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CHAPTER 9. ELECTROMAGNETIC WAVES
189
(b)
(c) Shake it around in a circle, instead of up and down.
Problem 9.9
⇣ ! ⌘
!
!
x̂; n̂ = ẑ. k · r =
x̂ · (x x̂ + y ŷ + z ẑ) =
x; k ⇥ n̂ =
c
c
c
⇣!
⌘
⇣!
⌘
E0
E(x, t) = E0 cos
x + !t ẑ; B(x, t) =
cos
x + !t ŷ.
c
c
c
(a) k =
x̂ ⇥ ẑ = ŷ.
◆
x̂ + ŷ + ẑ
x̂ ẑ
p
; n̂ = p . (Since n̂ is parallel to the x z plane, it must have the form ↵ x̂ + ẑ;
3
2
p
since n̂ · k = 0, = ↵; and since it is a unit vector, ↵ = 1/ 2.)
(b) k =
!
c
✓
x̂ ŷ ẑ
!
!
1
1
k · r = p (x̂ + ŷ + ẑ) · (x x̂ + y ŷ + z ẑ) = p (x + y + z); k̂ ⇥ n̂ = p 1 1 1 = p ( x̂ + 2 ŷ
3c
3c
6 1 0 1
6

✓
◆
!
x̂ ẑ
p
E(x, y, z, t) = E0 cos p (x + y + z) !t
;
3c
2

✓
E0
!
x̂ + 2ŷ
p
B(x, y, z, t) =
cos p (x + y + z) !t
c
3c
6
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ẑ
◆
.
ẑ).
190
CHAPTER 9. ELECTROMAGNETIC WAVES
Problem 9.10
I
1.3 ⇥ 103
P = =
= 4.3 ⇥ 10 6 N/m2 . For a perfect reflector the pressure is twice as great:
c
3.0 ⇥ 108
8.6 ⇥ 10 6 N/m2 . Atmospheric pressure is 1.03 ⇥ 105 N/m2 , so the pressure of light on a reflector is
(8.6 ⇥ 10
6
)/(1.03 ⇥ 105 ) = 8.3 ⇥ 10
11
atmospheres.
Problem 9.11 The fields are E(z, t) = E0 cos(kz
(a) The electric force is Fe = qE = qE0 cos(kz
qE0
v=
x̂
m
But vave = C = 0, so v =
(b) The magnetic force is
Fm = q(v ⇥ B) = q
✓
qE0
m!
Z
cos(kz
qE0
sin(kz
m!
◆✓
E0
c
◆
1
E0 cos(kz
c
dv
!t) x̂ = ma = m , so
dt
!t) x̂, B(z, t) =
qE0
sin(kz
m!
!t) dt =
!t) ŷ, with ! = ck.
!t) x̂ + C.
!t) x̂.
sin(kz
!t) cos(kz
q 2 E02
sin(kz
m!c
!t)(x̂ ⇥ ŷ) =
!t) cos(kz
!t) ẑ.
Z T
q 2 E02
ẑ
sin(kz !t) cos(kz !t) dt, where T = 2⇡/! is the
m!c
0
⇤
1 ⇥ 2
=
sin (kz 2⇡) sin2 (kz) = 0, so (Fm )ave = 0.
2!
(c) The (time) average force is (Fm )ave =
1
sin2 (kz !t)
2!
(d) Adding in the damping term,
period. The integral is
F = qE
mv = qE0 cos(kz
T
0
mv = m
!t)x̂
dv
dv
qE0
)
+ v=
cos(kz
dt
dt
m
dv
= A! sin(kz
dt
in, and using the trig identity cos u = cos ✓ cos(u + ✓) + sin ✓ sin(u + ✓),
The steady state solution has the form v = A cos(kz
A! sin(kz
!t + ✓) + A cos(kz
!t + ✓) =
!t + ✓) x̂,
qE0
[cos ✓ cos(kz
m
!t) x̂.
!t + ✓) x̂. Putting this
!t + ✓) + sin ✓ sin(kz
!t + ✓)] .
Equating like terms:
A! =
qE0
qE0
!
sin ✓, A =
cos ✓ ) tan ✓ = , A2 (! 2 +
m
m
2
)=
✓
qE0
m
◆2
)A=
So
v=
qE
p 0
m !2 +
2
cos(kz
!t + ✓)x̂, ✓ ⌘ tan
1
(!/ ); Fm =
q2 E 2
p 0
mc ! 2 +
2
cos(kz
qE
p 0
m !2 +
2
!t + ✓) cos(kz
To calculate the time average, write cos(kz !t + ✓) = cos ✓ cos(kz !t) sin ✓ sin(kz
know that the average of cos(kz !t) sin(kz !t) is zero, so
Z T
q 2 E02
p
(Fm )ave =
ẑ cos ✓
cos2 (kz !t) dt.
mc ! 2 + 2
0
.
!t) ẑ.
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191
CHAPTER 9. ELECTROMAGNETIC WAVES
The integral is T /2 = ⇡/!, and cos ✓ = /
p
!2 +
2
(see figure), so (Fm )ave =
⇡ q 2 E02
m!c(! 2 +
2)
ẑ.
w
q
g
Problem 9.12
hf gi =
1
T
Z
T
0
ab
=
2T
Z
0
a cos(k · r
T
[cos(2k · r
!t +
a )b cos(k
2!t +
a
+
b)
·r
!t +
+ cos(
b ) dt
a
b )]
dt =
ab
cos(
2T
b )T
a
=
1
ab cos(
2
a
b ).
Meanwhile, in the complex notation: f˜ = ãeik·r !t) , g̃ = b̃eik·r✓ !t) , where
ã = aei a , b̃ = bei b . So
◆
1˜⇤
1
1
1
1˜⇤
1
f g̃ = ãei(k·r !t) b̃⇤ e i(k·r !t) = ãb̃⇤ = abei( a b ) , Re
f g̃
= ab cos( a
qed
b ) = hf gi.
2
2
2
2
2
2
Problem 9.13
✓
Tij = ✏0 Ei Ej
1
2
ij E
2
◆
1
+
µ0
✓
Bi Bj
1
2
ij B
2
◆
.
With the fields in Eq. 9.48, E has only an x component, and B only a y component. So all the “o↵-diagonal”
(i 6= j) terms are zero. As for the “diagonal” elements:
✓
◆
✓
◆
✓
◆
1 2
1
1 2
1
1 2
2
Txx = ✏0 Ex Ex
E +
B
=
✏0 E
B
= 0.
2
µ0
2
2
µ0
✓
◆
✓
◆
✓
◆
1 2
1
1 2
1
1 2
Tyy = ✏0
E +
By By
B
=
✏0 E 2 +
B
= 0.
2
µ0
2
2
µ0
✓
◆
✓
◆
1 2
1
1 2
Tzz = ✏0
E +
B
= u.
2
µ0
2
So Tzz =
✏0 E02 cos2 (kz
!t + ) (all other elements zero).
The momentum of these fields is in the z direction, and it is
being transported in the z direction, so yes, it does make sense
that Tzz should be the only nonzero element in Tij . According
\$
to Sect. 8.2.3, T · da is the rate at which momentum crosses
an area da. Here we have no momentum crossing areas oriented in the x or y direction; the momentum per unit time
per unit area flowing across a surface oriented in the z direction is Tzz = u = gc (Eq. 9.59), so p = gcA t, and hence
p/ t = gcA = momentum per unit time crossing area A.
Evidently momentum flux density = energy density. X
Problem 9.14
R=
✓
E0 R
E0 I
◆2
(Eq. 9.86) ) R =
✓
1
1+
◆2
(Eq. 9.82), where
⌘
µ1 v1
✏2 v2
. T =
µ2 v2
✏1 v1
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✓
E0T
E0I
◆2
(Eq. 9.87)
192
) T =
T +R=
✓
2
1+
◆2
CHAPTER 9. ELECTROMAGNETIC WAVES
✏2 v2
µ1 ✏2 µ2 v2
µ1
(Eq. 9.82). [Note that
=
=
✏1 v1
µ2 ✏1 µ1 v1
µ2
⇥
1
4 + (1
2
(1 + )
⇤
)2 =
1
(4 + 1
(1 + )2
2 +
2
)=
✓
v1
v2
◆2
v2
µ1 v1
=
= .]
v1
µ2 v2
1
(1 + 2 +
(1 + )2
2
) = 1. X
Problem 9.15
Equation 9.78 is replaced by Ẽ0I x̂ + Ẽ0R n̂R = Ẽ0T n̂T , and Eq. 9.80 becomes Ẽ0I ŷ Ẽ0R (ẑ ⇥ n̂R ) =
Ẽ0T (ẑ ⇥ n̂T ). The y component of the first equation is Ẽ0R sin ✓R = Ẽ0T sin ✓T ; the x component of the
second is Ẽ0R sin ✓R =
Ẽ0T sin ✓T . Comparing these two, we conclude that sin ✓R = sin ✓T = 0, and hence
✓R = ✓T = 0. qed
Problem 9.16
Aeiax + Beibx = Ceicx for all x, so (using x = 0), A + B = C.
Di↵erentiate: iaAeiax + ibBeibx = icCeicx , so (using x = 0), aA + bB = cC.
Di↵erentiate again: a2 Aeiax b2 Beibx = c2 Ceicx , so (using x = 0), a2 A + b2 B = c2 C.
2
a A + b2 B = c(cC) = c(aA + bB); (A + B)(a2 A + b2 B) = (A + B)c(aA + bB) = cC(aA + bB);
a2 A2 + b2 AB + a2 AB + b2 B 2 = (aA + bB)2 = a2 A2 + 2abAB + b2 B 2 , or (a2 + b2 2ab)AB = 0, or
(a b)2 AB = 0. But A and B are nonzero, so a = b. Therefore (A + B)eiax = Ceicx .
a(A + B) = cC, or aC = cC, so (since C 6= 0) a = c. Conclusion: a = b = c. qed
Problem
9.17
8
9
< ẼI = Ẽ0I ei(kI ·r !t) ŷ,
=
1
i(kI ·r !t)
Ẽ0 e
( cos ✓1 x̂ + sin ✓1 ẑ); ;
: B̃I =
v1 I
8
9
< ẼR = Ẽ0R ei(kR ·r !t) ŷ,
=
1
Ẽ0 ei(kR ·r !t) (cos ✓1 x̂ + sin ✓1 ẑ); ;
: B̃R =
v1 R
8
9
< ẼT = Ẽ0T ei(kT ·r !t) ŷ,
=
1
Ẽ0 ei(kT ·r !t) ( cos ✓2 x̂ + sin ✓2 ẑ); ;
: B̃T =
v2 T
8
> (i) ✏1 E1? = ✏2 E2? , (iii) Ek1 = Ek2 ,
<
Boundary conditions:
>
k
k
: (ii) B ? = B ? ,
(iv) µ11 B1 = µ12 B2 .
1
2
sin ✓2
v2
= . [Note: kI · r !t = kR · r
sin ✓1
v1
exponential factors in applying the boundary conditions.]
Law of refraction:
!t = kT · r
!t, at z = 0, so we can drop all
Boundary condition (i): 0 = 0 (trivial). Boundary condition (iii): Ẽ0I + Ẽ0R = Ẽ0T .
✓
◆
1
1
1
v1 sin ✓2
Boundary condition (ii):
Ẽ0I sin ✓1 + Ẽ0R sin ✓1 = Ẽ0T sin ✓2 ) Ẽ0I + Ẽ0R =
Ẽ0T .
v1
v1
v2
v2 sin ✓1
But the term in parentheses is 1,by the law of refraction, so this is the same as (iii).
1 1
1
1
Boundary condition (iv):
Ẽ0I ( cos ✓1 ) + Ẽ0R cos ✓1 =
Ẽ0 ( cos ✓2 ) )
µ1 v1
v1
µ2 v2 T
✓
◆
µ1 v1 cos ✓2
cos ✓2
µ1 v1
Ẽ0I Ẽ0R =
Ẽ0T . Let ↵ ⌘
; ⌘
. Then Ẽ0I Ẽ0R = ↵ Ẽ0T .
µ2 v2 cos ✓1
cos ✓1
µ2 v2
✓
◆
2
Solving for Ẽ0R and Ẽ0T : 2Ẽ0I = (1 + ↵ )Ẽ0T ) Ẽ0T =
Ẽ0I ;
1+↵
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193
CHAPTER 9. ELECTROMAGNETIC WAVES
◆
✓
◆
2
1+↵
1 ↵
Ẽ0I ) Ẽ0R =
Ẽ0I .
1+↵
1+↵
1+↵
Since ↵ and are positive, it follows that 2/(1 + ↵ ) is positive, and hence the transmitted wave is in phase
✓
◆
2
with the incident wave, and the (real) amplitudes are related by E0T =
E0I . The reflected wave is
1+↵
Ẽ0R = Ẽ0T
Ẽ0I =
✓
in phase if ↵ < 1 and 180 out of phase if ↵ > 1; the (real) amplitudes are related by E0R =
1 ↵
1+↵
E0 I .
These are the Fresnel equations for polarization
perpendicular to the plane of incidence.
q
p
2
1 sin2 ✓/ 2
sin2 ✓
To construct the graphs, note that ↵ =
=
, where ✓ is the angle of incidence,
cos ✓
cos ✓
p
2
2.25 sin ✓
so, for = 1.5, ↵ =
. [In the figure, the minus signs on the vertical axis should be decimal
cos ✓
points.]
Isqthere a Brewster’s angle? Well, E0R = 0 would mean that ↵ = 1, and hence that
✓ ◆2
✓
◆2
1 (v2 /v1 )2 sin2 ✓
1
µ2 v2
v2
µ2 v2
↵=
= =
, or 1
sin2 ✓ =
cos2 ✓, so
cos ✓
µ1 v1
v1
µ1 v1
✓ ◆2
⇥ 2
⇤
v2
1=
sin ✓ + (µ2 /µ1 )2 cos2 ✓ . Since µ1 ⇡ µ2 , this means 1 ⇡ (v2 /v1 )2 , which is only true for optically
v1
indistinguishable media, in which case there is of course no reflection—but that would be true at any angle,
not just at a special “Brewster’s angle”. [If µ2 were substantially di↵erent from µ1 , and the relative velocities
were just right, it would be possible to get a Brewster’s angle for this case, at
✓
v1
v2
◆2
=1
cos ✓ +
2
✓
µ2
µ1
◆2
cos2 ✓ ) cos2 ✓ =
(v1 /v2 )2
(µ2 /µ1 )2
1
(µ2 ✏2 /µ1 ✏1 ) 1
(✏2 /✏1 )
=
=
2
1
(µ2 /µ1 )
1
(µ2 /µ1 )
(µ1 /µ2 )
.
(µ1 /µ2 )
But the media would be very peculiar.]
By the same token, R is either always 0, or always ⇡, for a given interface—it does not switch over as you
change ✓, the way it does for polarization in the plane of incidence. In particular, if = 3/2, then ↵ > 1, for
↵ =
In general, for
p
2.25 sin2 ✓
> 1 if 2.25
cos ✓
> 1, ↵ > 1, and hence
sin2 ✓ > cos2 ✓, or 2.25 > sin2 ✓ + cos2 ✓ = 1. X
< 1, ↵ < 1, and ✓R = 0.◆
2
1
At normal incidence, ↵ = 1, so Fresnel’s equations reduce to E0T =
E0 I ; E 0 R =
1+
1+
consistent with Eq. 9.82.
R
= ⇡. For
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E0I ,
194
CHAPTER 9. ELECTROMAGNETIC WAVES
Reflection and Transmission coefficients: R =
T =
✏2 v2
↵
✏1 v1
✓
E0T
E0 I
◆2
✓
= ↵
R+T =
◆2
2
1+↵
(1
✓
E0 R
E0I
◆2
=
✓
1 ↵
1+↵
◆2
.
Referring to Eq. 9.116,
.
↵ )2 + 4↵
(1 + ↵ )2
=
1
2↵ + ↵2 2 + 4↵
(1 + ↵ )2
=
(1 + ↵ )2
= 1. X
(1 + ↵ )2
Problem 9.18
Equation
p 9.106 ) = 2.42; Eq. 9.110 )
1 (sin ✓/2.42)2
↵=
.
cos ✓
✓
◆
E0 R
↵
(a) ✓ = 0 ) ↵ = 1. Eq. 9.109 )
=
E0I
↵+
1 2.42
1.42
=
=
= 0.415;
3.42
✓
◆1 + 2.42
E0T
2
2
=
=
= 0.585.
E0I
↵+
3.42
(b) Equation 9.112 ) ✓B = tan 1 (2.42) = 67.5 .
(c) E0R = E0T ) ↵
= 2 ) ↵ = + 2 = 4.42;
(4.42)2 cos2 ✓ = 1 sin2 ✓/(2.42)2 ;
(4.42)2 (1 sin2 ✓) = (4.42)2 (4.42)2 sin2 ✓
= 1 0.171 sin2 ✓; 19.5 1 = (19.5 0.17) sin2 ✓;
18.5 = 19.3 sin2 ✓; sin2 ✓ = 18.5/19.3 = 0.959;
sin ✓ = 0.979; ✓ = 78.3 .
Problem 9.19
(a) Equation 9.120 ) ⌧ = ✏/ . Now ✏ = ✏0 ✏r (Eq. 4.34), ✏r ⇠
= n2 (Eq. 9.70), and for glass the index
2
12
of refraction is typically around 1.5, so ✏ ⇡ (1.5) ⇥ 8.85 ⇥ 10
= 2 ⇥ 10 11 C2 /N m2 , while = 1/⇢ ⇡
12
1
11
12
10
(⌦ m) (Table 7.1). Then ⌧ = (2⇥10 )/10
= 20 s. (But the resistivity of glass varies enormously
from one type to another, so this answer could be o↵ by a factor of 100 in either direction.)
(b) For silver, ⇢ = 1.59 ⇥ 10 8 (Table 7.1), and ✏ ⇡ ✏0 , so !✏ = 2⇡ ⇥ 1010 ⇥ 8.85 ⇥ 10 12 = 0.56.
Since = 1/⇢ = 6.25 ⇥ 107
!✏, the skin depth (Eq. 9.128) is
1
d= ⇠
=

r
2
! µ
=
r
2⇡ ⇥
1010
2
⇥ 6.25 ⇥ 107 ⇥ 4⇡ ⇥ 10
7
= 6.4 ⇥ 10
7
m = 6.4 ⇥ 10
4
mm.
I’d plate silver to a depth of about 0.001 mm; there’s no point in making it any thicker, since the fields don’t
penetrate much beyond this anyway.
8
7
6
12
5
(c) For copper, Table 7.1 gives
r = 1/(1.68 ⇥ 10 ) = 6 ⇥ 10 , !✏0 = (2⇡ ⇥ 10 ) ⇥ (8.85 ⇥ 10 ) = 6 ⇥ 10 .
! µ
Since
!✏, Eq. 9.126 ) k ⇡
, so (Eq. 9.129)
2
= 2⇡
r
2
= 2⇡
! µ0
r
2
2⇡ ⇥ 106 ⇥ 6 ⇥ 107 ⇥ 4⇡ ⇥ 10
7
= 4 ⇥ 10
4
m = 0.4 mm.
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195
CHAPTER 9. ELECTROMAGNETIC WAVES
!
!
=
k
2⇡
From Eq. 9.129, the propagation speed is v =
= ⌫ = (4 ⇥ 10
4
) ⇥ 106 = 400 m/s. In vacuum,
c
3 ⇥ 108
=
= 300 m; v = c = 3 ⇥ 108 m/s. (But really, in a good conductor the skin depth is so small,
⌫
106
compared to the wavelength, that the notions of “wavelength” and “propagation speed” lose their meaning.)
=
Problem 9.20
(a) Use the binomial expansion for the square root in Eq. 9.126:
⇠
=!
r

✏µ
1 ⇣ ⌘2
1+
2
2 ✏!
1/2
1
=!
r
✏µ 1
p
=
2 2 ✏!
2
r
µ
.
✏
r
1 ⇠ 2 ✏
. qed
=

µ
8
< ✏ = ✏r ✏0 = 80.1 ✏0 (Table 4.2),
For pure water, µ = µ0 (1 + m ) = µ0 (1 9.0 ⇥ 10 6 ) ⇠
= µ0 (Table 6.1),
:
= 1/(2.5 ⇥ 105 ) (Table 7.1).
r
(80.1)(8.85 ⇥ 10 12 )
5
So d = (2)(2.5 ⇥ 10 )
= 1.19 ⇥ 104 m.
4⇡ ⇥ 10 7
(b) In this case ( /✏!)2 dominates, so (Eq. 9.126) k ⇠
= , and hence (Eqs. 9.128 and 9.129)
2⇡ ⇠ 2⇡
=
= 2⇡d, or d =
. qed
=
k

2⇡
r
r r
r
✏µ
!µ
(1015 )(4⇡ ⇥ 10 7 )(107 )
1
1
⇠
Meanwhile  = !
=
=
= 8 ⇥ 107 ; d =
=
=
2 ✏!
2
2

8 ⇥ 107
1.3 ⇥ 10 8 = 13 nm. So the fields do not penetrate far into a metal—which is what accounts for their opacity.
So (Eq. 9.128) d =
(c) Since k ⇠
= , as we found in (b), Eq. 9.134 says = tan 1 (1) = 45 . qed
r
r
r
B0 ⇠
µ
B0
(107 )(4⇡ ⇥ 10
Meanwhile, Eq. 9.137 says
=
. For a typical metal, then,
=
= ✏µ
E0
✏!
!
E0
1015
10 7 s/m. (In vacuum, the ratio is 1/c = 1/(3 ⇥ 108 ) = 3 ⇥ 10
about 100 times larger in a metal.)
Problem 9.21✓
◆

1
1
1
(a) u =
✏E 2 + B 2 = e 2z ✏E02 cos2 (kz
2
µ
2
over a full cycle, using hcos2 i = 12 and Eq. 9.137:
1
hui = e
2
2z

✏ 2
1 2
1
E +
B = e
2 0 2µ 0
4
r
⇣
⌘2
2z
"
✏E02
!t +
1+
=
humag i
B 2 /µ
1
= 02 =
µ✏
huelec i
E0 ✏
µ✏
E)
r
1+
⇣
✏!
2z
⌘2
✏E02
=
r
=
s/m, so the magnetic field is comparatively
+
1 2
B cos2 (kz
µ 0
#
r
⇣ ⌘2
1 2
1
+ E0 ✏µ 1 +
= e
µ
✏!
4
2 k2
1
, so hui = e
✏!
✏µ ! 2
4
magnetic contribution to the electric contribution is
But Eq. 9.126 ) 1 +
9
7)
2z
!t +
✏E02
2 k2
k2
=
E2e
2
✏µ !
2µ! 2 0
1+
⇣
✏!
⌘2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
"
1+
2z
> 1. qed
E
+ ) . Averaging
r
1+
⇣
✏!
⌘2
#
.
. So the ratio of the
196
1
cos(kz !t+ E ) cos(kz !t+ E + ) ẑ; hSi =
E0 B0 e 2z cos ẑ. [The
2µ
R 2⇡
1
average of the product of the cosines is (1/2⇡) 0 cos ✓ cos(✓+ ) d✓ = (1/2) cos .] So I =
E0 B0 e 2z cos =
2µ
✓
◆
1 2 2z K
k
E0 e
cos , while, from Eqs. 9.133 and 9.134, K cos = k, so I =
E 2 e 2z . qed
2µ
!
2µ! 0
(b) S =
1
1
(E⇥B) = E0 B0 e
µ
µ
CHAPTER 9. ELECTROMAGNETIC WAVES
2z
Problem 9.22
!
!
2
2
Ẽ0R
1 ˜
1 ˜
1 ˜⇤
µ1 v1
According to Eq. 9.147, R =
=
=
, where ˜ =
k̃2
⇤
˜
˜
˜
µ2 !
Ẽ0I
1+
1+
1+
µ1 v1
=
(k2 + i2 ) (Eqs. 9.125 and 9.146). Since silver is a good conductor (
✏!), Eq. 9.126 reduces to
µ2 !
r
r
r
r
r
!µ2
µ1 v1
!µ2
⇠ k2 =
⇠ ! ✏2 µ2
2 =
=
, so ˜ =
(1 + i) = µ1 v1
(1 + i).
2
✏2 !
2
µ2 !
2 s
2µ2 !
r
r
r
µ0
(6 ⇥ 107 )(4⇡ ⇥ 10 7 )
Let ⌘ µ1 v1
= µ0 c
=c
= (3 ⇥ 108 )
= 29. Then
2µ2 !
2µ0 !
2!
(2)(4 ⇥ 1015 )
✓
◆✓
◆
1
i
1
+i
(1
)2 + 2
R=
=
= 0.93. Evidently 93% of the light is reflected.
1+ +i
1+
i
(1 + )2 + 2
Problem 9.23
p
(a) We are told that v = ↵ , where ↵ is a constant. But = 2⇡/krand v = !/k, so
p
p
p
d!
1
1
2⇡
1 p
1
! = ↵k 2⇡/k = ↵ 2⇡k. From Eq. 9.150, vg =
= ↵ 2⇡ p = ↵
= ↵
= v, or v = 2vg .
dk
2
k
2
2
2 k
i(px Et)
p
E
p2
~k 2
!
E
p
~k
(b)
= i(kx !t) ) k = , ! =
=
=
. Therefore v = =
=
=
;
~
~
~
2m~
2m
k
p
2m
2m
vg =
d!
2~k
~k
p
1
=
=
=
. So v = vg . Since p = mvc (where vc is the classical speed of the particle), it
dk
2m
m
m
2
follows that vg (not v) corresponds to the classical veloctity.
Problem 9.24
s
✓
◆
1 qd
1 q2
q2
2
E =
) F = qE =
x = kspring x = m!0 x (Eq. 9.151). So !0 =
.
3
3
4⇡✏0 a
4⇡✏0 a
4⇡✏0 ma3
s
!0
1
(1.6 ⇥ 10 19 )2
⌫0 =
=
= 7.16 ⇥ 1015 Hz. This is ultraviolet.
2⇡
2⇡ 4⇡(8.85 ⇥ 10 12 )(9.11 ⇥ 10 31 )(0.5 ⇥ 10 10 )3
From Eqs. 9.173 and 9.174,
⇢
0
s #
nq 2 f
N = # of molecules per unit volume = Avogadro
=
22.4 liters
A=
,
2
2m✏0 !0
f = # of electrons per molecule = 2 (for H2 ).
6.02⇥1023
22.4⇥10 3
= 2.69 ⇥ 1025 ,
(2.69 ⇥ 1025 )(1.6 ⇥ 10 19 )2
= 4.2 ⇥ 10 5 (which is about 1/3 the actual value);
(9.11 ⇥ 10 31 )(8.85 ⇥ 10 12 )(4.5 ⇥ 1016 )2
✓
◆2 ✓
◆2
2⇡c
2⇡ ⇥ 3 ⇥ 108
B =
=
= 1.8 ⇥ 10 15 m2 (which is about 1/4 the actual value).
!0
4.5 ⇥ 1016
=
So even this extremely crude model is in the right ball park.
protected under all copyright laws as they currently exist. No portion of this material may be
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197
CHAPTER 9. ELECTROMAGNETIC WAVES
Problem 9.25
N q2
(!02 ! 2 )
Equation 9.170 ) n = 1 +
. Let the denominator ⌘ D. Then
2m✏0 [(!02 ! 2 )2 + 2 ! 2 ]
⇢
⇤
⇥
⇤
dn
N q2
2! (!02 ! 2 ) ⇥
2
=
2(!02 ! 2 )( 2!) + 2 2! = 0 ) 2!D = (!02 ! 2 ) 2(!02 ! 2 )
2!;
d!
2m✏0
D
D2
2
(!02 ! 2 )2 + 2 ! 2 = 2(!02 ! 2 )2
(!02 ! 2 ), or (!02 ! 2 )2 = 2 (! 2 + !02 ! 2 ) = 2 !02 ) (!02 ! 2 ) = ±!0 ;
p
! 2 = ! 2 ⌥ !0 , ! = !0 1 ⌥ /!0 ⇠
/2, and the
= !0 (1 ⌥ /2!0 ) = !0 ⌥ /2. So !2 = !0 + /2, !1 = !0
0
width of the anomalous region is
! = !2
!1 = .
N q2
,
so
at
the
maximum
(!
=
!
),
↵
=
.
0
max
! 2 )2 + 2 ! 2
m✏0 c
✓
◆
N q2 !2
!2
At !1 and !2 , ! 2 = !02 ⌥ !0 , so ↵ =
=
↵
. But
max
m✏0 c 2 !02 + 2 ! 2
! 2 + !02
✓
◆
✓
◆
✓
◆
!2
!02 ⌥ !0
1 (1 ⌥ /!0 ) ⇠ 1
1
1
⇠
⇠
=
=
1⌥
1±
1⌥
=
=
= .
! 2 + !02
2!02 ⌥ !0
2 (1 ⌥ /2!0 )
2
!0
2!0
2
2!0
2
1
⇠
So ↵ = 2 ↵max at !1 and !2 . qed
From Eq. 9.171, ↵ =
2
2
Nq !
m✏0 c (!02
Problem 9.26
Equation 9.170 for a single resonance:
ck
=1+
!
✓
N q2
2m✏0
◆

(!02 ! 2 )
a! 2 (! 2 ! 2 )
)
ck
=
!
1 + 2 0 20 2
2
2
2
2
2
(!0 ! ) + !
(!0 ! ) + 2 ! 2
where a ⌘
N q2
.
2m✏0 !02
From Eq. 9.150, the group velocity is vg = d!/dk, so
c
dk
=c
vg
d!

= 1+
a!02 (!02 ! 2 )
[(! 2
+ !a!02 0
2
2
2
2
! ) + !
2
! 2 )[(!02 ! 2 )2 + 2 ! 2 ]
2 (!0
! 2 )2 +
2! 2 [(!02
[(!02
2 2
(! 2 ! 2 )3 + 2! 2 (!02 ! 2 )2
! (!02
= 1 + a!02 0
2
2
2
2
[(!0 ! ) + ! 2 ]2
2 2
(! 2 ! 2 )2
!
= 1 + a!02 (!02 + ! 2 ) 20
.
[(!0 ! 2 )2 + 2 ! 2 ]2
= 1 + a!0
⇢
(! 2
vg = c 1 + a!02 (!02 + ! 2 ) 20
[(!0
(a) For
! 2 ]( 2!) (!02 ! 2 )[2(!02 ! 2 )( 2!) + 2 (2!)]
[(!02 ! 2 )2 + 2 ! 2 ]2
! 2 )2 + 2 ! 2 ] + 4! 2 (!02 ! 2 )2 2 2 ! 2 (!02 ! 2 )
! 2 )2 + 2 ! 2 ]2
!2 ) 2 2 !4
2
(!02
! 2 )2
! 2 )2 +
2
!2
2 ! 2 ]2
1
.
= 0 and a = 0.003:

vg
(! 2 + ! 2 )
= 1 + a!02 20
c
(!0 ! 2 )2
1

1+x
) y = 1 + 0.003
(1 x)2
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
1
.
198
CHAPTER 9. ELECTROMAGNETIC WAVES
1.0
0.8
0.6
0.4
0.2
0.0
(b) For
0.5
1.0
1.5
2.0
= (0.1)!0 and a = 0.003:
⇢
(1 x)2 0.01x
y = 1 + 0.003(1 + x)
[(1 x)2 + 0.01x]2
1
.
2.5
2.0
1.5
1.0
0.0
0.5
Problem 9.27
(a) From Eqs. 9.176 and 9.177, r ⇥ Ẽ =
In the terminology of Eq. 9.178:
@ Ẽz
@ Ẽy
@ Ẽ0z
(r ⇥ Ẽ)x =
=
@y
@z
@y
ik Ẽ0y
1.0
@ B̃
= i! B̃0 ei(kz
@t
!
@ Ẽz
=
@x
ik Ẽ0x
@ Ẽ0z
@x
@ Ẽy
(r ⇥ Ẽ)z =
@x
@ Ẽx
=
@y
@ Ẽ0y
@x
@ Ẽ0x
@y
@ B̃z
@y
@ B̃y
=
@z
@ B̃0z
@y
ik B̃0y
@ B̃x
(r ⇥ B̃)y =
@z
@ B̃z
=
@x
ik B̃0x
@ B̃0z
@x
@ B̃y
(r ⇥ B̃)z =
@x
@ B̃x
=
@y
@ B̃0y
@x
@ B̃0x
@y
2.0
!t)
; r ⇥ B̃ =
@Ez
@y
1 @ Ẽ
=
c2 @t
ei(kz
!t)
.
So (ii)
ei(kz
!t)
.
So (iii) ikEx
ei(kz
!
!t)
.
So (i)
ei(kz
!t)
.
So (v)
ei(kz
!t)
.
So (vi) ikBx
@Bz
=
@x
@By
@x
@Bx
=
@y
!
@ Ẽx
(r ⇥ Ẽ)y =
@z
(r ⇥ B̃)x =
1.5
!
!
!
@Ey
@x
@Bz
@y
i!
Ẽ0 ei(kz
c2
!t)
.
ikEy = i!Bx .
@Ez
= i!By .
@x
@Ex
= i!Bz .
@y
ikBy =
i!
Ex .
c2
i!
Ey .
c2
i!
Ez .
c2
@Ez
@Bz
This confirms Eq. 9.179. Now multiply (iii) by k, (v) by !, and subtract: ik 2 Ex k
!
+ i!kBy =
@x
@y
✓
◆
✓
◆
i! 2
!2
@Ez
@Bz
i
@Ez
@Bz
ik!By + 2 Ex ) i k 2
Ex = k
+!
, or (i) Ex =
k
+!
.
c
c2
@x
@y
(!/c)2 k 2
@x
@y
✓ 2
◆
2
@Ez
@Bz
i!
!
2
Multiply (ii) by k, (vi) by !, and add: k
ik 2 Ey +i!kBx !
= i!kBx
E
)
i
k
Ey =
y
@y
@x
c2
c2
ei(kz
!t)
.
So (iv)
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 9. ELECTROMAGNETIC WAVES
◆
@Ez
@Bz
!
.
k2
@y
@x
! @Ez
!k
@Bz
!2
!k
Multiply (ii) by !/c2 , (vi) by k, and add: 2
i 2 Ey + ik 2 Bx k
= i 2 Bx i 2 Ey )
c
@y
c
@x
c
c
✓
◆
✓
◆
2
!
@B
!
@E
i
@B
!
@E
z
z
z
z
i k2
Bx = k
, or (iii) Bx =
k
.
c2
@x
c2 @y
(!/c)2 k 2
@x
c2 @y
!k
! @Ez
@Bz
!2
i!k
Multiply (iii) by !/c2 , (v) by k, and subtract: i 2 Ex
k
+ ik 2 By = i 2 By + 2 Ex )
2
c
c
@x
@y
c
c
✓
◆
✓
◆
!2
! @Ez
@Bz
i
@Bz
! @Ez
2
i k
By = 2
+k
, or (iv) By =
k
+ 2
.
c2
c @x
@y
(!/c)2 k 2
@y
c @x
This completes the confirmation of Eq. 9.180.
!
@ Ẽ0y
@ Ẽx
@ Ẽy
@ Ẽz
@ Ẽ0x
@Ex
@Ey
(b) r · Ẽ =
+
+
=
+
+ ik Ẽ0z ei(kz !t) = 0 )
+
+ ikEz = 0.
@x
@y
@z
@x
@y
@x
@y
✓ 2
◆
✓
◆
i
@ Ez
@ 2 Bz
i
@ 2 Ez
@ 2 Bz
k
+
!
+
k
!
+ ikEz = 0,
Using Eq. 9.180,
(!/c)2 k 2
@x2
@[email protected]
(!/c)2 k 2
@2y
@[email protected]
⇤
@ 2 Ez
@ 2 Ez ⇥
or
+ 2 + (!/c)2 k 2 Ez = 0.
2
@x
@ y
@Bx
@By
Likewise, r · B̃ = 0 )
+
+ ikBz = 0 )
@x
@y
✓ 2
◆
✓ 2
◆
i
@ Bz
! @ 2 Ez
i
@ Bz
! @Ez
k
+
k
+ 2
+ ikBz = 0 )
(!/c)2 k 2
@x2
c2 @[email protected]
(!/c)2 k 2
@y 2
c @[email protected]
2
2
⇤
@ Bz
@ Bz ⇥
+ 2 + (!/c)2 k 2 Bz = 0.
2
@x
@ y
This confirms Eqs. 9.181. [You can also do it by putting Eq. 9.180 into Eq. 9.179 (i) and (iv).]
k
@Ez
@Bz
i
+!
, or (ii) Ey =
@y
@x
(!/c)2
✓
199
k
Problem 9.28
Here Ez = 0 (TE) and !/c = k (n = m = 0), so Eq. 9.179(ii) ) Ey = cBx , Eq. 9.179(iii) ) Ex = cBy ,
⇣
⌘
⇣
⇣
@Bz
!
! ⌘
@Bz
! ⌘
Eq. 9.179(v) )
= i kBy
E
=
i
kB
B
=
0,
Eq.
9.179(vi)
)
=
i
kB
+
Ey =
x
y
y
x
@y
c2
c
@x
c2
⇣
⌘
!
@Bz
@Bz
i kBx
Bx = 0. So
=
= 0, and since Bz is a function only of x and y, this says Bz is in fact
c
@x
@y
I
Z
@B
a constant (as Eq. 9.186 also suggests). Now Faraday’s law (in integral form) says
E · dl =
· da,
@t
H
R
@B
and Eq. 9.176 )
= i!B, so E · dl = i! B · da. Applied to a cross-section of the waveguide this gives
@t Z
I
i(kz !t)
E · dl = i!e
Bz da = i!Bz ei(kz !t) (ab) (since Bz is constant, it comes outside the integral). But
if the boundary is just inside the metal, where E = 0, it follows that Bz = 0. So this would be a TEM mode,
which we already know cannot exist for this guide.
Problem 9.29
1
c
c
Here a = 2.28 cm and b = 1.01 cm, so ⌫10 =
!10 =
= 0.66 ⇥ 1010 Hz; ⌫20 = 2
= 1.32 ⇥ 1010 Hz;
2⇡
2a
2a
r
c
c
c
c
1
1
10
10
10
⌫30 = 3
= 1.97 ⇥ 10 Hz; ⌫01 =
= 1.49 ⇥ 10 Hz; ⌫02 = 2
= 2.97 ⇥ 10 Hz; ⌫11 =
+ 2 =
2
2a
2b
2b
2 a
b
1.62 ⇥ 1010 Hz. Evidently just four modes occur: 10, 20, 01, and 11.
To get only one mode you must drive the waveguide at a frequency between ⌫10 and ⌫20 :
c
0.66 ⇥ 1010 < ⌫ < 1.32 ⇥ 1010 Hz.
= , so 10 = 2a; 20 = a. 2.28 cm < < 4.56 cm.
⌫
protected under all copyright laws as they currently exist. No portion of this material may be
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200
CHAPTER 9. ELECTROMAGNETIC WAVES
Problem 9.30
1
From Prob. 9.11, hSi =
Re(Ẽ ⇥ B̃⇤ ). Here (Eq. 9.176) Ẽ = Ẽ0 ei(kz
2µ0
the TEmn mode (Eqs. 9.180 and 9.186)
✓
◆
⇣ m⇡x ⌘
⇣ n⇡y ⌘
ik
m⇡
⇤
Bx =
B
sin
cos
;
0
(!/c)2 k 2
a
a
b
✓
◆
⇣ m⇡x ⌘
⇣ n⇡y ⌘
ik
n⇡
By⇤ =
B
cos
sin
;
0
(!/c)2 k 2
b
a
b
⇣ m⇡x ⌘
⇣ n⇡y ⌘
Bz⇤ = B0 cos
cos
;
a ✓
◆b
⇣ m⇡x ⌘
⇣ n⇡y ⌘
i!
n⇡
Ex =
B
cos
sin
;
0
(!/c)2 k 2
b
a
b
✓
◆
⇣ m⇡x ⌘
⇣ n⇡y ⌘
i!
m⇡
B
sin
Ey =
cos
;
0
(!/c)2 k 2
a
a
b
Ez = 0.
!t)
, B̃⇤ = B̃⇤0 e
i(kz !t)
⇣ m⇡x ⌘
⇣ n⇡y ⌘
, and, for
So
hSi =
Z
Ra
0
hSi · da =
!k⇡ 2 B02
[(!/c)2
2
k2 ]
⇣ ⌘
n 2
b
1
!k⇡ 2 B02
ab
8µ0 [(!/c)2 k 2 ]2
sin2 (m⇡x/a) dx =
Similarly,
Ra
0
cos2
⇣ m⇡x ⌘
⇣ ⌘
m 2
a
cos2 (m⇡x/a) dx = a/2;
a
+
Rb
0
sin2
⇣ n ⌘2
⇣ n⇡y ⌘
b
⇣ m ⌘2
+
a
a
cos2
b
sin2 (n⇡y/b) dy =
Rb
0
cos2 (n⇡y/b) dy = b/2.]
✓
◆
1
1
⇤
⇤
hui =
✏0 Ẽ · Ẽ +
B̃ · B̃
4
µ0

⇣ n ⌘2
⇣
⌘
⇣
⌘ ⇣ m ⌘2
⇣
⌘
⇣
⌘
✏0
! 2 ⇡ 2 B02
2 n⇡y
2 m⇡x
2 m⇡x
2 n⇡y
=
cos
sin
+
sin
cos
4 [(!/c)2 k 2 ]2
b
a
b
a
a
b
⇢
⇣
⌘
⇣
⌘
1
m⇡x
n⇡y
+
B02 cos2
cos2
4µ0
a
b
⇣ ⌘
⇣ m⇡x ⌘
⇣
⌘ ⇣ m ⌘2
⇣
⌘
⇣
⌘
k 2 ⇡ 2 B02
n 2
2 n⇡y
2 m⇡x
2
2 n⇡y
+
cos
sin
+
sin
cos
2
b
a
b
a
a
b
[(!/c)2 k 2 ]
Z
ab
hui da =
4
(
o
ẑ .
[In the last step I used
.
b
sin2
.
)
⇣ ⌘
⇣ ⌘
✏0
! 2 ⇡ 2 B02
n 2 ⇣ m ⌘2
B02
1
k 2 ⇡ 2 B02
n 2 ⇣ m ⌘2
+
+
+
+
.
4 [(!/c)2 k 2 ]2
b
a
4µ0
4µ0 [(!/c)2 k 2 ]2
b
a
⇥
These results can be ⇥simplified, using Eq.
9.190
to
write
(!/c)2
⇤
and Eq. 9.188 to write (m/a)2 + (n/b)2 = (!mn /⇡c)2 :
Z
hSi · da =
!kabc2 2
B0 ;
2
8µ0 !mn
Z
hui da =
⇤
k 2 = (!mn /c)2 , ✏0 µ0 = 1/c2 to eliminate ✏0 ,
! 2 ab
B02 .
2
8µ0 !mn
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201
CHAPTER 9. ELECTROMAGNETIC WAVES
Evidently
R
hSi · da
energy per unit time
kc2
cp 2
= R
=
=
!
energy per unit length
!
!
hui da
2
!mn
= vg (Eq. 9.192). qed
Problem 9.31
Following Sect. 9.5.2, the problem is to solve Eq. 9.181 with Ez 6= 0, Bz = 0, subject to the boundary
conditions 9.175. Let Ez (x, y) = X(x)Y (y); as before, we obtain X(x) = A sin(kx x) + B cos(kx x). But the
boundary condition requires Ez = 0 (and hence X = 0) when x = 0 and x = a, so B = 0 and kx = m⇡/a.
But this time m = 1, 2, 3, . . . , but not zero, since m = 0 would kill X entirely. The same goes for Y (y). Thus
⇣ m⇡x ⌘
⇣ n⇡y ⌘
Ez = E0 sin
sin
with n, m = 1, 2, 3, . . . .
a
b
p
The rest is the same as for TE waves: !mn = c⇡ (m/a)2 + (n/b)2 is the cuto↵ frequency, the wave
p
p
velocity is v = c/ 1 (!mn /!)2 , and
the
group
velocity
is
v
=
c
1 (!mn /!)2 . The lowest TM mode is
g
p
2
2
11, with cuto↵ frequency
!11 = c⇡ (1/a) + (1/b) . So the ratio of the lowest TM frequency to the lowest
p
p
c⇡ (1/a)2 + (1/b)2
TE frequency is
=
1 + (a/b)2 .
(c⇡/a)
Problem 9.32
1 @
1 @
@Es ˆ 1 @Es
E0 k sin(kz !t) ˆ ?
(a) r·E =
(sEs ) = 0 X; r·B =
(B ) = 0 X; r⇥E =
ẑ =
=
s @s
s@
@z
s @
s
@B
E0 ! sin(kz !t) ˆ
@B
1 @
E0 k sin(kz !t) ?
=
X (since k = !/c); r ⇥ B =
ŝ +
(sB ) ẑ =
ŝ =
@t
c
s
@z
s @s
c
s
1 @E
E0 ! sin(kz !t)
= 2
ŝ X. Boundary conditions: E k = Ez = 0 X; B ? = Bs = 0 X.
c2 @t
c
s
(b) To determine , use Gauss’s law for a cylinder of radius s and length dz:
cos(kz !t)
1
1
E · da = E0
(2⇡s) dz = Qenc =
dz ) = 2⇡✏0 E0 cos(kz !t).
s
✏0
✏0
To determine I, use Ampére’s law for a circle of radius s (note that the displacement current through this
I
E0 cos(kz !t)
2⇡E0
loop is zero, since E is in the ŝ direction):
B·dl =
(2⇡s) = µ0 Ienc ) I =
cos(kz !t).
c
s
µ0 c
I
The charge and current on the outer conductor are precisely the opposite of these, since E = B = 0 inside
the metal, and hence the total enclosed charge and current must be zero.
Problem 9.33
Z 1
Z 1
f˜(z, 0) =
Ã(k)eikz dk ) f˜(z, 0)⇤ =
Ã(k)⇤ e ikz dk. Let l ⌘ k; then f˜(z, 0)⇤ =
1
Z 1
Z 1
Z11
⇤ ilz
⇤ ilz
Ã( l) e ( dl) =
Ã( l) e dl =
Ã( k)⇤ eikz dk (renaming the dummy variable l ! k).
1
1
1
h
i 1h
i Z 1 1h
i
⇤
˜
˜
˜
f (z, 0) = Re f (z, 0) =
f (z, 0) + f (z, 0) =
Ã(k) + Ã( k)⇤ eikz dk. Therefore
2
1 2
Z 1
i
1h
1
⇤
ikz
Ã(k) + Ã( k) =
f (z, 0)e
dz.
2
2⇡
1
Z 1
Z 1
˙
˙
Meanwhile, f˜(z, t) =
Ã(k)( i!)ei(kz !t) dk ) f˜(z, 0) =
[ i! Ã(k)]eikz dk.
1
1
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202
CHAPTER 9. ELECTROMAGNETIC WAVES
(Note that ! = |k|v, here, so it does not come outside the integral.)
˙
f˜(z, 0)⇤ =
=
Z
Z
1
1
1
[i! Ã(k)⇤ ]e
ikz
dk =
Z
[i|k|v Ã( k)⇤ ]eikz dk =
1
[i|k|v Ã(k)⇤ ]e
1
Z
ikz
dk =
Z
1
[i|l|v Ã( l)⇤ ]eilz ( dl)
1
1
[i! Ã( k)⇤ ]eikz dk.
h
i 1h
i Z 1 1
˙
˙
˙
f˙(z, 0) = Re f˜(z, 0) =
f˜(z, 0) + f˜(z, 0)⇤ =
[ i! Ã(k) + i! Ã( k)⇤ ]eikz dk.
2
1 2
1
1
Z 1
i
1h
1
i ˙
Ã(k) Ã( k)⇤ =
f (z, 0) e
2
2⇡
!
1
1
Z 1
1
i
Adding these two results, we get Ã(k) =
f (z, 0) + f˙(z, 0) e ikz dz. qed
2⇡
!
1
i! h
Ã(k)
2
Z
i
1
Ã( k)⇤ =
2⇡
1
f˙(z, 0)e
ikz
dz, or
ikz
dz.
Problem 9.34
(a) Since (E ⇥ B) points in the direction of propagation, B =
(b) From Eq. 7.63, K ⇥ ( ẑ) =
E0
[cos(kz
c
!t) + cos(kz + !t)] ŷ.
1
E0
2E0
B=
[2 cos(!t)] ŷ, K =
cos(!t) x̂.
µ0
µ0 c
µ0 c
(c) The force per unit area is f = K ⇥ Bave =
average of cos2 (!t) is 1/2, so
2E02
[cos(!t) x̂] ⇥ [cos(!t) ŷ] = 2✏0 E02 cos2 (!t) ẑ. The time
µ0 c2
fave = ✏0 E02 .
This is twice the pressure in Eq. 9.64, but that was for a perfect absorber, whereas this is a perfect reflector.
Problem 9.35
(a) (i) Gauss’s law: r · E =
1 @E
= 0. X
r sin ✓ @
@B
1
@
1 @
= r⇥E=
(sin ✓E ) r̂
(rE ) ✓ˆ
@t
r sin ✓ @✓
r @r

✓
◆

✓
◆
1
@
sin2 ✓
1
1 @
1
ˆ
=
E0
cos u
sin u r̂
E0 sin ✓ cos u
sin u ✓.
r sin ✓ @✓
r
kr
r @r
kr
@
@
But
cos u = k sin u;
sin u = k cos u.
@r
✓ @r
◆
✓
◆
1 E0
1
1
1
1
ˆ
=
2 sin ✓ cos ✓ cos u
sin u r̂
E0 sin ✓
k sin u + 2 sin u
cos u ✓.
r sin ✓ r
kr
r
kr
r
Integrating with respect to t, and noting that
B=
2E0 cos ✓
!r2
✓
sin u +
Z
cos u dt =
1
sin u and
!
Z
sin u dt =
1
cos u, we obtain
!
◆
✓
◆
1
E0 sin ✓
1
1
ˆ
cos u r̂ +
k cos u + 2 cos u + sin u ✓.
kr
!r
kr
r
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203
CHAPTER 9. ELECTROMAGNETIC WAVES
(iii) Divergence of B:
1 @
1
@
r2 Br +
(sin ✓B✓ )
2
r @r
r sin ✓ @✓

✓
◆

✓
◆
1 @ 2E0 cos ✓
1
1
@ E0 sin2 ✓
1
1
= 2
sin u +
cos u +
k cos u + 2 cos u + sin u
r @r
!
kr
r sin ✓ @✓
!r
kr
r
✓
◆
1 2E0 cos ✓
1
1
= 2
k cos u
cos u
sin u
r
!
kr2
r
✓
◆
1 2E0 sin ✓ cos ✓
1
1
+
k cos u + 2 cos u + sin u
r sin ✓
!r
kr
r
r·B =
2E0 cos ✓
=
!r2
✓
k cos u
1
cos u
kr2
1
sin u
r
1
1
k cos u + 2 cos u + sin u
kr
r
◆
= 0. X
(iv) Ampére/Maxwell:

1 @
@Br ˆ
r⇥B =
(rB✓ )
r @r
@✓
⇢ 
✓
◆

✓
◆
1 @ E0 sin ✓
1
1
@ 2E0 cos ✓
1
=
k cos u + 2 cos u + sin u
sin u +
cos u
r @r
!
kr
r
@✓
!r2
kr
✓
◆
E0 sin ✓
2
1
1
k
2
2
2
=
k sin u
cos u
sin u
sin u + cos u + 2 sin u + 3 cos u ˆ
!r
kr3
r2
r2
r
r
kr
✓
◆
✓
◆
k E0 sin ✓
1
1
E
sin
✓
1
0
=
k sin u + cos u ˆ =
k sin u + cos u ˆ.
!
r
r
c
r
r
✓
◆
⌘
1 @E
1 E0 sin ✓ ⇣
!
1
!
E
sin
✓
1
0
ˆ
= 2
! sin u +
cos u
= 2
k sin u + cos u ˆ
c2 @t
c
r
kr
c k
r
r
✓
◆
1 E0 sin ✓
1
=
k sin u + cos u ˆ = r ⇥ B. X
c
r
r
(b) Poynting Vector:
✓
◆
✓
◆
1
E0 sin ✓
1
2E0 cos ✓
1
(E ⇥ B) =
cos u
sin u
sin
u
+
cos
u
✓ˆ
µ0
µ0 r
kr
!r2
kr
✓
◆
E0 sin ✓
1
1
+
k cos u + 2 cos u + sin u ( r̂)
!r
kr
r
⇢

2
E sin ✓ 2 cos ✓
1
1
= 0 2
sin u cos u + (cos2 u sin2 u)
sin u cos u ✓ˆ
µ0 !r
r
kr
k2 r2
✓
◆
1
1
1
1
1
2
sin ✓
k cos2 u + 2 cos2 u + sin u cos u + sin u cos u
sin
u
cos
u
sin
u
r̂
kr
r
r
k2 r3
kr2
⇢
✓
◆
E02 sin ✓ 2 cos ✓
1
1
1
sin u cos u + (cos2 u sin2 u) ✓ˆ
µ0 !r2
r
k2 r2
kr
=
✓
◆
2
1
1
+ sin ✓
+ 2 3 sin u cos u + k cos2 u + 2 (sin2 u cos2 u) r̂ .
r
k r
kr
S=
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ˆ
204
CHAPTER 9. ELECTROMAGNETIC WAVES
Averaging over a full cycle, using hsin u cos ui = 0, hsin2 ui = hcos2 ui = 12 , we get the intensity:
I = hSi =
E02 sin ✓
µ0 !r2
✓
◆
k
E 2 sin2 ✓
sin ✓ r̂ = 0
r̂.
2
2µ0 cr2
It points in the r̂ direction, and falls o↵ as 1/r2 , as we would expect for a spherical wave.
Z
Z
Z ⇡
E02
sin2 ✓ 2
E02
4⇡ E02
3
(c) P = I · da =
r
sin
✓
d✓
d
=
2⇡
sin
✓
d✓
=
.
2µ0 c
r2
2µ0 c
3 µ0 c
0
Problem 9.36
z<0:
(
ẼI (z, t) = ẼI ei(k1 z !t) x̂, B̃I (z, t) = v11 ẼI ei(k1 z
ẼR (z, t) = ẼR ei( k1 z !t) x̂, B̃R (z, t) = v11 ẼR ei(
0<z<d:
(
Ẽr (z, t) = Ẽr ei(k2 z
Ẽl (z, t) = Ẽl ei( k2 z
z>d:
n
ẼT (z, t) = ẼT ei(k3 z
k
k
!t)
x̂, B̃r (z, t) =
x̂, B̃l (z, t) =
!t)
!t)
k
x̂, B̃T (z, t) =
!t)
ŷ
k1 z !t)
1
i(k2 z !t)
ŷ
v2 Ẽr e
1
i( k2 z !t)
Ẽ
e
v2 l
1
i(k3 z !t)
v3 ẼT e
ŷ.
ŷ.
ŷ.
k
Boundary conditions: E1 = E2 , B1 = B2 , at each boundary (assuming µ1 = µ2 = µ3 = µ0 ):
z=0:
z=d:
8
>
< ẼI + ẼR = Ẽr + Ẽl ;
>
: 1 ẼI
v1
1
1
ẼR = Ẽr
v1
v2
8
ik2 d
+ Ẽl e
>
< Ẽr e
>
: 1 Ẽr eik2 d
v2
ik2 d
1
Ẽl e
v2
1
Ẽl ) ẼI
v2
ẼR = (Ẽr
Ẽl ), where
⌘ v1 /v2 .
= ẼT eik3 d ;
ik2 d
=
1
ẼT eik3 d ) Ẽr eik2 d
v3
Ẽl e
ik2 d
= ↵ẼT eik3 d , where ↵ ⌘ v2 /v3 .
We have here four equations; the problem is to eliminate ẼR , Ẽr , and Ẽl , to obtain a single equation for
ẼT in terms of ẼI .
Add the first two to eliminate ẼR :
2ẼI = (1 + )Ẽr + (1
)Ẽl ;
Add the last two to eliminate Ẽl :
2Ẽr eik2 d = (1 + ↵)ẼT eik3 d ;
Subtract the last two to eliminate Ẽr : 2Ẽl e ik2 d = (1 ↵)Ẽ T eik3 d .
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205
CHAPTER 9. ELECTROMAGNETIC WAVES
Plug the last two of these into the first:
1
1
2ẼI = (1 + ) e ik2 d (1 + ↵)ẼT eik3 d + (1
) eik2 d (1 ↵)ẼT eik3 d
2
2
⇥
⇤
4ẼI = (1 + ↵)(1 + )e ik2 d + (1 ↵)(1
)eik2 d ẼT eik3 d
⇥
⇤
= (1 + ↵ ) e ik2 d + eik2 d + (↵ + ) e ik2 d eik2 d ẼT eik3 d
= 2 [(1 + ↵ ) cos(k2 d)
i(↵ + ) sin(k2 d)] ẼT eik3 d .
✓
◆
v3 ✏3 ET20
v3 µ0 ✏3 |ẼT |2
v1 |ẼT |2
|ẼT |2
Now the transmission coefficient is T =
=
=
=↵
, so
2
v1 ✏1 EI0
v1 µ0 ✏1 |ẼI |2
v3 |ẼI |2
|ẼI |2
T
1
1
↵
1
=
4↵
1
=
4↵
1
=
4↵
=
2
|ẼI |2
1 1
=
[(1 + ↵ ) cos(k2 d) i(↵ + ) sin(k2 d)] eik3 d
↵ 2
|ẼT |2
⇥
⇤
(1 + ↵ )2 cos2 (k2 d) + (↵ + )2 sin2 (k2 d) . But cos2 (k2 d) = 1
⇥
(1 + ↵ )2 + (↵2 + 2↵ +
⇥
(1 + ↵ )2
But n1 =
(1
↵2 )(1
1
2
2
2↵
⇤
) sin2 (k2 d) .
c
c
c
n3
, n2 = , n3 = , so ↵ =
,
v1
v2
v3
n2

1
(n2
=
(n1 + n3 )2 + 1
4n1 n3
n22 )(n23
n22
n22 )
↵2
=
2
sin2 (k2 d).
⇤
) sin2 (k2 d)
n2
.
n1
sin2 (k2 d) .
Problem 9.37
T = 1 ) sin kd = 0 )p
kd = 0, ⇡, 2⇡ . . .. The minimum (nonzero) thickness is d p
= ⇡/k. But k = !/v =
p
2⇡⌫/v = 2⇡⌫n/c, and n = ✏µ/✏0 µ0 (Eq. 9.69), where (presumably) µ ⇡ µ0 . So n = ✏/✏0 = ✏r , and hence
8
⇡c
c
3 ⇥ 10
p
d=
= 9.49 ⇥ 10 3 m, or 9.5 mm.
p = p =
2⇡⌫ ✏r
2⌫ ✏r
2(10 ⇥ 109 ) 2.5
Problem 9.38
From Eq. 9.199,
T
⇢
1
[(16/9) (9/4)][1 (9/4)]
[(4/3) + 1]2 +
sin2 (3!d/2c)
4(4/3)(1)
(9/4)

3 49 ( 17/36)( 5/4)
49
85
=
+
sin2 (3!d/2c) =
+
sin2 (3!d/2c).
16 9
(9/4)
48 (48)(36)
48
T =
.
49 + (85/36) sin2 (3!d/2c)
1
=
48
48
= 0.935; Tmax =
= 0.980. Not much
49 + (85/36)
49
variation, and the transmission is good (over 90%) for all frequencies. Since Eq. 9.199 is unchanged when you
switch 1 and 3, the transmission is the same either direction, and the fish sees you just as well as you see it.
Since sin2 (3!d/2c) ranges from 0 to 1, Tmin =
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206
CHAPTER 9. ELECTROMAGNETIC WAVES
Problem 9.39
i(kT ·r
(a) Equation 9.91 ) ẼT (r, t) = Ẽ0T eq
kT (x sin ✓T + z cos ✓T ) = xkT sin ✓T + izkT
!t)
sin ✓T
2
; kT · r = kT (sin ✓T x̂ + cos ✓T ẑ) · (x x̂ + y ŷ + z ẑ) =
1 = kx + iz, where
⇣ !n ⌘ n
!n1
2
1
sin ✓I =
sin ✓I ,
c
n2
c
q
q
!n2
!
1=
(n1 /n2 )2 sin2 ✓I 1 =
n21 sin2 ✓I
c
c
ẼT (r, t) = Ẽ0T e z ei(kx !t) . qed
k ⌘ kT sin ✓T =
q
 ⌘ kT sin2 ✓T
2
2
↵
. Here is real (Eq. 9.106) and ↵ is purely imaginary (Eq. 9.108); write ↵ = ia,
↵+
✓
◆✓
◆
ia
ia
a2 + 2
with a real: R =
= 2
= 1.
ia +
ia +
a + 2
(b) R =
Ẽ0R
Ẽ0I
n22 . So
=
(c) From Prob. 9.17, E0R =
1 ↵
1+↵
E0I , so R =
1 ↵
1+↵
2
=
2
1 ia
1 + ia
=
(1 ia )(1 + ia )
= 1.
(1 + ia )(1 ia )
(d) From the solution to Prob. 9.17, the transmitted wave is
Ẽ(r, t) = Ẽ0T ei(kT ·r
!t)
ŷ,
B̃(r, t) =
Using the results in (a): kT · r = kx + iz, sin ✓T =
Ẽ(r, t) = Ẽ0T e
z i(kx !t)
e
ŷ,
1
Ẽ0 ei(kT ·r
v2 T
!t)
( cos ✓T x̂ + sin ✓T ẑ).
ck
c
, cos ✓T = i
:
!n2
!n2
B̃(r, t) =
1
Ẽ0 e
v2 T
z i(kx !t)
e
✓
i
◆
c
ck
x̂ +
ẑ .
!n2
!n2
We may as well choose the phase constant so that Ẽ0T is real. Then
E(r, t) = E0 e z cos(kx !t) ŷ;
1
c
B(r, t) =
E0 e z
Re {[cos(kx !t) + i sin(kx !t)] [ i x̂ + k ẑ]}
v2
!n2
1
= E0 e z [ sin(kx !t) x̂ + k cos(kx !t) ẑ] . qed
!
(I used v2 = c/n2 to simplfy B.)
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CHAPTER 9. ELECTROMAGNETIC WAVES
⇤
@ ⇥
E0 e z cos(kx !t) = 0. X
@y


@ E0 z
@ E0 z
(ii) r · B =
e
 sin(kx !t) +
e
k cos(kx !t)
@x !
@z !
⇤
E0 ⇥ z
=
e
k cos(kx !t) e z k cos(kx !t) = 0. X
!
x̂
ŷ
ẑ
@Ey
@Ey
(iii) r ⇥ E = @/@x @/@y @/@z =
x̂ +
ẑ
@z
@x
0
Ey
0
(e)
(i) r · E =
= E0 e z cos(kx !t) x̂ E0 e z k sin(kx !t) ẑ.
@B
E0 z
=
e
[ ! cos(kx !t) x̂ + k! sin(kx !t) ẑ]
@t
!
= E0 e z cos(kx !t) x̂ kE0 e z sin(kx !t) ẑ = r ⇥ E. X
◆
✓
x̂
ŷ
ẑ
@Bx
@Bz
(iv) r ⇥ B = @/@x @/@y @/@z =
ŷ
@z
@x
Bx
0
Bz

E0 2 z
E0 z 2
E0
=
 e
sin(kx !t) +
e
k sin(kx !t) ŷ = (k 2 2 ) e z sin(kx !t) ŷ.
!
!
!
⇣ ! ⌘2 ⇥
⇣ n ! ⌘2
⇤
2
Eq. 9.202 ) k 2 2 =
n21 sin2 ✓I (n1 sin ✓I )2 + (n2 )2 =
= ! 2 ✏2 µ2 .
c
c
= ✏2 µ2 !E0 e
@E
µ2 ✏2
= µ2 ✏2 E0 e
@t
z
z
sin(kx
!t) ŷ.
!t) ŷ = r ⇥ B X.
! sin(kx
(f)
S=
=
1
1 E02
(E ⇥ B) =
e
µ2
µ2 !
E02
e
µ2 !
2z
⇥
k cos2 (kx
2z
x̂
0
 sin(kx
!t) x̂
ŷ
cos(kx
!t)
0
 sin(kx
!t)
!t) cos(kx
ẑ
0
k cos(kx
!t)
⇤
!t) ẑ .
E02 k 2z
e
x̂. On average,
2µ2 !
then, no energy is transmitted in the z direction, only in the x direction (parallel to the interface). qed
Averaging over a complete cycle, using hcos2 i = 1/2 and hsin cosi = 0, hSi =
Problem 9.40
Look for solutions of the form E = E0 (x, y, z)e i!t , B = B0 (x, y, z)e i!t , subject to the boundary condik
?
tions
equations, in the form of Eq. 9.177, give
⇢ E = 0, B = 0 at all surfaces. Maxwell’s
r · E = 0 ) r · E0 = 0; r ⇥ E = @B
)
r ⇥ E0 = i!B0 ;
@t
1 @E
r · B = 0 ) r · B0 = 0; r ⇥ B = c2 @t ) r ⇥ B0 = i!
c2 E0 .
From
now
on
I’ll
leave
o↵
the
subscript
(0).
The
problem
is
to
solve
the (time independent) equations
⇢
r · E = 0; r ⇥ E = i!B;
r · B = 0; r ⇥ B = i!
c2 E.
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208
CHAPTER 9. ELECTROMAGNETIC WAVES
From r ⇥ E = i!B it follows that I can get B once I know E, so I’ll
on the latter for the moment.
✓ concentrate
◆
2
i!
!
r ⇥ (r ⇥ E) = r(r · E) r2 E = r2 E = r ⇥ (i!B) = i!
E
=
E. So
c2
c2
⇣ ! ⌘2
⇣ ! ⌘2
⇣ ! ⌘2
r 2 Ex =
Ex ; r 2 Ey =
Ey ; r 2 E z =
Ez . Solve each of these by separation of variables:
c
c
c
⇣ ! ⌘2
2
2
2
d X
d Y
d Z
1 d2 X 1 d2 Y 1 d2 Z
Ex (x, y, z) = X(x)Y (y)Z(z) ) Y Z 2 +ZX 2 +XY 2 =
XY Z, or
+
+
=
dx
dy
dz
c
X dx2 Y dy 2 Z dz 2
d2 X
d2 Y
d2 Z
2
(!/c) . Each term must be a constant, so
= kx2 X,
= ky2 Y,
= kz2 Z, with
2
2
dx
dy
dz 2
2
kx2 + ky2 + kz2 = (!/c) . The solution is
Ex (x, y, z) = [A sin(kx x) + B cos(kx x)][C sin(ky y) + D cos(ky y)][E sin(kz z) + F cos(kz z)].
But Ek = 0 at the boundaries ) Ex = 0 at y = 0 and z = 0, so D = F = 0, and Ex = 0 at y = b and z = d, so
ky = n⇡/b and kz = l⇡/d, where n and l are integers. A similar argument applies to Ey and Ez . Conclusion:
Ex (x, y, z) = [A sin(kx x) + B cos(kx x)] sin(ky y) sin(kz z),
Ey (x, y, z) = sin(kx x)[C sin(ky y) + D cos(ky y)] sin(kz z),
Ez (x, y, z) = sin(kx x) sin(ky y)[E sin(kz z) + F cos(kz z)],
where kx = m⇡/a. (Actually, there is no reason at this stage to assume that kx , ky , and kz are the same for
all three components, and I should really affix a second subscript (x for Ex , y for Ey , and z for Ez ), but in a
moment we shall see that in fact they do have to be the same, so to avoid cumbersome notation I’ll assume
they are from the start.)
Now r·E = 0 ) kx [A cos(kx x) B sin(kx x)] sin(ky y) sin(kz z)+ky sin(kx x)[C cos(ky y) D sin(ky y)] sin(kz z)+
kz sin(kx x) sin(ky y)[E cos(kz z) F sin(kz z)] = 0. In particular, putting in x = 0, kx A sin(ky y) sin(kz z) = 0,
and hence A = 0. Likewise y = 0 ) C = 0 and z = 0 ) E = 0. (Moreover, if the k’s were not equal for di↵erent
components, then by Fourier analysis this equation could not be satisfied (for all x, y, and z) unless the other
three constants were also zero, and we’d be left with no field at all.) It follows that (Bkx + Dky + F kz ) = 0
(in order that r · E = 0), and we are left with
E = B cos(kx x) sin(ky y) sin(kz z) x̂ + D sin(kx x) cos(ky y) sin(kz z) ŷ + F sin(kx x) sin(ky y) cos(kz z) ẑ,
with kx = (m⇡/a), ky = (n⇡/b), kz = (l⇡/d) (l, m, n all integers), and Bkx + Dky + F kz = 0.
Or:
The corresponding magnetic field is given by B = (i/!)r ⇥ E:
✓
◆
i @Ez
@Ey
i
Bx =
=
[F ky sin(kx x) cos(ky y) cos(kz z)
!
@y
@z
!
✓
◆
i @Ex
@Ez
i
By =
=
[Bkz cos(kx x) sin(ky y) cos(kz z)
!
@z
@x
!
✓
◆
i @Ey
@Ex
i
Bz =
=
[Dkx cos(kx x) cos(ky y) sin(kz z)
!
@x
@y
!
B =
i
(F ky
!
i
(Dkx
!
Dkz ) sin(kx x) cos(ky y) cos(kz z) x̂
i
(Bkz
!
Dkz sin(kx x) cos(ky y) cos(kz z)] ,
F kx cos(kx x) sin(ky y) cos(kz z)] ,
Bky cos(kx x) cos(ky y) sin(kz z)] .
F kx ) cos(kx x) sin(ky y) cos(kz z) ŷ
Bky ) cos(kx x) cos(ky y) sin(kz z) ẑ.
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CHAPTER 9. ELECTROMAGNETIC WAVES
209
These automatically satisfy the boundary condition B ? = 0 (Bx = 0 at x = 0 and x = a, By = 0 at y = 0 and
y = b, and Bz = 0 at z = 0 and z = d).
As a check, let’s see if r · B = 0 :
r·B =
=
i
i
(F ky Dkz )kx cos(kx x) cos(ky y) cos(kz z)
(Bkz F kx )ky cos(kx x) cos(ky y) cos(kz z)
!
!
i
(Dkx Bky )kz cos(kx x) cos(ky y) cos(kz z)
!
i
(F kx ky Dkx kz + Bkz ky F kx ky + Dkx kz Bky kz ) cos(kx x) cos(ky y) cos(kz z) = 0. X
!
The boxed equations satisfy all of Maxwell’s equations, and they meet the boundary conditions. For TE
modes, we pick Ez = 0, so F = 0 (and hence Bkx + Dky = 0, leaving only the overall amplitude undetermined,
for given l, m, and n); for TM modes we want Bz = 0 (so Dkx Bky = 0, again leaving only one amplitude
undetermined, since Bkx + Dky + F kz = 0). In either case (TElmn or TMlmn ), the frequency is given by
p
⇥
⇤
! 2 = c2 (kx2 + ky2 + kz2 ) = c2 (m⇡/a)2 + (n⇡/b)2 + (l⇡/d)2 , or ! = c⇡ (m/a)2 + (n/b)2 + (l/d)2 .
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210
CHAPTER 10. POTENTIALS AND FIELDS
Chapter 10
Potentials and Fields
Problem 10.1
22 V +
@L
= r2 V
@t
22 A
rL = r2 A
@2V
@
@2V
@
+
(r
·
A)
+
µ
✏
= r2 V + (r · A) =
0
0
2
2
@t
@t
@t
@t
✓
◆
@2A
@V
µ0 ✏0 2
r r · A + µ0 ✏0
= µ0 J. X
@t
@t
µ0 ✏0
Problem 10.2
◆
Z ✓
1
1 2
(a) W =
✏0 E 2 +
B d⌧. At t1 = d/c, x
2
µ0
At T2 = (d + h)/c, ct2 = d + h:
E=
so B 2 =
1 2
E , and
c2
Therefore
1
µ2 k 2
W (t2 ) = (2✏0 ) 0
2
4
✓
µ0 k
(d + h
2
✏0 E 2 +
(d+h)
Z
1 2
B
µ0
(d + h
◆
d = ct1 , so E = 0, B = 0, and hence W (t1 ) = 0.
x) ẑ,
B=
✓
= ✏0 E 2 +
1 µ0 k
(d + h
c 2
1 1 2
E
µ0 ✏0 c2
✏0 µ20 k 2 lw
x) dx (lw) =
4
2
d
(b) S(x) =
1
⇢. X
✏0

◆
x) ŷ,
= 2✏0 E 2 .
(d + h
3
1
1 2
1 2
µ0 k 2
(E ⇥ B) =
E [ ẑ ⇥ (±ŷ)] = ±
E x̂ = ±
(ct
4c
µ0
µ0 c
µ0 c
x)3
d+h
=
d
✏0 µ20 k 2 lwh3
.
12
|x|)2 x̂
(plus sign for x > 0, as here). For |x| > ct, S = 0.
So the energy per unit time entering the box in this time interval is
Z
dW
µ0 k 2 lw
= P = S(d) · da =
(ct d)2 .
dt
4c
Note that no energy flows out the top, since S(d + h) = 0.
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211
CHAPTER 10. POTENTIALS AND FIELDS
Zt2
(c) W =
µ0 k 2 lw
P dt =
4c
t1
(d+h)/c
Z
(ct
d)2 dt =
d/c

µ0 k 2 lw (ct d)3
4c
3c
(d+h)/c
µ0 k 2 lwh3
.
12c2
=
d/c
Since 1/c2 = µ0 ✏0 , this agrees with the answer to (a).
Problem 10.3
(a)
E=
rV
@A
1 q
=
r̂.
@t
4⇡✏0 r2
B = r⇥A = 0.
This is a funny set of potentials for a stationary point charge q at the origin. (V =
1 q
, A = 0 would, of
4⇡✏0 r
course, be the customary choice.) Evidently ⇢ = q 3 (r); J = 0.
(b)
@
=0
@t
V =V
0
✓
1 q
4⇡✏0 r
◆
1 q
=
; A0 = A + r =
4⇡✏0 r
1 qt
r̂ +
4⇡✏0 r2
✓
1
qt
4⇡✏0
◆✓
◆
1
r̂ = 0.
r2
This gauge function transforms the “funny” potentials into the “ordinary” potentials of a stationary point
charge.
Problem 10.4
@A
= A0 cos(kx !t) ŷ( !) = A0 ! cos(kx
@t
@
B = r⇥A = ẑ
[A0 sin(kx !t)] = A0 k cos(kx !t) ẑ.
@x
E=
rV
!t) ŷ,
Hence r·E = 0 X, r·B = 0 X.
r⇥E = ẑ
so r⇥E =
@
[A0 ! cos(kx
@x
!t)] =
A0 !k sin(kx
!t) ẑ,
@B
=
@t
A0 !k sin(kx
!t) ẑ,
@B
X.
@t
r⇥B =
So r⇥B = µ0 ✏0
ŷ
@
[A0 k cos(kx
@x
!t)] = A0 k 2 sin(kx
!t) ŷ,
@E
= A0 ! 2 sin(kx
@t
@E
provided k 2 = µ0 ✏0 ! 2 , or, since c2 = 1/µ0 ✏0 , ! = ck.
@t
Problem 10.5
@V
Ex. 10.1: r·A = 0;
= 0.
Both Coulomb and Lorentz.
@t
✓ ◆
qt
r̂
qt 3
@V
Prob. 10.3: r·A =
r·
=
(r);
= 0.
Neither.
4⇡✏0
r2
✏0
@t
@V
Prob. 10.4: r·A = 0;
= 0.
Both.
@t
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reproduced, in any form or by any means, without permission in writing from the publisher.
!t) ŷ.
212
Problem 10.6
Suppose r·A 6=
CHAPTER 10. POTENTIALS AND FIELDS
@V
@V
. (Let r·A + µ0 ✏0
=
@t
@t 0
@V
that A0 and V 0 (Eq. 10.7) do obey r·A0 = µ0 ✏0
.
@t
µ0 ✏0
r·A0 + µ0 ✏0
—some known function.) We want to pick
@V 0
@V
= r·A + r2 + µ0 ✏0
@t
@t
µ0 ✏0
@2
=
@t2
such
+ 22 .
This will be zero provided we pick for the solution to 22 =
, which by hypothesis (and in fact) we know
how to solve.
Rt
We could always find a gauge in which V 0 = 0, simply by picking = 0 V dt0 . We cannot in general pick
A = 0—this would make B = 0. [Finding such a gauge function would amount to expressing A as r , and
we know that vector functions cannot in general be written as gradients—only if they happen to have curl
zero, which A (ordinarily) does not.]
Problem 10.7
(a) Using Eq. 1.99,
r·J=
(b) From Eq. 10.10,
q̇
r·
4⇡
✓
r̂
r2
V (r, t) =
◆
=
q
4⇡✏0
Z
q̇
4⇡ 3 (r) =
4⇡
3
(r0 )
r
d⌧ 0 =
q̇ 3 (r) =
@⇢
. X
@t
1 q(t)
.
4⇡✏0 r
By symmetry, B = 0 (what direction could it point?), so r ⇥ A = 0, r · A = 0, and A ! 0 at infinity.
Evidently A = 0.
(c) E =
@A
1 q(t)
=
r̂.
B = 0. Checking Maxwell’s equations:
@t
4⇡✏0 r2
✓ ◆
q
r̂
q
q 3 (r)
⇢
r·
=
4⇡ 3 (r) =
= . X
2
4⇡✏0
r
4⇡✏0
✏0
✏0
0. X
@B
0=
. X
@t
✓
◆
✓
◆
@E
q̇ 1
1 q̇
0; µ0 J + µ0 ✏0
= µ0
r̂
+
µ
✏
r̂
= 0. X
0
0
@t
4⇡ r2
4⇡✏0 r2
rV
r·E =
r·B =
r⇥E =
r⇥B =
[Note that the displacement current exactly cancels the conduction current. Physically, this configuration is a
point charge at the origin that is changing with time as current flows in symmetrically (from infinity).]
Problem 10.8
Noting the A is independent of t and B is independent of r, use Eq. 10.19:

dA
@A
1
=
+ (v · r)A = (v · r)
(r ⇥ B)
dt
@t
2

1
@
@
@
=
vx
+ vy
+ vz
[(yBz zBy ) x̂ + (zBx xBz ) ŷ + (xBy yBx ) ẑ]
2
@x
@y
@z
1
=
[vx ( Bz ŷ + By ẑ) + vy (Bz x̂ Bx ẑ) + vz ( By x̂ + Bx ŷ)]
2
1
1
=
[(vy Bz vz By ) x̂ + (vz Bx vx Bz ) ŷ + (vx By vy Bx ) ẑ] =
(v ⇥ B). X
2
2
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213
CHAPTER 10. POTENTIALS AND FIELDS
Equation 10.20 says
d
dp
(p + qA) =
dt
dt
q
(v ⇥ B) = qr(v · A) =
2
q
r[v · (r ⇥ B)],
2
or
dp
q
q
= (v ⇥ B)
r[r · (B ⇥ v)].
dt
2
2
Now, for a vector c that is independent of position,
✓
◆
@
@
@
r(r · c) = x̂
+ ŷ
+ ẑ
(xcx + ycy + zcz ) = cx x̂ + cy ŷ + cz ẑ) = c.
@x
@y
@z
In this case c = (B ⇥ v) =
(v ⇥ B), so
dp
q
q
= (v ⇥ B) + (v ⇥ B) = q(v ⇥ B). X
dt
2
2
Problem 10.9
d
d
(T + qV ) =
dt
dt
✓
1
mv 2
2
◆
+q

dV
dv
@V
= mv ·
+q
+ (v · r)V
dt
dt
@t
=v·F+q

@V
+ (v · r)V .
@t
But Eq. 10.17 says
v · F = qv ·
so

rV


@A
@V
q (v · r)V + v ·
+q
+ (v · r)V
@t
@t
d
(T + qV ) =
dt
Problem 10.10
From the product rule:
✓ ◆
✓
◆
J
1
1
r·
=
(r·J) + J · r
,
But r
1
r
=
r0
1
r
r·
r
r
, since
r
✓
J
r
◆
=
r
=r
1
r
But
r·J =
and

@A
q v · rV + v ·
,
@t
@A
+ v ⇥ (r ⇥ A) =
@t
0
r·
✓
J
r
=q
◆
=
✓
@V
@t
1
r
@A
v·
@t
◆
✓
(r ·J) + J · r
0
0
1
r
◆
r0 . So
(r·J)
✓
◆
1
1
1
J · r0
=
(r·J) +
(r0 ·J)
r
r
r
r0 ·
@Jx
@Jy
@Jz
@Jx @tr
@Jy @tr
@Jz @tr
+
+
=
+
+
,
@x
@y
@z
@tr @x
@tr @y
@tr @z
@tr
=
@x
@
[q(V
@t
=
[email protected]
,
c @x
@tr
=
@y
[email protected]
,
c @y
@tr
=
@z
[email protected]
,
c @z
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
✓
J
r
◆
.
.
v · A)] .
214
so
r·J =
Similarly,
CHAPTER 10. POTENTIALS AND FIELDS

1 @Jx @ r
@Jy @ r
@Jz @ r
+
+
c @tr @x
@tr @y
@tr @z
=
1 @J
· (r r ).
c @tr
@⇢ 1 @J
· (r0 r ).
@t
c @tr
[The first term arises when we di↵erentiate with respect to the explicit r0 , and use the continuity equation.]
thus
✓ ◆


✓ ◆
✓ ◆
J
1
1 @J
1
@⇢ 1 @J
J
1 @⇢
J
r·
=
· (r0 r ) +
· (r0 r )
r·
=
r0 ·
r
r
r
r
r @t
r
c @tr
@t
c @tr
r0 ·J =
(the other two terms cancel, since r r = r0 r ). Therefore:

✓ ◆

Z
Z
Z
I
µ0
@
⇢
J
@
1
⇢
µ0
J
0
r·A =
r·
r d⌧
r d⌧ = µ0 ✏0 @t 4⇡✏0 r d⌧ 4⇡ r · da.
4⇡
@t
@V
The last term is over the suface at “infinity”, where J = 0, so it’s zero. Therefore r·A = µ0 ✏0
.X
@t
Problem 10.11
(a) As in Ex. 10.2, for t < s/c, A = 0; for t > s/c,
8 p
p 2 2
p 2 2 9
(ct) s
(ct)2 s2
(ct) s
>
>
p
>
>
Z
Z
Z
=
⇣µ ⌘
k(t
s2 + z 2 /c)
µ0 k <
dz
1
0
p
p
A(s, t) =
ẑ 2
dz =
ẑ t
dz
>
4⇡
2⇡ >
c
s2 + z 2
s2 + z 2
>
>
:
;
0
0
0
"
!
#
p
✓
◆
ct + (ct)2 s2
µ0 k
1p
=
ẑ t ln
(ct)2 s2 . Accordingly,
2⇡
s
c
(
!
p
ct + (ct)2 s2
µ0 k
E(s, t) =
ẑ ln
+
2⇡
s
!✓ ◆
!
)
s
1
1
2c2 t
1
2c2 t
p
p
t
c+ p
s
2 (ct)2 s2
2c (ct)2 s2
ct + (ct)2 s2
(
!
)
p
ct + (ct)2 s2
µ0 k
ct
ct
p
=
ẑ ln
+p
2⇡
s
(ct)2 s2
(ct)2 s2
!
p
ct + (ct)2 s2
µ0 k
=
ln
ẑ
(or zero, for t < s/c).
2⇡
s
@A
=
@t
B(s, t) =
=
=
@Az ˆ
@s

8
>
! s 1 p ( 2s)
>
2
(ct)2 s2
µ0 k <
s
p
t
2
2⇡ >
s2
>
: ct + (ct)
(
)
µ0 k
ct2
s
ˆ=
p
+ p
2⇡ s (ct)2 s2
c (ct)2 s2
ct
s2
p
(ct)2
s2
9
>
>
=
1
( 2s)
ˆ
p
2c (ct)2 s2 >
>
;
µ0 k ( c2 t2 + s2 ) ˆ
µ0 k p
p
=
(ct)2
2⇡ sc (ct)2 s2
2⇡sc
s2 ˆ.
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215
CHAPTER 10. POTENTIALS AND FIELDS
(b) A(s, t) =
µ0
ẑ
4⇡
Z
1
q0 (t
1
r
r
/c)
dz. But
A(s, t) =
Now z =
Now (t
p
r
r
/c) = c ( r
E(s, t) =
=
p
s2 + z 2 , so the integrand is even in z:
⇣ µ q ⌘ Z 1 (t
0 0
ẑ 2
4⇡
0
r
r
/c)
dz.
r d r , and z = 0 ) r = s, z = 1 ) r = 1. So:
1 2r dr
p
=p 2
2 r 2 s2
r s2
✓
◆
Z 1
r pr d r .
µ0 q0
1
A(s, t) =
ẑ
t
r
2⇡
c
r 2 s2
s
Z 1
µ0 q0
(r
ct)
p
ct) (Ex. 1.15); therefore A =
ẑ c
d r , so
2
2⇡
r s2
s
s2 ) dz =
2
A(s, t) =
r
µ0 q0 c
1
p
2⇡
(ct)2
B(s, t) =
ẑ (or zero, if ct < s);
s2
✓
◆
µ0 q0 c
1
2c2 t
µ0 q0 c3 t
@A
=
ẑ
=
ẑ (or zero, for t < s/c);
@t
2⇡
2 [(ct)2 s2 ]3/2
2⇡[(ct)2 s2 ]3/2
✓
◆
@Az ˆ
µ0 q0 c
1
2s
µ0 q0 cs
ˆ=
ˆ (or zero, for t < s/c).
=
@t
2⇡
2 [(ct)2 s2 ]3/2
2⇡[(ct)2 s2 ]3/2
Problem 10.12
µ0
A=
4⇡
Z
I(tr )
r
µ0 k
dl =
4⇡
Z
(t
r
r
/c)
µ0 k
dl =
4⇡
⇢ Z
dl
t
r
( Z
Z
Z b
R
µ0 kt 1
1
But for the complete loop, dl = 0, so A =
dl +
dl + 2 x̂
4⇡
a 1
b 2
a
R
circle), 2 dl = 2b x̂ (outer circle), so

µ0 kt 1
1
µ0 kt
A=
(2a) + ( 2b) + 2 ln(b/a) x̂ ) A =
ln(b/a) x̂,
4⇡ a
b
2⇡
E=
Z
1
dl .
c
)
R
dx
. Here 1 dl = 2a x̂ (inner
x
@A
=
@t
µ0 k
ln(b/a) x̂.
2⇡
The changing magnetic field induces the electric field. Since we only know A at one point (the center), we
can’t compute r ⇥ A to get B.
Problem 10.13
In this case ⇢(r,
˙ t) = ⇢(r,
˙ 0) and J̇(r, t) = 0, so Eq. 10.36 )

Z
r (Eq. 10.18), so
1
⇢(r0 , 0) + ⇢(r
˙ 0 , 0)tr
⇢(r
˙ 0 , 0)
r̂
E(r, t) =
+
d⌧ 0 , but tr = t
2
r
4⇡✏0
cr
c
Z  0
Z
1
⇢(r , 0) + ⇢(r
˙ 0 , 0)t ⇢(r
˙ 0 , 0)( r /c) ⇢(r
˙ 0 , 0)
1
⇢(r0 , t)
0
0
r̂
=
+
d⌧
=
r2
r2
r 2 r̂ d⌧ . qed
4⇡✏0
cr
4⇡✏0
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216
CHAPTER 10. POTENTIALS AND FIELDS
Problem 10.14
In this approximation we’re dropping the higher derivatives of J, so J̇(tr ) = J̇(t), and Eq. 10.38 )

Z
r 0
r
µ0
1
0
0
0
B(r, t) =
r 2 J(r , t) + (tr t)J̇(r , t) + c J̇(r , t) ⇥ r̂ d⌧ , but tr t = c (Eq. 10.25), so
4⇡
Z
µ0
J(r0 , t) ⇥ r̂
=
d⌧ 0 . qed
r2
4⇡
Problem 10.15
At time t the charge is at r(t) = a[cos(!t) x̂ + sin(!t) ŷ], so v(t) = !a[ sin(!t) x̂ p
+ cos(!t) ŷ]. Therefore
r = z ẑ a[cos(!tr ) x̂ + sin(!tr ) ŷ], and hence r 2 = z 2 + a2 (of course), and r = z 2 + a2 .
r̂
·v =
1
r (r
· v) =
1
r
!a [ sin(!tr ) cos(!tr ) + sin(!tr ) cos(!tr )] = 0, so
2
✓
1
r̂
·v
c
◆
= 1.
Therefore
V (z, t) =
1
q
q!a
p
p
; A(z, t) =
[ sin(!tr ) x̂ + cos(!tr ) ŷ), where tr = t
4⇡✏0 z 2 + a2
4⇡✏0 c2 z 2 + a2
p
z 2 + a2
.
c
Problem 10.16
Term under square root in (Eq. 10.49) is:
I = c4 t2 2c2 t(r · v) + (r · v)2 + c2 r2 c4 t2 v 2 r2 + v 2 c2 t2
= (r · v)2 + (c2 v 2 )r2 + c2 (vt)2 2c2 (r · vt). put in vt = r R2 .
= (r · v)2 + (c2 v 2 )r2 + c2 (r2 + R2 2r · R) 2c2 (r2 r · R) = (r · v)2
r2 v 2 + c2 R2 .
but
(r · v)2
r2 v 2 = ((R + vt) · v)2 (R + vt)2 v 2
= (R · v)2 + v 4 t2 + 2(R · v)v 2 t R2 v 2 2(R · v)tv 2
= (R · v)2 R2 v 2 = R2 v 2 cos2 ✓ R2 v 2 = R2 v 2 1
=
v 2 t2 v 2
cos2 ✓
R2 v 2 sin2 ✓.
Therefore
I=
✓
R2 v 2 sin2 ✓ + c2 R2 = c2 R2 1
Hence
V (r, t) =
1
q
4⇡✏0 R 1
q
v2
c2
sin2 ✓
◆
v2
2
sin
✓
.
c2
. qed
Problem 10.17
Once seen, from a given point x, the particle will forever remain in view—to disappear it would have to
travel faster than light.
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CHAPTER 10. POTENTIALS AND FIELDS
217
Problem 10.18
First calculateptr : tr = t |r w(tr )|/c ) p
c(tr t) = x
b2 + c2 t2r ) c(tr t) + x = b2 + c2 t2r ;
2 2
2
2 2
c tr 2c tr t + c t + 2xctr 2xct + x2 = b2 + c2 t2r ;
2ctr (x ct) + (x2 2xct + c2 t2 ) = b2 ;
b2 (x ct)2
2ctr (x ct) = b2 (x ct)2 , or tr =
.
2c(x ct)
1
qc
Now V (x, t) =
, and r c r · v = r (c v); r = c(t tr ).
4⇡✏0 ( r c r · v)
1
1
c2 tr
c2 tr
c2 tr + c(x ct) c2 tr
c(x ct)
v= p
2c2 tr =
=
; (c v) =
=
;
2 b2 + c2 t2r
c(tr t) + x
ctr + (x ct)
ctr + (x ct)
ctr + (x ct)
ct)
c2 (t tr )(x ct)
b2 (x ct)2
b2 + (x ct)2
r c r ·v = c(tct +tr )c(x
=
; ctr +(x ct) =
+(x ct) =
;
(x ct)
ctr + (x ct)
2(x ct)
2(x ct)
r
2
2
2
2
2 2
2
2ct(x ct) b + (x ct)
(x ct)(x + ct) b
(x
c t
b )
t tr =
=
=
. Therefore
ct)
2c(x ct)
2c(x ct)
 2c(x
1
b2 + (x ct)2
1
2c(x ct)
b2 + (x ct)2
=
=
.
2
2
2
r c r ·v
2(x ct)
c (x ct) [2ct(x ct) b + (x ct) ]
c(x ct) [2ct(x ct) b2 + (x ct)2 ]
The term in square brackets simplifies to (2ct + x ct)(x ct) b2 = (x + ct)(x ct) b2 = x2 c2 t2 b2 .
q
4⇡✏0 (x
Meanwhile
So V (x, t) =
A =
=
b2 + (x ct)2
.
ct)(x2 c2 t2 b2 )
 2
V
c2 tr
V
b
(x ct)2
2(x ct)
q
v
=
x̂
=
2
2
c
ctr + (x ct) c
2c(x ct)
b2 + (x ct)2 4⇡✏0 (x
q
4⇡✏0 c (x
b2 (x ct)2
x̂.
ct)(x2 c2 t2 b2 )
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b2 + (x ct)2
x̂
ct)(x2 c2 t2 b2 )
218
CHAPTER 10. POTENTIALS AND FIELDS
Problem 10.19
From Eq.
c(t tr ) = r ) c2 (t ✓tr )2 = r ◆2 = r · r . Di↵erentiate with respect to t:
✓ 10.44, ◆
@tr
@r
@tr
@r
2c2 (t tr ) 1
= 2r ·
, or c r 1
= r ·
. Now r = r w(tr ), so
@t
@t
@t
@t
✓
◆
@r
@w
@w @tr
@tr
@tr
@tr
@tr
r · v) =
=
=
= v
;
cr 1
= r ·v
;
cr =
(c r
@t
@t
@tr @t
@t
@t
@t
@t
@tr
@tr
cr
( r · u) (Eq. 10.71), and hence
=
. qed
r
@t
@t
·u
v
Now Eq. 10.47 says A(r, t) = 2 V (r, t), so
c
✓
◆
✓
◆
@A
1 @v
@V
1 @v @tr
@V
= 2
V +v
= 2
V +v
@t
c
@t
@t
c
@tr @t
@t

1
@tr 1
qc
1
qc @
= 2 a
+v
( r c r · v)
c
@t 4⇡✏0 r · u
4⇡✏0 ( r · u)2 @t

✓
◆
1 qc
a @tr
v
@r
@r
@v
= 2
c
·v r ·
.
c 4⇡✏0 r · u @t
( r · u)2
@t
@t
@t
✓
◆
@r
@tr
@r
@tr
But r = c(t tr ) )
=c 1
, r = r w(tr ) )
= v
(as above), and
@t
@t
@t
@t
@v
@v @tr
@tr
=
=a
.
@t
@tr @t
@t
⇢
 ✓
◆
q
@tr
@tr
@tr
@tr
2
r
r
=
a(
·
u)
v
c
1
+ v2
·a
2
4⇡✏0 c( r · u)
@t
@t
@t
@t
⇢
⇥
⇤
q
@t
r
=
c2 v + ( r · u)a + (c2 v 2 + r · a)v
4⇡✏0 c( r · u)2
@t
⇢
⇥
⇤ cr
q
=
c2 v + ( r · u)a + (c2 v 2 + r · a)v
2
r ·u
4⇡✏0 c( r · u)
⇥ 2
⇤
q
2
2
=
c v( r · u) + c r ( r · u)a + c r (c
v + r · a)v
3
4⇡✏0 c( r · u)

✓
◆
r a + r (c2 v2 + r · a)v . qed
qc
1
r
r
=
(
c
·
v)
v
+
4⇡✏0 ( r c r · v)3
c
c
Problem 10.20
r ⇥(c2 v2 )u + r ⇥ (u ⇥ a)⇤. Here
q
E=
4⇡✏0 ( r · u)3
v = v x̂, a = a x̂, and, for points to the right, r̂ = x̂.
So u = (c v) x̂, u ⇥ a = 0, and r · u = r (c v).
✓
◆
r
q
q
q
1 (c + v)(c v)2
1
c+v
2
2
E=
(c
v )(c v) x̂ =
x̂ =
x̂;
4⇡✏0 r 3 (c v)3
4⇡✏0 r 2
(c v)3
4⇡✏0 r 2 c v
1
B = r̂ ⇥ E = 0. qed
c
For field points to the left, r̂ = x̂ and u = (c + v) x̂, so r · u = r (c + v), and
✓
◆
r
q
q 1
c v
2
2
E=
(c
v
)(c
+
v)
x̂
=
x̂; B = 0.
4⇡✏0 r 3 (c + v)3
4⇡✏0 r 2 c + v
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219
CHAPTER 10. POTENTIALS AND FIELDS
Problem 10.21
By Gauss’s law (in integral form) the answer has to be q/✏0 .
b
1
q(1 v 2 /c2 ) R
E=
(Eq. 10.75), so
2
4⇡✏0 (1 vc2 sin2 ✓)3/2 R2
I
Z
Z ⇡
q(1 v 2 /c2 )
R2 sin ✓ d✓ d
q(1 v 2 /c2 )
E·da =
=
2⇡
2
4⇡✏0
4⇡✏0
R2 (1 vc2 sin2 ✓)3/2
0 (1
Let u ⌘ cos ✓, so du = sin ✓d✓, sin2 ✓ = 1 u2 .
I
Z
Z
q(1 v 2 /c2 ) 1
du
q(1 v 2 /c2 ) ⇣ c ⌘3
E·da =
=
v2
v 2 2 3/2
2✏0
2✏0
v
1 [1
c2 + c2 u ]
The integral is:
c2
v2
1
u
q
c2
v2
I
+1
1 + u2
E·da =
q(1
=
1
2
c2
v2
1
c
v
=
⇣ v ⌘3
c
v 2 /c2 ) ⇣ c ⌘3 ⇣ v ⌘3
2✏0
v
c (1
(1
sin ✓ d✓
.
2
v2
3/2
c2 sin ✓)
1
du
c2
v2
1
1 + u2
3/2
.
2
. So
v 2 /c2 )
2
q
=
.
v 2 /c2 )
✏0
X
Problem 10.22
Z
R̂
dx
.
⇥
2
4⇡✏0
R 1 (v/c)2 sin2 ✓⇤3/2
The horizontal components cancel; the vertical component of R̂ is sin ✓ (see diagram). Here d = R sin ✓, so
1
sin2 ✓
x
d
=
;
= cot ✓, so dx = d( csc2 ✓) d✓ =
d✓;
2
2
R
d
d
sin2 ✓
(a) E =
(1
v /c )
2
2
1
d sin2 ✓
d✓
dx
=
d✓ =
. Thus
2
2
2
R
d
sin ✓ d
✓ ◆Z ⇡
ŷ
sin ✓
2
E=
(1 v 2 /c2 )
⇥
⇤3/2 d✓. Let z ⌘ cos ✓, so sin ✓ = 1
4⇡✏0
d
0
1 (v/c)2 sin2 ✓
Z
(1 v 2 /c2 ) ŷ 1
1
=
dz
3/2
2
4⇡✏0 d
1 [1
(v/c) + (v/c)2 z 2 ]
"
#
+1
(1 v 2 /c2 ) ŷ
1
z
p
=
4⇡✏0 d
(v/c)3 (c2 /v 2 1) (c/v)2 1 + z 2
1
=
(1
v 2 /c2 ) c
4⇡✏0 d
v (1
2
1
p
v 2 /c2 ) (c/v)2
1
(b) B = 2 (v ⇥ E) for each segment dq =
c
same formula holds for the total field:
B=
1+1
ŷ =
z2.
1 2
ŷ (same as for a line charge at rest).
4⇡✏0 d
dx. Since v is constant, it comes outside the integral, and the
1
1
1 2
1 2
µ0 2 v
(v ⇥ E) = 2 v
(x̂ ⇥ ŷ) = µ0 ✏0 v
ẑ =
ẑ.
c2
c 4⇡✏0 d
4⇡✏0 d
4⇡ d
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220
But v = I, so B =
CHAPTER 10. POTENTIALS AND FIELDS
µ0 2I ˆ
(the same as we got in magnetostatics, Eq. 5.39 and Ex. 5.7).
4⇡ d
Problem 10.23
w(t) = R[cos(!t) x̂ + sin(!t) ŷ];
v(t) = R![ sin(!t) x̂ + cos(!t) ŷ];
a(t) = R! 2 [cos(!t) x̂ + sin(!t) ŷ] =
r = w(tr );
r = R;
tr = t R/c;
r̂ = [cos(!tr ) x̂ + sin(!tr ) ŷ];
r
! 2 w(t);
u = c r̂
v(tr ) = c[cos(!tr ) x̂ + sin(!tr ) ŷ] !R[ sin(!tr ) x̂ + cos(!tr ) ŷ]
= {[c cos(!tr ) !R sin(!tr )] x̂ + [c sin(!tr ) + !R cos(!tr )] ŷ} ;
⇥ (u ⇥ a) = ( r · a)u ( r · u)a; r · a = w · ( ! 2 w) = ! 2 R2 ;
⇥
⇤
r · u = R c cos2 (!tr ) !R sin(!tr ) cos(!tr ) + c sin2 (!tr ) + !R sin(!tr ) cos(!tr ) = Rc;
v 2 = (!R)2 . So (Eq. 10.72):
⇤
q
R ⇥
q cu Ra
u c2 ! 2 R2 + u(!R)2 a(Rc) =
4⇡✏0 (Rc)3
4⇡✏0 (Rc)2
q
1
=
[c2 cos(!tr ) !Rc sin(!tr )] x̂ [c2 sin(!tr ) + !Rc cos(!tr )] ŷ
4⇡✏0 (Rc)2
E=
+ R2 ! 2 cos(!tr ) x̂ + R2 ! 2 sin(!tr ) ŷ
=
⇥ 2 2
q
1
! R
2
4⇡✏0 (Rc)
⇤
⇥
c2 cos(!tr ) + !Rc sin(!tr ) x̂ + ! 2 R2
c2 sin(!tr )
⇤
!Rc cos(!tr ) ŷ .
1
r̂ ⇥ E = 1 r̂ x Ey r̂ y Ex ẑ
c
c
⇥
⇤
1 q
1
=
cos(!tr ) ! 2 R2 c2 sin(!tr ) !Rc cos(!tr )
c 4⇡✏0 (Rc)2
⇥
⇤
sin(!tr ) ! 2 R2 c2 cos(!tr ) + !Rc sin(!tr ) ẑ
⇤
q
1 ⇥
q
1
q
!
=
!Rc cos2 (!tr ) !Rc sin2 (!tr ) ẑ =
!Rc ẑ =
ẑ.
2
3
4⇡✏0 R c
4⇡✏0 R2 c3
4⇡✏0 Rc2
B =
Notice that B is constant in time.
To obtain the field at the center of a circular ring of charge, let q ! (2⇡R); for this ring to carry current
2⇡I 1
I, we need I = v = !R, so = I/!R, and hence q ! (I/!R)(2⇡R) = 2⇡I/!. Thus B =
ẑ, or,
4⇡✏0 Rc2
µ0 I
since 1/c2 = ✏0 µ0 , B =
ẑ, the same as Eq. 5.41, in the case z = 0.
2R
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221
CHAPTER 10. POTENTIALS AND FIELDS
Problem 10.24
( , t) = 0 | sin(✓/2)|, where ✓ =
!t. So the (retarded) scalar potential at the center is (Eq. 10.26)
Z
Z 2⇡
1
1
!tr )/2]|
0 |sin[(
V (t) =
dl0 =
r
4⇡✏0
4⇡✏0 0
a
Z 2⇡
2⇡
0
0
=
sin(✓/2) d✓ =
[ 2 cos(✓/2)]
4⇡✏0 0
4⇡✏0
0
=
0
4⇡✏0
[2
( 2)] =
0
⇡✏0
.
(Note: at fixed tr , d = d✓, and it goes through one full cycle of or ✓.)
Meanwhile I( , t) = v = 0 !a |sin[(
!t)/2]| ˆ. From Eq. 10.26 (again)
Z
Z
µ0
I 0
µ0 2⇡ 0 !a |sin[(
!tr )/2]| ˆ
A(t) =
dl =
r
4⇡
4⇡ 0
a
But tr = t a/c is again constant, for the integration, and ˆ = sin x̂ + cos ŷ.
Z
µ0 0 !a 2⇡
=
|sin[(
!tr )/2]| ( sin x̂ + cos ŷ) d . Again, switch variables to ✓ =
!tr ,
4⇡
0
and integrate from ✓ = 0 to ✓ = 2⇡ (so we don0 t have to worry about the absolute value).
Z
µ0 0 !a 2⇡
=
sin(✓/2) [ sin(✓ + !tr ) x̂ + cos(✓ + !tr ) ŷ] d✓. Now
4⇡
0
Z
2⇡
0
1
sin (✓/2) sin(✓ + !tr ) d✓ =
2
=
2⇡
2⇡
[cos (✓/2 + !tr )
cos (3✓/2 + !tr )] d✓
0

1
=
2 sin (✓/2 + !tr )
2
=
Z
Z
sin (✓/2) cos(✓ + !tr ) d✓ =
0
=
=
=
A(t) =
µ0
0 !a
4⇡
2⇡
2
sin (3✓/2 + !tr )
3
0
1
1
sin(⇡ + !tr ) sin(!tr )
sin(3⇡ + !tr ) + sin(!tr )
3
3
2
4
2 sin(!tr ) + sin(!tr ) =
sin(!tr ).
3
3
Z 2⇡
1
[ sin (✓/2 + !tr ) + sin (3✓/2 + !tr )] d✓
2 0

2⇡
1
2
2 cos (✓/2 + !tr )
cos (3✓/2 + !tr )
2
3
0
1
1
cos(⇡ + !tr ) cos(!tr )
cos(3⇡ + !tr ) + cos(!tr )
3
3
2
4
2 cos(!tr ) + cos(!tr ) =
cos(!tr ). So
3
3
✓ ◆
4
[sin(!tr ) x̂
3
cos(!tr ) ŷ] =
µ0
0 !a
{sin[!(t
3⇡
a/c)] x̂
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cos[!(t
a/c)] ŷ} .
222
CHAPTER 10. POTENTIALS AND FIELDS
Problem 10.25
1
r·A=
1 @V
c2 @t
Problem 10.26
⇢(r, t) =
8
<
Q
3Q
=
, (r < R = vt),
(4/3)⇡R3
4⇡v 3 t3
:
0,
(otherwise).
For a point at the center, tr = t r/c, so
8
3Q
3Q
<
=
4⇡(v/c)3 (ct
⇢(r, tr ) = 4⇡v 3 (t r/c)3
:
0,
✓
vt
,
r < v(t r/c) ) r <
r)3
1 + v/c
(otherwise).
◆
,
Let a ⌘ vt/(1 + v/c); then
Z a
Z a
3Q
1
3Q
r2
2
4⇡r
dr
=
dr
Qe↵ =
3
3
3
4⇡(v/c) 0 (ct r)
(v/c) 0 (ct r)3

a
3Q
2ct
(ct)2
=
ln(ct
r)
+
(v/c)3
(ct r) 2(ct r)2 0

3Q
2ct
(ct)2
1
=
ln(ct
a)
+
ln(ct) 2 +
(v/c)3
(ct a) 2(ct a)2
2
"
✓
◆
✓
◆
✓
◆2 #
3Q
3
ct
ct
1
ct
=
+ ln
2
+
.
(v/c)3 2
ct a
ct a
2 ct a
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CHAPTER 10. POTENTIALS AND FIELDS
Now, ct
223
vt
ct ⇣
v v⌘
ct
ct
v
=
1+
=
, so
= 1 + , and hence
1 + v/c
1 + v/c
c
c
1 + v/c
ct a
c

⇣
⌘
⇣
⌘
⇣
⌘
3Q
3
v
v
1
v 2
=
+ ln 1 +
2 1+
+
1+
3
(v/c) 2
c
c
2
c

 ⇣
⇣
⌘
3Q
3
v
v 1 v 1 ⇣ v ⌘2
3Q
v ⌘ v 1 ⇣ v ⌘2
=
+
ln
1
+
2
2
+
+
+
=
ln
1
+
+
.
(v/c)3 2
c
c
2
c
2 c
(v/c)3
c
c
2 c
a = ct
Qe↵
If ✏ ⌧ 1, then ln(1 + ✏) = ✏ 12 ✏2 + 13 ✏3 14 ✏4 + . . ., so for v ⌧ c,

✓
3Q
v 1 ⇣ v ⌘2 1 ⇣ v ⌘3 1 ⇣ v ⌘4 v 1 ⇣ v ⌘2
Qe↵ ⇡
+
+
=
Q
1
(v/c)3 c
2 c
3 c
4 c
c
2 c
3v
4c
◆
. X
Problem 10.27
Using Product Rule #5, Eq. 10.50 )
⇥
⇤ 1/2
µ0
r·A =
qcv · r (c2 t r · v)2 + (c2 v 2 )(r2 c2 t2 )
4⇡
⇢
⇤ 3/2 ⇥ 2
⇤
µ0 qc
1⇥ 2
=
v·
(c t r · v)2 + (c2 v 2 )(r2 c2 t2 )
r (c t r · v)2 + (c2 v 2 )(r2 c2 t2 )
4⇡
2
⇤ 3/2
µ0 qc ⇥ 2
=
(c t r · v)2 + (c2 v 2 )(r2 c2 t2 )
v·
2(c2 t r · v)r(r · v) + (c2 v 2 )r(r2 ) .
8⇡
Product Rule #4 )
r(r · v) = v ⇥ (r ⇥ r) + (v · r)r, but r ⇥ r = 0,
✓
◆
@
@
@
(v · r)r = vx
+ vy
+ vz
(x x̂ + y ŷ + z ẑ) = vx x̂ + vy ŷ + vz ẑ = v, and
@x
@y
@z
r(r2 ) = r(r · r) = 2r ⇥ (r ⇥ r) + 2(r · r)r = 2r. So
⇤ 3/2
⇥
⇤
µ0 qc ⇥ 2
(c t r · v)2 + (c2 v 2 )(r2 c2 t2 )
v · 2(c2 t r · v)v + (c2 v 2 )2r
8⇡
⇤ 3/2
µ0 qc ⇥ 2
=
(c t r · v)2 + (c2 v 2 )(r2 c2 t2 )
(c2 t r · v)v 2 (c2 v 2 )(r · v) .
4⇡
But the term in curly brackets is : c2 tv 2 v 2 (r · v) c2 (r · v) + v 2 (r · v) = c2 (v 2 t r · v).
µ0 qc3
(v 2 t r · v)
=
.
4⇡ [(c2 t r · v)2 + (c2 v 2 )(r2 c2 t2 )]3/2
r·A =
Meanwhile, from Eq. 10.49,
✓
◆
⇤ 3/2
1
1 ⇥ 2
µ0 ✏0
qc
(c t r · v)2 + (c2 v 2 )(r2 c2 t2 )
⇥
4⇡✏0
2
⇤
@ ⇥ 2
(c t r · v)2 + (c2 v 2 )(r2 c2 t2 )
@t
⇤ 3/2 ⇥ 2
µ0 qc ⇥ 2
=
(c t r · v)2 + (c2 v 2 )(r2 c2 t2 )
2(c t r · v)c2 + (c2
8⇡
µ0 qc3
(c2 t r · v c2 t + v 2 t)
=
= r · A. X
4⇡ [(c2 t r · v)2 + (c2 v 2 )(r2 c2 t2 )]3/2
@V
µ0 ✏0
=
@t
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⇤
v 2 )( 2c2 t)
224
CHAPTER 10. POTENTIALS AND FIELDS
Problem 10.28
(a) F2 =
q1 q2
1
x̂.
4⇡✏0 (b2 + c2 t2 )
(This is just Coulomb’s law, since q1 is at rest.)

Z
1
q1 q2 1
q1 q2 ⇥
q1 q2 1
1
1
(b) I2 =
dt
=
tan
(ct/b)
=
tan
2 + c2 t2 )
4⇡✏0
(b
4⇡✏
bc
4⇡✏
0
0 bc
1
1
q1 q2 h ⇡ ⇣ ⇡ ⌘i
q1 q2 ⇡
=
=
.
4⇡✏0 bc 2
2
4⇡✏0 bc
✓
◆
q2 1 c v
(c) From Prob. 10.20, E =
x̂. Here x
4⇡✏0 x2 c + v
and v are to be evaluated at the
p retarded time tr , which is
given by c(t tr ) = x(tr ) = b2 + c2 t2r ) c2 t2 2cttr +
c2 t2 b2
c2 t2r = b2 + c2 t2r ) tr =
. Note: As we found
2c2 t
in Prob. 10.17, q2 first “comes into view” (for q1 ) at time
t = 0. Before that it can exert no force on q1 , and there is
no retarded time. From the graph of tr versus t we see that
tr ranges all the way from 1 to 1 while t > 0.
1
(1)
tan
1
⇤
( 1)
2c2 t2 c2 t2 + b2
b2 + c2 t2
1
2c2 t
c2 t
x(tr ) = c(t tr ) =
=
(for t > 0). v(t) = p
=
, so
2ct
2ct
2 b2 + c2 t2
x
✓ 2 2
◆
✓
◆
✓
◆
c t
b2
2ct
c2 t2 b2
v(tr ) =
=
c
(for t > 0). Therefore
2t
b2 + c2 t2
c2 t2 + b2
c v
(c2 t2 + b2 ) (c2 t2 b2 )
2b2
b2
q2
4c2 t2
b2
= 2 2
= 2 2 = 2 2 (for t > 0). E =
x̂ )
2
2
2
2
2
2
2
2
c+v
(c t + b ) + (c t
b )
2c t
c t
4⇡✏0 (b + c t ) c2 t2
8
t < 0;
< 0,
q1 q2
4b2
F1 =
x̂, t > 0.
:
4⇡✏0 (b2 + c2 t2 )2
Z
q1 q2 2 1
1
(d) I1 =
4b
dt. The integral is
2 + c2 t2 )2
4⇡✏0
(b
0
✓ 2 ◆
Z 1
Z 1
1
1
1
1
c
t
1
1 ⇣ ⇡c ⌘
⇡
dt
=
+
dt
=
=
.
4
2
2
2
4
2
2
2
2
2
2
2
c 0 [(b/c) + t ]
c
2b
(b/c) + t 0
[(b/c) + t )]
2c b
2b
4cb3
0
q1 q2 ⇡
So I1 =
.
4⇡✏0 bc
(e) F1 6= F2 , so Newton’s third law is not obeyed. On the other hand, I1 = I2 in this instance, which
suggests that the net momentum delivered from (1) to (2) is equal and opposite to the net momentum delivered
from (2) to (1), and hence that the total mechanical momentum is conserved. (In general the fields might carry
o↵ some momentum, leaving the mechanical momentum altered; but that doesn’t happen in the present case.)
Problem 10.29
The electric field of q1 at q2 [Eq. 10.75, with ✓ = 45 and R = ( vt x̂ + vt ŷ)] is
E1 =
q1
1 v 2 /c2
1
p
( x̂ + ŷ).
2
2
3/2
4⇡✏0 (1 v /2c ) 2 2(vt)2
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225
CHAPTER 10. POTENTIALS AND FIELDS
The magnetic field [Eq. 10.76, with v1 =
B1 =
1
(v1 ⇥ E) =
c2
v x̂] is
v
(x̂ ⇥ E) =
c2
v q1
1 v 2 /c2
1
p
ẑ.
c2 4⇡✏0 (1 v 2 /2c2 )3/2 2 2(vt)2
The force on q2 is therefore [Lorentz force law with v2 =
v ŷ]
✓
q1 q2
1 v 2 /c2
1
p
F2 = q2 (E1 + v2 ⇥ B1 ) =
4⇡✏0 (1 v 2 /2c2 )3/2 2 2(vt)2
◆
v2
x̂ + ŷ + 2 x̂ .
c
The electric field of q2 at q1 is reversed, E2 = E1 ; so is the magnetic field B2 = B1 . The electric force
is also reversed, but the magnetic force now points in the y direction instead of the x direction. So the force
on q1 is
F1 =
q1 q2
1 v 2 /c2
1
p
4⇡✏0 (1 v 2 /2c2 )3/2 2 2(vt)2
✓
x̂
ŷ +
◆
v2
ŷ
.
c2
The forces are equal in magnitude, but not opposite in direction. No, Newton’s third law is not obeyed.
Problem 10.30
c(t
t1 ) = vt1 , t = t1 (1 + v/c), t1 =
x1 = vt1 =
vt
.
1 + v/c
x2 = vt2 + L =
V (0, t) =
If L ⌧ vt, V =
V =
4⇡✏0 ( r
q
r
1
4⇡✏0
Z
t
.
1 + v/c
t2 ) = vt2 + L, t
c(t
v(t L/c)
vt
+L=
1 + v/c
x2
x1
x
dx =
4⇡✏0
ln
✓
x2
x1
· v/c)
(In the limit L ! 0, tr = t1 = t2 .)
t L/c
.
1 + v/c
vL/c + L + vL/c
vt + L
=
.
1 + v/c
1 + v/c
◆
✓
◆
✓ ◆
L
L
q
1
ln 1 +
⇡
=
.
4⇡✏0
vt
4⇡✏0 vt
4⇡✏0 vt
. Here ( r
L/c = t2 (1 + v/c), t2 =
=
4⇡✏0
ln
✓
vt + L
vt
◆
.
The Liénard-Wiechert potential is
t
· v/c) = vtr + v 2 tr /c = v(1 + v/c)tr = v(1 + v/c)
= vt.
1 + v/c
✓ ◆
q
1
So V =
. X
4⇡✏0 vt
r
Problem 10.31
1
1
(E ⇥ B); B = 2 (v ⇥ E) (Eq. 10.76).
µ0
c
⇥
⇤
1
So S =
[E ⇥ (v ⇥ E)] = ✏0 E 2 v (v · E)E .
2
µ0 c
R
The power crossing the plane is P = S · da,
S=
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226
CHAPTER 10. POTENTIALS AND FIELDS
and da = 2⇡r dr x̂ (see diagram). So
P = ✏0
Z
(E 2 v
= 2⇡✏0 v
Z
Ex2 v)2⇡r dr; Ex = E cos ✓, so E 2
E 2 sin2 ✓ r dr. From Eq. 10.75, E =
✓
Ex2 = E 2 sin2 ✓.
q 1
4⇡✏0 2 R2 ⇥1
R̂
(v/c)2
sin ✓
2
⇤3/2 where
1
⌘p
1
v 2 /c2
.
◆2
Z 1
q
1
r sin2 ✓
1
1
cos ✓
= 2⇡✏0 v
=
.
⇥
⇤3 dr. Now r = a tan ✓ ) dr = a 2 d✓;
2
2
4⇡✏0
cos ✓
R
a
0
R4 1 (v/c)2 sin ✓
Z ⇡/2
v q2 1
sin3 ✓ cos ✓
2
=
⇥
⇤3 d✓. Let u ⌘ sin ✓, so du = 2 sin ✓ cos ✓ d✓.
4
2
2
2
2 4⇡✏0 a 0
1 (v/c) sin ✓
✓ 4◆
Z
1
vq 2
u
vq 2
vq 2
=
du
=
=
.
3
2
4
2
4
2
16⇡✏0 a
16⇡✏0 a
2
32⇡✏0 a2
(v/c) u]
0 [1
Problem 10.32
(a) F12 (t) =
1 q1 q2
ẑ.
4⇡✏0 (vt)2
(b) From Eq. 10.75, with ✓ = 180 , R = vt, and R̂ =
F21 (t) =
ẑ:
1 q1 q2 (1 v 2 /c2 )
ẑ.
4⇡✏0
(vt)2
No, Newton’s third law does not hold: F12 6= F21 ,
because of the extra factor (1
v 2 /c2 ).
R
(c) From Eq. 8.28, p = ✏0 (E⇥B) d⌧ . Here E = E1 +E2 , whereas B = B2 , so E⇥B = (E1 ⇥B2 )+(E2 ⇥B2 ).
But the latter,
when integrated over all space, is independent of time. We want only the time-dependent part:
Z
1 q1
1
p(t) = ✏0 (E1 ⇥ B2 ) d⌧. Now E1 =
r̂, while, from Eq. 10.76, B2 = 2 (v ⇥ E2 ), and (Eq. 10.75)
4⇡✏0 r2
c
q2
(1 v 2 /c2 )
R̂
r sin ✓
E2 =
. But R = r vt; R2 = r2 + v 2 t2 2rvt cos ✓; sin ✓0 =
. So
4⇡✏0 (1 v 2 sin2 ✓0 /c2 )3/2 R2
R
q2
(1 v 2 /c2 )
(r vt)
E2 =
. Finally, noting that v ⇥ (r vt) = v ⇥ r = vr sin ✓ ˆ, we get
4⇡✏0 [1 (vr sin ✓/Rc)2 ]3/2 R3
Z
2 2
q2 (1 v 2 /c2 )
vr sin ✓
1
r sin ✓ (r̂ ⇥ ˆ)
ˆ. So p(t) = ✏0 q1 q2 (1 v /c )v
B2 =
.
2
2
2
3/2
4⇡✏0 c
4⇡✏0
4⇡✏0 c
r [R2 (vr sin ✓/c)2 ]3/2
[R2 (vr sin ✓/c)2 ]
But r̂ ⇥ ˆ = ✓ˆ = (cos ✓ cos x̂ + cos ✓ sin ŷ sin ✓ ẑ), and the x and y components integrate to zero, so:
q1 q2 v(1 v 2 /c2 ) ẑ
p(t) =
(4⇡c)2 ✏0
q1 q2 v(1 v 2 /c2 ) ẑ
=
8⇡c2 ✏0
Z
Z
sin2 ✓
r [r2 + (vt)2
2rvt cos ✓
3/2
(vr sin ✓/c)2 ]
r sin3 ✓
[r2 + (vt)2
2rvt cos ✓
3/2
(vr sin ✓/c)2 ]
r2 sin ✓ dr d✓ d
dr d✓.
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227
CHAPTER 10. POTENTIALS AND FIELDS
I’ll do the r integral first. According to the CRC Tables,

Z 1
1
x
2(bx + 2a)
2
b
2a
p
p
p
dx =
=
2
2 )3/2
2
2
4ac
b
c
a
(a
+
bx
+
cx
(4ac b ) a + bx + cx 0
0
p
p
p
2
2
(2 ac b)
2
p
p
= p
b 2 ac = p
= p 2 ac + b
2
c(4ac b )
c (2 ac b) (2 ac + b)
c
In this case x = r, a = (vt)2 , b =
2vt cos ✓, and c = 1
2
 q
(v/c)2 sin2 ✓ 2vt 1 (v/c)2 sin2 ✓
q
1 (v/c)2 sin2 ✓ + cos ✓
q
=
⇥
vt 1 (v/c)2 sin2 ✓ 1 (v/c)2 sin2 ✓
So
=
q
1
But
Z ⇡
0
q1 q2 v(1 v 2 /c2 ) ẑ
p(t) =
8⇡c2 ✏0
vt(1
8
Z
q1 q2 ẑ < ⇡
=
sin ✓ d✓ +
8⇡c2 ✏0 t : 0
R⇡
2vt cos ✓
c
v
Z
0
q
vt 1
⇡
1
q
(v/c)2 sin2 ✓
1
v 2 /c2 )
2
41 + q
1
(v/c)2 sin2 ✓
cos ✓
(v/c)2 sin2 ✓
2
3
1 4
cos ✓
5 sin3 ✓ d✓
1+ q
2
2
2
0 sin ✓
1 (v/c) sin ✓
9
=
cos ✓ sin ✓
q
d✓ .
;
(c/v)2 sin2 ✓
Z
.
(v/c)2 sin2 ✓. So the r integral is
1
⇤ = vt sin2 ✓(1
cos2 ✓
1
v 2 /c2 )
1
cos ✓
3
5.
⇡
sin ✓ d✓ = 2. In the second integral let u ⌘ cos ✓, so du = sin ✓ d✓:
Z 1
cos ✓ sin ✓
u
q
p
d✓ =
du = 0 (the integrand is odd, and the interval is even).
2
(c/v)
1 + u2
1
(c/v)2 sin2 ✓
µ0 q1 q2
Conclusion: p(t) =
ẑ (plus a term constant in time).
4⇡t
(d)
✓
◆
1 q1 q2
1 q1 q2 (1 v 2 /c2 )
q1 q2
v2
q1 q2
µ0 q1 q2
F12 + F21 =
ẑ
ẑ
=
1
1
+
ẑ =
ẑ =
ẑ.
4⇡✏0 v 2 t2
4⇡✏0
v 2 t2
4⇡✏0 v 2 t2
c2
4⇡✏0 c2 t2
4⇡t2
dp
µ0 q1 q2
=
ẑ = F12 + F21 . qed
dt
4⇡t2
0
Since q1 is at rest, and q2 is moving at constant velocity, there must be another force (Fmech ) acting on
them, to balance F12 + F21 ; what we have found is that Fmech = dpem /dt, which means that the impulse
imparted to the system by the external force ends up as momentum in the fields. [For further discussion of
this problem see J. J. G. Scanio, Am. J. Phys. 43, 258 (1975).]
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228
CHAPTER 10. POTENTIALS AND FIELDS
Problem 10.33
Maxwell’s equations with magnetic charge read:
@B
,
@t
@E
(ii) r · B = µ0 ⇢m , (iv) r ⇥ B = µ0 Je + µ0 ✏0
,
@t
(i) r · E =
1
✏ 0 ⇢e ,
(iii) r ⇥ E =
µ0 Jm
where ⇢e is the electric charge density, ⇢m is the magnetic charge density, Je is the electric current density,
and Jm is the magnetic current density.
If there are only electric charges and currents, then the usual potential formulation applies (Section 10.1.1),
with E = rVe @Ae /@t, B = r ⇥ Ae . If there are only magnetic charges and currents,
(i) r · E = 0,
(iii) r ⇥ E =
@B
,
@t
µ0 Jm
(ii) r · B = µ0 ⇢m , (iv) r ⇥ B = µ0 ✏0
@E
.
@t
This time equation (i) says that E can be expressed as the curl of a vector potential: E = r ⇥ Am , and
plugging this into (iv) yields r ⇥ (B + µ0 ✏0 @Am /@t) = 0, which tells us that the quantity in parentheses can
be represented as the gradient of a scalar: B = rVm µ0 ✏0 @Am /@t. [The signs of Vm and Am are arbitrary,
but I think this choice yields the most symmetrical formulation.] Putting these into (ii) and (iii) yields
✓
◆
✓
◆
@
@Vm
@Vm
22 Vm = µ0 ⇢m µ0 ✏0
r · Am +
, 22 Am = µ0 Jm + r r · Am +
.
@t
@t
@t
The “Lorenz gauge” for the magnetic potentials, r · Am = @Vm /@t, kills the terms in parentheses.
In general, if there are both electric and magnetic charges and currents present, the total fields are the sums
(by the superposition principle):
E=
@Ae
@t
rVe
B=
r ⇥ Am ,
rVm
µ0 ✏0
@Am
+ r ⇥ Ae .
@t
If we choose to work in the Lorenz gauge:
r · Ae =
µ0 ✏0
@Ve
,
@t
r · Am =
@Vm
,
@t
Maxwell’s equations become (in terms of the potentials)
22 V e =
1
⇢e ,
✏0
22 V m =
µ0 ⇢m ,
22 A e =
µ0 Je ,
22 A m =
µ0
4⇡
Z
⇢m (r0 , tr )
µ0
Am (r, t) =
4⇡
Z
Jm (r0 , tr )
µ0 Jm ,
and the retarded solutions are
Ve (r, t) =
1
4⇡✏0
µ0
Ae (r, t) =
4⇡
Z
Z
⇢e (r0 , tr )
r
Je (r0 , tr )
r
d⌧ 0 ,
0
d⌧ ,
Vm (r, t) =
r
r
d⌧ 0 ,
d⌧ 0 .
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229
CHAPTER 10. POTENTIALS AND FIELDS
Problem 10.34
The retarded potentials are
V (r, t) =
1
4⇡✏0
Z
⇢(r0 , tr )
r
d⌧ 0 ,
A(r, t) =
µ0
4⇡
Z
J(r0 , tr )
r
d⌧ 0 .
We want a source localized at the origin, so we expand in powers of r0 , keeping terms up to first order:
r2
r
t
= (r r0 ) · (r r0 ) = r2 2r · r0 + (r0 )2 ,
✓
◆
r · r0
⇡r 1
,
r2
✓
◆
1
1
r · r0
⇡
1+ 2
,
r
r
r
r ⇡ t r + r · r0 = t + r · r0 ,
0
c
c
rc
rc
where t0 ⌘ t
r/c is the retarded time for a source at the origin. Thus (Taylor expanding about t0 )
✓
◆
r ⇡ ⇢(r0 , t ) + r · r0 ⇢(r
⇢(r0 , tr ) = ⇢ r0 , t
˙ 0 , t0 ),
0
c
rc
✓
◆
⇢(r0 , tr )
1
r · r0
r · r0
1
r · r0
r · r0
=
1+ 2
⇢(r0 , t0 ) +
⇢(r
˙ 0 , t0 ) ⇡ ⇢(r0 , t0 ) + 3 ⇢(r0 , t0 ) + 2 ⇢(r
˙ 0 , t0 ),
r
r
r
rc
r
r
r c
and hence
 Z
Z
Z
1
1
r
r
0
0
0
0
0
V (r, t) =
⇢(r , t0 ) d⌧ + 3 · r ⇢(r , t0 ) d⌧ + 2 · r0 ⇢(r
˙ 0 , t0 ) d⌧ 0 .
4⇡✏0 r
r
r c
The first integral is the total charge of the dipole, which is zero; the second integral is the dipole moment, and
the third is the time derivative of the dipole moment:
V (r, t) =
i
1 r̂ h
r
·
p(t
)
+
ṗ(t
)
.
0
0
4⇡✏0 r2
c
By the same reasoning, the vector potential is
 Z
Z
Z
µ0 1
1
1
A(r, t) =
J(r0 , t0 ) d⌧ 0 + 3 (r · r0 )J(r0 , t0 ) d⌧ 0 + 2
(r · r0 )J̇(r0 , t0 ) d⌧ 0 .
4⇡ r
r
r c
Z
From Eq. 5.31
J(r0 , t) d⌧ 0 = ṗ(t), and J = ⇢v0 = ⇢(dr0 /dt) is already first order in r0 , so the second and
third integrals are second order, and we drop them.
A(r, t) =

µ0 ṗ(t0 )
.
4⇡
r
In calculating the fields, remember that the dipole moments themselves depend on r, through their argument
(t0 = t r/c). For a function of t0 alone,
✓
◆
df
1
1 ˙
rf (t0 ) =
rt0 = f˙
rr =
f r̂,
dt0
c
c
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230
CHAPTER 10. POTENTIALS AND FIELDS
h
i
r̂ d
r
. For a structure of the form f (t0 ) + f˙(t0 )
c dt
c
✓
◆
h
i
r
r̂ ˙ r
r̂ ¨
f˙
r ¨
r f (t0 ) + f˙(t0 ) =
f+
f + r̂ =
f.
c
c
c
c
c
c2
so you simply replace r by
Using the product rule for r(A · B):
⇢
✓
◆ ✓
◆h
⇣
h
⌘ r̂
1
r̂
r i⌘ h
r i
r̂
r̂
r i ⇣h
r i
⇥
r
⇥
p
+
+
·
r
p
+
rV =
ṗ
+
p
+
ṗ
⇥
r
⇥
ṗ
+
p
+
ṗ
·
r
4⇡✏0 r2
c
c
r2
r2
c
c
r2
⇢

⇣
⌘
⇣
⌘
⇣h
i
⇣
h
i⌘⌘
1
r̂
r
r̂
r
1
r
r
=
⇥
⇥ p̈ + 0 + 2 ·
p̈ + 3 p + ṗ
3r̂ r̂ · p + ṗ
2
2
2
4⇡✏0 r
c
r
c
r
c
c
⇢
1
r̂(r̂ · p̈) [p + (r/c)ṗ] 3r̂(r̂ · [p + (r/c)ṗ])
=
+
.
4⇡✏0
rc2
r3
The others are easier:

µ0 1
r⇥A =
(r ⇥ ṗ)
4⇡ r
✓ ◆
@A
µ0 p̈
=
.
@t
4⇡ r
✓
◆
 ✓
◆ ✓
◆
1
µ0 1
r̂
r̂
ṗ ⇥ r
=
⇥ p̈ + ṗ ⇥ 2
=
r
4⇡ r
c
r
µ0
4⇡
⇢
r̂ ⇥ [(ṗ + (r/c)p̈]
r2
The fields are therefore
E(r, t) =
rV
@A
=
@t
B(r, t) = r ⇥ A =
µ0
4⇡
µ0
4⇡
⇢
⇢
p̈
r̂(r̂ · p̈)
[p + (r/c)ṗ]
+ c2
r
3r̂(r̂ · [p + (r/c)ṗ])
r3
,
r̂ ⇥ [ṗ + (r/c)p̈]
r2
(with all dipole moments evaluated at the retarded time t0 = t
r/c).
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,
231
Chapter 11
Problem 11.1
µ0 p0 ! 1
ˆ so
From Eq. 11.17, A =
sin[!(t r/c)](cos ✓ r̂ sin ✓ ✓),
4⇡ r
⇢


µ0 p0 ! 1 @
1
@
1
21
r·A =
r sin[!(t r/c)] cos ✓ +
sin2 ✓ sin[!(t r/c)]
2
4⇡
r @r
r
r sin ✓ @✓
r
⇢
⌘
µ0 p0 ! 1 ⇣
!r
2 sin ✓ cos ✓
=
sin[!(t r/c)]
cos[!(t r/c)] cos ✓
sin[!(t r/c)]
4⇡
r2
c
r2 sin ✓
⇢
✓
◆
p0 !
1
!
= µ0 ✏0
sin[!(t r/c)] +
cos[!(t r/c)] cos ✓ .
2
4⇡✏0 r
rc
Meanwhile, from Eq. 11.12,
⇢
@V
p0 cos ✓
!2
!
=
cos[!(t r/c)]
sin[!(t r/c)]
@t
4⇡✏0 r
c
r
⇢
p0 !
1
!
sin[!(t r/c)] +
=
cos[!(t r/c)] cos ✓. So r · A =
4⇡✏0 r2
rc
Problem 11.2
! p0 · r̂
sin[!(t
4⇡✏0 c r
Eq. 11.14: V (r, t) =
r/c)]. Eq. 11.17: A(r, t) =
Now p0 ⇥ r̂ = p0 sin ✓ ˆ and r̂ ⇥ (p0 ⇥ r̂) = p0 sin ✓(r̂ ⇥ ˆ) =
Eq. 11.18: E(r, t) =
Eq. 11.21: hSi =
µ0 ! r̂ ⇥ (p0 ⇥ r̂)
cos[!(t
4⇡
r
2
µ0 ✏0
@V
. qed
@t
µ0 ! p0
sin[!(t
4⇡ r
r/c)].
ˆ so
p0 sin ✓ ✓,
r/c)]. Eq. 11.19: B(r, t) =
µ0 ! 2 (p0 ⇥ r̂)
cos[!(t
4⇡c
r
r/c)].
µ0 ! 4 (p0 ⇥ r̂)2
r̂.
32⇡ 2 c
r2
Problem 11.3
P = I 2 R = q02 ! 2 sin2 (!t)R (Eq. 11.15) ) hP i = 12 q02 ! 2 R. Equate this to Eq. 11.22:
1 2 2
µ0 q02 d2 ! 4
µ0 d2 ! 2
2⇡c
q0 ! R =
) R=
; or, since ! =
,
2
12⇡c
6⇡c
✓ ◆2
✓ ◆2
✓ ◆2
µ0 d2 4⇡ 2 c2
2
d
2
d
d
7
8
2
R=
=
⇡µ
c
=
⇡(4⇡
⇥
10
)(3
⇥
10
)
=
80⇡
⌦ = 789.6(d/ )2 ⌦.
0
2
6⇡c
3
3
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232
For the wires in an ordinary radio, with d = 5 ⇥ 10 2 m and (say)
which is negligible compared to the Ohmic resistance.
= 103 m, R = 790(5 ⇥ 10
) = 2 ⇥ 10
5 2
6
⌦,
Problem 11.4
By the superposition principle, we can add the✓potentials of the
◆ two dipoles. Let’s first express V (Eq. 11.14)
p0 !
z
sin[!(t r/c)]. That’s for an oscillating dipole
in Cartesian coordinates: V (x, y, z, t) =
4⇡✏0 c x2 + y 2 + z 2
along the z axis. For one along x or y, we just change z to x or y. In the present case,
p = p0 [cos(!t) x̂ + cos(!t
sin[!(t
r/c)] ! sin[!(t
p0 !
4⇡✏0 c
V =
=
A =
⇢
⇡/2) ŷ], so the one along y is delayed by a phase angle ⇡/2:
⇡/2] =
r/c)
cos[!(t
x
sin[!(t
x2 + y 2 + z 2
r/c)]
p0 ! sin ✓
{cos sin[!(t
4⇡✏0 c r
r/c)]
µ0 p0 !
{sin[!(t
4⇡r
cos[!(t
r/c)] x̂
r/c)] (just let !t ! !t
y
cos[!(t
x2 + y 2 + z 2
sin cos[!(t
r/c)]} .
⇡/2). Thus
r/c)]
Similarly,
r/c)] ŷ} .
We could get the fields by di↵erentiating these potentials, but I prefer to work with Eqs. 11.18 and 11.19,
ˆ and cos ✓ = z/r, Eq. 11.18 can be written
using superposition. Since ẑ = cos ✓ r̂ sin ✓ ✓,
⇣
µ0 p0 ! 2
z ⌘
E=
cos[!(t r/c)] ẑ
r̂ . In the case of the rotating dipole, therefore,
4⇡r
r
E=
B =
S=
µ0 p0 ! 2 n
cos[!(t
4⇡r
1
(r̂ ⇥ E).
c
⇣
r/c)] x̂
x ⌘
r̂ + sin[!(t
r
1
1
1 ⇥ 2
(E ⇥ B) =
[E ⇥ (r̂ ⇥ E)] =
E r̂
µ0
µ0 c
µ0 c
E2 =
✓
where a ⌘ x̂
µ0 p0 ! 2
4⇡r
◆2
a2 cos2 [!(t
(x/r)r̂ and b ⌘ ŷ
⇣
r/c)] ŷ
y ⌘o
r̂ ,
r
⇤
E2
(E · r̂)E =
r̂ (notice that E · r̂ = 0). Now
µ0 c
r/c)] + b2 sin2 [!(t
r/c)] + 2(a · b) sin[!(t
r/c)] cos[!(t
r/c)] ,
(y/r)r̂. Noting that x̂ · r = x and ŷ · r = y, we have
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233
a2 = 1 +
x2
r2
E =
✓
2
=
=
=
=
2
x2
=1
r2
x2 2
; b =1
r2
y2
; a·b=
r2
yx
rr
x y xy
+ 2 =
rr
r
xy
.
r2
◆2 ⇢✓
◆
✓
◆
µ0 p0 ! 2
x2
y2
2
1
cos [!(t r/c)] + 1
sin2 [!(t r/c)]
4⇡r
r2
r2
o
xy
2 2 sin[!(t r/c)] cos[!(t r/c)]
r
✓
◆2 ⇢
µ0 p0 ! 2
1
1
x2 cos2 [!(t r/c)] + 2xy sin[!(t r/c)] cos[!(t r/c)] + y 2 sin2 [!(t
4⇡r
r2
✓
◆2 ⇢
µ0 p0 ! 2
1
2
1
(x cos[!(t r/c)] + y sin[!(t r/c)])
4⇡r
r2
But x = r sin ✓ cos and y = r sin ✓ sin .
✓
◆2 n
o
µ0 p0 ! 2
2
1 sin2 ✓ (cos cos[!(t r/c)] + sin sin[!(t r/c)])
4⇡r
✓
◆2 n
o
µ0 p0 ! 2
2
1 (sin ✓ cos[!(t r/c)
]) .
4⇡r
S=
hSi =
P =
=
r/c)]
✓
◆2 n
o
µ0 p0 ! 2
2
1 (sin ✓ cos[!(t r/c)
]) r̂.
c
4⇡r
✓
◆2 
µ0 p0 ! 2
1
1
sin2 ✓ r̂.
c
4⇡r
2
✓
◆2 Z
✓
◆
Z
µ0 p0 ! 2
1
1
2
hSi · da =
1
sin ✓ r2 sin ✓ d✓ d
c
4⇡
r2
2
Z ⇡
✓
◆
Z
µ0 p20 ! 4
1 ⇡ 3
µ0 p20 ! 4
1 4
µ0 p20 ! 4
2⇡
sin ✓ d✓
sin ✓ d✓ =
2
·
=
.
2
16⇡ c
2 0
8⇡c
2 3
6⇡c
0
This is twice the power radiated by either oscillating dipole alone (Eq. 11.22). In general, S =
1
(E ⇥ B) =
µ0
1
1
[(E1 + E2 ) ⇥ (B1 + B2 )] =
[(E1 ⇥ B1 ) + (E2 ⇥ B2 ) + (E1 ⇥ B2 ) + (E2 ⇥ B1 )] = S1 + S2 + cross terms.
µ0
µ0
In this particular case the fields of 1 and 2 are 90 out of phase, so the cross terms go to zero in the time
averaging, and the total power radiated is just the sum of the two individual powers.
Problem 11.5
Go back to Eq. 11.33:
A=
µ0 m0
4⇡
✓
sin ✓
r
◆⇢
1
cos[!(t
r
r/c)]
!
sin[!(t
c
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r/c)]
ˆ.
234
Since V = 0 here,
✓
◆⇢
µ0 m0 sin ✓
1
!
E=
( !) sin[!(t r/c)]
! cos[!(t
4⇡
r
r
c
✓
◆⇢
µ0 m0 ! sin ✓
1
!
=
sin[!(t r/c)] + cos[!(t r/c)] ˆ.
4⇡
r
r
c
@A
=
@t
r/c)]
ˆ
1
@
1 @
B = r⇥A=
(A sin ✓) r̂
(rA ) ✓ˆ
r
sin
✓
@✓
r @r
⇢

µ0 m0
1 2 sin ✓ cos ✓ 1
!
=
cos[!(t r/c)]
sin[!(t r/c)] r̂
4⇡
r sin ✓
r
r
c

sin ✓
1
!
! ⇣ !⌘
cos[!(t
r/c)]
+
sin[!(t
r/c)]
cos[!(t r/c)] ✓ˆ
r
r2
rc
c
c

µ0 m0 n 2 cos ✓ 1
!
cos[!(t r/c)]
sin[!(t r/c)] r̂
4⇡
r2  r
c
=
⇣ ! ⌘2
o
sin ✓
1
!
cos[!(t r/c)] +
sin[!(t r/c)] +
cos[!(t r/c)] ✓ˆ .
2
r
r
rc
c
These are precisely the fields we studied in Prob. 9.35, with A !
the solution to that problem) is
µ0 m0 ! 2
. The Poynting vector (quoting
4⇡c
✓
◆
✓
◆
µ0 m20 ! 3 sin ✓ n 2 cos ✓
c2
c
S=
1
sin u cos u +
cos2 u sin2 u ✓ˆ
2 r2
16⇡ 2 c2✓ r2
r
!
!r
◆
o
2
c2
!
c
sin ✓
+ 2 3 sin u cos u + cos2 u + 2 sin2 u cos2 u r̂ ,
r
! r
c
!r
where u ⌘
!(t
r/c). The intensity is hSi =
µ0 m20 ! 4 sin2 ✓
r̂, the same as Eq. 11.39.
32⇡ 2 c3 r2
Problem 11.6
I 2 R = I02 R cos2 (!t) ) hP i =
1 2
µ0 m20 ! 4
µ0 ⇡ 2 b4 I02 ! 4
µ0 ⇡b4 ! 4
2⇡c
I0 R =
=
, so R =
; or, since ! =
,
3
3
2
12⇡c
12⇡c
6c3
µ0 ⇡b4 16⇡ 4 c4
8
R=
= ⇡ 5 µ0 c
3
4
6c
3
✓ ◆4
b
=
8 5
(⇡ )(4⇡ ⇥ 10
3
7
)(3 ⇥ 108 )(b/ )4 = 3.08 ⇥ 105 (b/ )4 ⌦.
Because b ⌧ , and R goes like the fourth power of this small number, R is typically much smaller than the
electric radiative resistance (Prob. 11.3). For the dimensions we used in Prob. 11.3 (b = 5 cm and = 103 m),
R = 3 ⇥ 105 (5 ⇥ 10 5 )4 = 2 ⇥ 10 12 ⌦, which is a millionth of the comparable electrical radiative resistance.
Problem 11.7
0
0
With ↵ = 90 , Eq. 7.68 ) E0 = cB, B0 = E/c, qm
= cqe ) m0 ⌘ qm
d = cqe d = cp0 . So
⇢
✓
◆
✓
◆
µ0 ( m0 /c)! 2 sin ✓
µ0 m0 ! 2 sin ✓
E0 = c
cos[!(t r/c)] ˆ =
cos[!(t r/c)] ˆ.
4⇡c
r
4⇡c
r
⇢
✓
◆
✓
◆
1
µ0 ( m0 /c)! 2 sin ✓
µ0 m0 ! 2 sin ✓
ˆ
B0 =
cos[!(t r/c)] ✓ˆ =
cos[!(t r/c)] ✓.
c
4⇡
r
4⇡c2
r
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235
These are identical to the fields of an Ampére dipole (Eqs. 11.36 and 11.37), which is consistent with our
general experience that the two models generate identical fields except right at the dipole (not relevant here,
since we’re in the radiation zone).
Problem 11.8
(a) The power radiated (Eq. 11.60) is
so
dWr
µ0 2
=
p̈ . In this case p = Qd, so p̈ = Q̈d = Q0
dt
6⇡c
dWr
µ0 (Q0 d)2 2t/RC
=
e
, and the total energy radiated is
dt
6⇡c (RC)4

Z
µ0 (Q0 d)2 1 2t/RC
µ0 (Q0 d)2
RC 2t/RC
Wr =
e
dt
=
e
6⇡c (RC)4 0
6⇡c (RC)4
2
1
✓
1
RC
◆2
e
t/RC
d,
µ0 (Q0 d)2 RC
µ0 (Q0 d)2
=
.
4
6⇡c (RC) 2
12⇡c (RC)3
=
0
The fraction of the original energy that is radiated is therefore
f=
Wr
µ0 (Q0 d)2 2C
µ0 d2
=
=
.
2
3
W0
12⇡c (RC) Q0
6⇡c R3 C 2
[Technically, Q̇(t) is discontinuous at t = 0, and Q̈ picks up a delta function. But any real circuit has some
(self-)inductance, which smoothes out the sudden change in Q̇.]
(b) With the parameters given,
f=
4⇡ ⇥ 10 7
10 8
6⇡(3 ⇥ 108 ) (106 )(10
24 )
Problem 11.9
p(t) = p0 [cos(!t) x̂ + sin(!t) ŷ] ) p̈(t) =
= 2.22 ⇥ 10
6
.
yes, this is negligible.
! 2 p0 [cos(!t) x̂ + sin(!t) ŷ] )
µ0 p20 ! 4 sin2 ✓
r̂. (This appears to disagree
16⇡ 2 c r2
with the answer to Prob. 11.4. The reason is that in Eq. 11.59 the polar axis is along the direction of p̈(t0 );
as the dipole rotates, so do the axes. Thus the angle ✓ here is not the same as in Prob. 11.4.) Meanwhile,
2
[p̈(t)] = ! 4 p20 [cos2 (!t) + sin2 (!t)] = p20 ! 4 . So Eq. 11.59 says S =
µ0 p20 ! 4
. (This does agree with Prob. 11.4, because we have now integrated over all angles,
6⇡c
and the orientation of the polar axis irrelevant.)
Eq. 11.60 says P =
Problem 11.10
At t = 0 the dipole moment of the ring is
Z
Z
p0 =
r dl = ( 0 sin )(b sin ŷ + b cos x̂)b d =
= b (⇡ ŷ + 0 x̂) = ⇡b
2
2
0
ŷ.
As it rotates (counterclockwise, say) p(t) = p0 [cos(!t) ŷ
Therefore (Eq. 11.60) P =
µ0 4 2
! (⇡b
6⇡c
2
0)
=
⇡µ0 ! 4 b4 20
6c
0b
2
✓ Z
ŷ
2⇡
sin2 d + x̂
0
sin(!t) x̂], so p̈ =
.
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Z
0
2⇡
sin cos d
◆
! 2 p, and hence (p̈)2 = ! 4 p20 .
236
Problem 11.11
Here V = 0 (since the ring is neutral), andI the current depends only on t (not on position), so the retarded
µ0
I(t r /c) 0
vector potential (Eq. 11.52) is A(r, t) =
dl . But in this case it does not suffice to replace
r
4⇡
r by r in the denominator—that
would
✓
◆ lead to Eq. 11.54, and hence to A = 0 (since p = 0). Instead, use
1 ⇠1
b
0
0
0ˆ
0
0
0
Eq. 11.30:
r = r 1 + r sin ✓ cos . Meanwhile, dl = b d = b( sin x̂ + cos ŷ) d , and
I(t
r
/c) ⇠
= I(t
r/c + (b/c) sin ✓ cos
0
) = I(t0 + (b/c) sin ✓ cos
0
˙ 0 ) b sin ✓ cos
)⇠
= I(t0 ) + I(t
c
0
(carrying all terms to first order in b). As always, t0 = t r/c. (From now on I’ll suppress the argument: I,
˙ etc. are all to be evaluated at t0 .) Then
I,
✓
◆✓
◆
I
µ0
1
b
b
A(r, t) =
1 + sin ✓ cos 0
I + I˙ sin ✓ cos 0 b( sin 0 x̂ + cos 0 ŷ) d 0
4⇡
r
r
c
Z 2⇡ 
µ0 b
b
b
⇠
I + I˙ sin ✓ cos 0 + I sin ✓ cos 0 ( sin 0 x̂ + cos 0 ŷ) d 0 .
=
4⇡r 0
c
r
Z 2⇡
Z 2⇡
Z 2⇡
Z 2⇡
0
0
0
0
0
0
0
But
sin d =
cos d =
sin cos d = 0, while
cos2 0 d 0 = ⇡.
0
0
0
0

⇣
µ0 b
b
b
µ0 b2
r ˙⌘
=
(⇡ ŷ) I˙ sin ✓ + I sin ✓ =
sin
✓
I
+
I ŷ.
4⇡r
c
r
4r2
c
In general (i.e. for points not on the x z plane) ŷ ! ˆ; moreover, in the radiation zone we are not interested
i sin ✓
µ0 b2 h ˙
ˆ.
in terms that go like 1/r2 , so A(r, t) =
I(t r/c)
4c
r
E(r, t) =
B(r, t) =
=
S=
P =
=
i sin ✓
µ0 b2 h ¨
ˆ.
I(t r/c)
4c
r
1
@
1 @
r⇥A=
(A sin ✓) r̂
(rA ) ✓ˆ
r sin ✓ @✓
r @r
"
#
✓
◆
µ0 b2
I˙ 1
1¨
1
µ0 b2 ¨sin ✓ ˆ
ˆ
2 sin ✓ cos ✓ r̂
I
sin ✓ ✓ =
I
✓.
4c r sin ✓ r
r
c
4c2
r
✓
◆2
1
1
µ0 b2 ¨sin ✓ ⇣ ˆ ˆ⌘
µ0 ⇣ 2 ¨⌘2 sin2 ✓
(E ⇥ B) =
I
⇥✓ =
b I
r̂.
µ0
µ0 c
4c
r
16c3
r2
✓ ◆
Z
Z
µ0 ⇣ 2 ¨⌘2
sin2 ✓ 2
µ0 ⇣ 2 ¨⌘2
4
µ0 ⇡ ⇣ 2 ¨⌘2
S · da =
b
I
r
sin
✓
d✓
d
=
b
I
(2⇡)
=
b I
16c3
r2
16c3
3
6c3
@A
=
@t
µ0 m̈2
.
6⇡c3
¨ 2 .)
(Note that m = I⇡b2 , so m̈ = I⇡b
Problem 11.12
p = ey ŷ, y = 12 gt2 , so p =
µ0
ge ŷ. Therefore (Eq. 11.60) :P =
(ge)2 . Now, the time
6⇡c
p
it takes to fall a distance h is given by h = 12 gt2 ) t = 2h/g, so the energy radiated in falling a distance h
1
2
2 get
ŷ; p̈ =
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237
µ0 (ge)2 p
2h/g. Meanwhile, the potential energy lost is Upot = mgh. So the fraction is
6⇡c
s
s
r
µ0 g 2 e2 2h 1
µ0 e2
2g
(4⇡ ⇥ 10 7 )(1.6 ⇥ 10 19 )2 (2)(9.8)
=
=
. =
= 2.76 ⇥ 10
6⇡c
g mgh
6⇡mc h
6⇡(9.11 ⇥ 10 31 )(3 ⇥ 108 )
(0.01)
is Urad = P t =
f=
Upot
22
.
Evidently almost all the energy goes into kinetic form (as indeed I assumed in saying y = 12 gt2 ).
Problem 11.13
dWr
µ0 q 2 a2
1 qQ
k
qQ
=
. Here F =
= ma, so a = 2 , where k ⌘
.
dt
6⇡c
4⇡✏0 x2
x
4⇡✏0 m
Z 2
Z
µ0 q 2
k
µ0 q 2 2 1 1 1
Wr =
dt
=
2k
dx,
4
6⇡c
x4
6⇡c
x0 x v
The power radiated (Eq. 11.70) is
where x0 is the distance of closest approach. Conservation of energy (ignoring radiative losses, as suggested)
says
1
1
1 qQ
qQ 1
2k
2k
mv 2 = mv 2 +
) v 2 = v02 2
= v02
, and x0 = 2
2 0
2
4⇡✏0 x
4⇡✏0 m x
x
v0
(note that q is at x0 when v = 0). So
Z
Z 1
µ0 q 2 2 1
1
µ0 q 2 2k 2
1
p
p
p
Wr =
2k
dx
=
2
4
4
6⇡c
6⇡c
2k
x
(1/x
x
v
2k/x
x0
x0
0)
0
(1/x)
dx =
µ0 q 2 2k 2 16
p
.
6⇡c 2k 15x5/2
0
(I used Mathematica to do the integral.) Simplifying,
Wr =
2µ0 q 2 v05
2µ0 q 2 v05 4⇡✏0 m
8qmv05
Wr
8qmv05 2
16q ⇣ v0 ⌘3
=
=
) f=
=
=
.
2
3
3
45⇡ck
45⇡c
qQ
45c Q
W0
45c Q mv0
45Q c
Problem 11.14
s
1 q2
v2
1 q2
F =
= ma = m ) v =
. At the beginning (r0 = 0.5 Å),
2
4⇡✏0 r
r
4⇡✏0 mr

v
(1.6 ⇥ 10 19 )2
=
12
c
4⇡(8.85 ⇥ 10 )(9.11 ⇥ 10 31 )(5 ⇥ 10
1/2
11 )
1
= 0.0075,
3 ⇥ 108
and when the radius is one hundredth of this v/c is only 10 times greater (0.075), so for most of the trip the
velocity is safely nonrelativistic.
✓ ◆2
✓
◆2
µ0 q 2 v 2
µ0 q 2
1 q2
From the Larmor formula, P =
=
(since a = v 2 /r), and P = dU/dt,
6⇡c
r
6⇡c 4⇡✏0 mr2
where U is the (total) energy of the electron:
✓
◆
1
1 q2
1
1 q2
1 q2
1 q2
2
U = Ukin + Upot = mv
=
=
.
2
4⇡✏0 r
2 4⇡✏0 r
4⇡✏0 r
8⇡✏0 r
So
dU
=
dt
1 q 2 dr
q2
=
P
=
8⇡✏0 r2 dt
6⇡✏0 c3
✓
1 q2
4⇡✏0 mr2
◆2
, and hence
dr
=
dt
1
3c
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✓
q2
2⇡✏0 mc
◆2
1
, or
r2
238
✓
◆2
✓
◆2 Z 0
✓
◆2
2⇡✏0 mc
2⇡✏0 mc
2⇡✏0 mc
2
2
dt = 3c
r dr ) t = 3c
r dr = c
r03
q2
q2
q2
r0

2
2⇡(8.85 ⇥ 10 12 )(9.11 ⇥ 10 31 )(3 ⇥ 108 )
8
= (3 ⇥ 10 )
(5 ⇥ 10 11 )3 = 1.3 ⇥ 10
(1.6 ⇥ 10 19 )2
11
s. (Not very long!)
Problem 11.15

d
sin2 ✓
According to Eq. 11.74, the maximum occurs at
= 0. Thus
d✓ (1
cos ✓)5
2 sin ✓ cos ✓
5 sin2 ✓( sin ✓)
= 0 ) 2 cos ✓(1
cos ✓) = 5 sin2 ✓ = 5 (1 cos2 ✓);
(1
cos ✓)5
(1
cos ✓)6
2 cos ✓ 2 cos2 ✓ = 5
5 cos2 ✓, or 3 cos2 ✓ + 2 cos ✓ 5 = 0. So
p
⌘
2 ± 4 + 60 2
1 ⇣ p
cos ✓ =
=
± 1 + 15 2 1 . We want the plus sign, since ✓m ! 90 (cos ✓m = 0) when
6
3
!
p
1 + 15 2 1
1
! 0 (Fig. 11.11): ✓max = cos
.
3
For v ⇡ c,
p
1 + 15
3
2
⇡ 1; write
1
!
=1
✏ (where ✏ ⌧ 1), and expand to first order in ✏:
hp
i 1
hp
i
1 ⇠
1 + 15(1 2✏) 1
= (1 + ✏)
3(1 ✏)
3
 ✓
h p
i 1
⇥p
⇤ 1
1
= (1 + ✏) 16 30✏ 1 = (1 + ✏) 4 1 (15✏/8) 1 = (1 + ✏) 4 1
3
3
3
✓
◆
1
15
5 ⇠
5
1
= (1 + ✏) 3
✏ = (1 + ✏)(1
✏) = 1 + ✏
✏=1
✏.
3
4
4
4
4
1
=
1 + 15(1
✏)2
2
2
Evidently ✓max ⇡ 0, so cos ✓max ⇠
= 1 14 ✏ ) ✓max
= 12 ✏, or ✓max ⇠
= 1 12 ✓max
=

2
(dP/d⌦|✓m )ur
sin ✓max
Let f ⌘
=
. Now sin2 ✓max ⇠
= ✏/2, and
(dP/d⌦|✓m )rest
(1
cos ✓max )5 ur
(1
cos ✓max ) ⇠
=1
1
=p
1
2
=p
1
(1
✏)(1
1
(1
✏)2
1
4 ✏)
⇠
=p
1
dP
q2
Equation 11.72 says
=
d⌦
16⇡ 2 ✏0
v = c r̂
1
4 ✏)
✓ ◆5
4
1
f=
(2
5
2
Problem 11.16
u = c r̂
✏/2
(1 ✏
=
So f =
=
(5✏/4)5
1
1
1
= p ) ✏ = 2 . Therefore
2
2✏
(1 2✏)
⇠
=1
vẑ )
r̂
·u=c
r̂
5
4 ✏.
1
) =
4
2 4
✓ ◆5
8
5
8
= 2.62
8
p
p
✏/2 =
(1
✓ ◆5
4
1
.
5
2✏4
◆
15
✏
16
)/2.
But
.
2
⇥ (u ⇥ a)
.
( r̂ · u)5
v( r̂ · ẑ) = c
Let
⌘ v/c.
⇣
v cos ✓ = c 1
⌘
v
cos ✓ = c(1
c
cos ✓);
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1
239
a · u = ac(x̂ ·
| r̂
r̂
)
av(x̂ · ẑ) = ac sin ✓ cos ;
u2 = u · u = c2
2cv( r̂ · ẑ) + v 2 = c2 + v 2
2cv cos ✓.
r̂
⇥ (u ⇥ a) = ( r̂ · a)u ( r̂ · u)a;
⇥ (u ⇥ a)|2 = ( r̂ · a)2 u2 2(u · a)( r̂ · a)( r̂ · u) + ( r̂ · u)2 a2
= (c2 + v 2 2cv cos ✓)(a sin ✓ cos )2 2(ac sin ✓ cos )(a sin ✓ cos )(c v cos ✓) + a2 c2 (1
⇥
⇤
= a2 c2 (1
cos ✓)2 + (sin2 ✓ cos2 )(c2 + v 2 2cv cos ✓ 2c2 + 2cv cos ✓
⇥
⇤
2
= a2 c2 (1
cos ✓)2 (1
)(sin ✓ cos )2 .
⇥
⇤
2
cos ✓)2 (1
) sin2 ✓ cos2
dP
µ0 q 2 a2 (1
=
.
d⌦
16⇡ 2 c
(1
cos ✓)5
cos ✓)2
The total power radiated (in all directions) is:
Z
Z Z ⇥
(1
cos ✓)2
dP
dP
µ0 q 2 a2
d⌦ =
sin ✓ d✓ d =
2
d⌦
d⌦
16⇡ c
(1
Z 2⇡
Z 2⇡
But
d = 2⇡ and
cos2 d = ⇡.
0
0
⇤
Z ⇡⇥
2
2(1
cos ✓)2 (1
) sin2 ✓
µ0 q 2 a2
=
⇡
sin ✓ d✓.
16⇡ 2 c
(1
cos ✓)5
0
P =
Z
Let w ⌘ (1
cos ✓). Then (1
(1
2w2
2
2
)
(
2
w)/ = cos ✓;
1 + 2w
w2 ) =
=
dw =
sin ✓ d✓ ) sin ✓ d✓ =
1
Z (1+ )
µ0 q 2 a2 1
1 ⇥
(1
3
5
16⇡c
(1
) w
Z
1
2 2
Int = (1
)
dw 2(1
w5

✓
◆
1
2 2
= (1
)
2(1
4w4
1+
1
1+
1
1+
1
1
(1 + )2
1
=
(1 + )3
1
=
(1 + )4
=
1
(1
)2
1
)3
(1
1
(1
)4
1 ⇥ 2
2w
2
1 ⇥
(1
2
2
=
=
=
(1
(1
(1
⇥
)
2 2
⇤
w)2 /
(1
2
+ (1
When ✓ = 0, w = (1
dw.
P =
1
w2
1
w3
1
w4
sin2 ✓ =
)
2
2
2(1
sin ✓ d✓ d .
, and the numerator becomes
2
)w + w2 (1
)w + (1 +
2
⇤
)w2 ;
2
⇤
)
); when ✓ = ⇡, w = (1 + ).
⇤
)w + (1 + 2 )w2 dw.
Z
Z
1
1
2
2
)
dw
+
(1
+
)
dw
4
w
w3
✓
◆
✓
◆
1
1
2
2
)
+
(1
+
)
3w3
2w2
)
2 2
⇤
) sin2 ✓ cos2
cos ✓)5
2
2(1
2 2
2(1
(1
The integral is
2
1+
.
1
2 + 2 ) (1 + 2 + 2 )
4
=
.
2
2
2 )2
(1 + ) (1
)
(1
3
3 +3 2
) (1 + 3 + 3 2 + 3 )
2 (3 + 2 )
=
.
2 )3
(1 + )3 (1
)3
(1
4 + 6 2 4 3 + 4 ) (1 + 4 + 6 2 + 4 3 + 4 )
8 (1 + 2 )
=
.
4
4
2 )4
(1 + ) (1
)
(1
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240
✓
◆
8 (1 + 2 )
1
2 (3 + 2 )
2
2(1
)
+ (1 +
2
4
2 )3
(1
)
3
(1

3
2
2
8
=
(1 + 2 )
(3 + 2 ) + (1 + 2 ) =
.
2
2
2 )2
(1
)
3
3 (1
Int = (1
P =
)
2 2
✓
1
4
µ0 q 2 a2 1 8
16⇡c 3 3 (1
◆
3
2 )2
=
µ0 q 2 a2
6⇡c
4
,
where
=p
1
1
2
2
)
✓
1
2
◆
4
(1
2 )2
.
Is this consistent with the Liénard formula (Eq. 11.73)? Here v ⇥ a = va(ẑ ⇥ x̂) = va ŷ, so
✓
◆
⇣v
⌘2
v2
1 2
µ0 q 2 6 a2
2 2
a2
⇥ a = a2 1
=
(1
)a
=
a
,
so
the
Liénard
formula
says
P
=
.X
2
2
c
c2
6⇡c
Problem 11.17
µ0 q 2
ȧ.
6⇡c
For circular motion, r(t) = R [cos(!t) x̂ + sin(!t) ŷ] , v(t) = ṙ = R! [ sin(!t) x̂ + cos(!t) ŷ] ;
(a) To counteract the radiation reaction (Eq. 11.80), you must exert a force Fe =
a(t) = v̇ =
R! 2 [cos(!t) x̂ + sin(!t) ŷ] =
Pe = Fe · v =
µ0 q 2 2 2
! v .
6⇡c
! 2 r; ȧ =
! 2 v. So Fe =
µ0 q 2 2 2
! v , and the two expressions agree.
6⇡c
µ0 q 2 a2
, and a2 = ! 4 r2 = ! 4 R2 = ! 2 v 2 , so
6⇡c
(b) For simple harmonic motion, r(t) = A cos(!t) ẑ; v = ṙ =
! 2 r; ȧ =
! 2 ṙ =
µ0 q 2 2
! v.
6⇡c
This is the power you must supply.
! 2 ṙ =
! 2 v. So Fe =
A! sin(!t) ẑ; a = v̇ =
A! 2 cos(!t) ẑ =
µ0 q 2 2
µ0 q 2 2 2
! v; Pe =
! v . But this time a2 = ! 4 r2 = ! 4 A2 cos2 (!t),
6⇡c
6⇡c
whereas ! 2 v 2 = ! 4 A2 sin2 (!t), so
µ0 q 2 4 2
µ0 q 2 4 2 2
! A cos2 (!t) 6= Pe =
! A sin (!t);
6⇡c
6⇡c
the power you deliver is not equal to the power radiated. However, since the time averages of sin2 (!t) and
cos2 (!t) are equal (to wit: 1/2), over a full cycle the energy radiated is the same as the energy input. (In the
mean time energy is evidently being stored temporarily in the nearby fields.)
(c) In free fall, v(t) = 12 gt2 ŷ; v = gt ŷ; a = g ŷ; ȧ = 0. So Fe = 0; the radiation reaction is zero, and
µ0 q 2 2
g . Evidently energy is being continuously extracted from
6⇡c
the nearby fields. This paradox persists even in the exact solution (where we do not assume v ⌧ c, as in the
Larmor formula and the Abraham-Lorentz formula)—see Prob. 11.34.
Problem 11.18
(a) From Eq. 11.80, Frad =
µ0 q 2
...
ȧ = m⌧ x (Eq. 11.82). The equation of motion is
6⇡c
F = mẍ = Fspring + Frad =
...
kx + m⌧ x, or ẍ + !02 x
p
...
⌧ x = 0, with !0 = k/m.
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241
...
Since the damping is small, it oscillates at the natural frequency !0 , and hence x =
2
2
or ẍ + ẋ + !0 x = 0, with = !0 ⌧. The solution with x(0) = 0 is (for ⌧ !0 )
x(t) = Ae
t/2
and x(t) =
v0
e
!0
sin(!0 t); v(t) =
t/2
2
Ae
t/2
sin(!0 t) + !0 Ae
t/2
cos(!0 t), so v(0) = A!0 = v0 ) A =
a=
!02 x =
!02
v0
e
!0
Averaging over a full cycle (holding e
Z
0
1
hP i dt =
v0
,
!0
sin(!0 t).
µ0 q 2 a2
. In this case (still assuming
6⇡c
According to the Larmor formula (Eq. 11.70), the power radiated is P =
⌧ !0 )
!02 ẋ, so ẍ+!02 ⌧ ẋ+!02 x = 0,
µ0 q 2 !02 v02
12⇡c

1
e
t
t
t/2
sin(!0 t),
constant) hP i =
1
0
=
P =
µ0 q 2
2
(!0 v0 ) e
6⇡c
µ0 q 2 !02 v02
e
6⇡c
t1
2
t
sin2 (!0 t).
, and the total energy radiated is
µ0 q 2 !02 v02
µ0 q 2 !02 v02
µ0 q 2 v02
=
=
2
12⇡c
12⇡c!0 ⌧
12⇡c
✓
6⇡mc
µ0 q 2
◆
=
1
mv 2 . X
2 0
(b) This “equivalent”
single oscillator has twice the charge, and twice the mass, so ⌧ (and hence ) is
R
doubled. Since hP i dt goes like q 2 / , it also doubles. The power radiated is indeed four times as great, but
the oscillations die away faster, and the total energy radiated is just twice as much as for one oscillator.
Problem 11.19
Z
Z
Z
F
dv
da
F
dv
da
1
(a) a = ⌧ ȧ +
)
=⌧
+
)
dt = ⌧
dt +
F dt.
m
dt
dt
m
dt
dt
m
2✏
[v(t0 + ✏) v(t0 ✏)] = ⌧ [a(t0 + ✏) a(t0 ✏)] + Fave , where Fave is the average force during the interm
val. But v is continuous, so as long as F is not a delta function, we are left (in the limit ✏ ! 0) with
[a(t0 + ✏) a(t0 ✏)] = 0. Thus a, too, is continuous.
Z
Z qed
da
da
1
da
1
t
(b) (i) a = ⌧ ȧ = ⌧
)
= dt )
=
dt ) ln a = + constant ) a(t) = Aet/⌧ , where A
dt
a
⌧
a
⌧
⌧
is a constant.
F
da
F
da
1
t
F
(ii) a = ⌧ ȧ +
)⌧
=a
)
= dt ) ln(a F/m) = + constant ) a
= Bet/⌧ )
m
dt
m
a F/m
⌧
⌧
m
F
a(t) =
+ Bet/⌧ , where B is some other constant.
m
(iii) Same as (i): a(t) = Cet/⌧ , where C is a third constant.
(c) At t = 0, A = F/m + B; at t = T , F/m + BeT /⌧ = CeT /⌧ ) C = (F/m)e
8
[(F/m) + B] et/⌧ ,
t  0;
>
>
>
>
>
>
h
i
<
(F/m) + Bet/⌧ ,
0  t  T;
a(t) =
>
>
>
>
h
i
>
>
: (F/m)e T /⌧ + B et/⌧ , t T.
To eliminate the runaway in region (iii), we’d need B = (F/m)e
(i), we’d need B = (F/m). Obviously, we cannot do both at once.
T /⌧
T /⌧
+ B.
So
; to avoid preacceleration in region
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242
(d) If we choose to eliminate the runaway, then
h
(i) v = (F/m) 1
e
T /⌧
iZ
8
h
>
(F/m)
1
>
>
>
>
>
<
h
a(t) = (F/m) 1
>
>
>
>
>
>
:
0,
h
et/⌧ dt = (F ⌧ /m) 1
e
T /⌧
e(t
i
et/⌧ , t  0;
T )/⌧
i
,
0  t  T;
t
e
T /⌧
i
T.
et/⌧ + D, where D is a constant determined by the
condition v( 1) =h0 ) D = 0. i
(ii) v = (F/m) t ⌧ e(t T )/⌧ + E, where E is a constant determined by the continuity of v at t = 0:
h
i
h
i
(F ⌧ /m) 1 e T /⌧ = (F/m) ⌧ e T /⌧ + E ) E = (F ⌧ /m).
(iii) v is a constant determined by the continuity of v at t = T : v = (F/m)[T + ⌧ ⌧ ] = (F/m)T.
(e)
8
h
i
T /⌧
>
(F
⌧
/m)
1
e
et/⌧ , t  0;
>
>
>
>
>
<
h
i
v(t) = (F/m) t + ⌧ ⌧ e(t T )/⌧ , 0  t  T ;
>
>
>
>
>
>
:
(F/m)T,
t T.
Problem 11.20

✓ ◆
µ0 (q/2)2
µ0 q 2
1
1
µ0 q 2
int
end
=
ȧ
+2
=
ȧ. X
6⇡c
6⇡c
2
4
6⇡c
(b) Following the suggested method:
✓ ◆
1
1
µ0 q 2 ȧ
F (q) = F int (q) + 2
F (q) ) F (q) = F int (q) ) F (q) = 2F int (q) =
. X
4
2
6⇡c
Z L ⇢Z y1
µ0
ȧ
2 dy2 2 dy1 . (Running the y2
12⇡c 0
0
integral up to y1 insures that y1
y2 , so we don’t count the
same pair twice. Alternatively, run both integrals from 0 to
L—intentionally double-counting—and divide the result by 2.)
end
=
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243
µ0 ȧ
=
(4
12⇡c
2
)
Z
L
y1 dy1 =
0
µ0 ȧ
(4
12⇡c
2
)
L2
µ0
µ0 q 2
=
( L)2 ȧ =
ȧ. X
2
6⇡c
6⇡c
Problem 11.21
(a) The total torque is twice the torque on +q; we might as well calculate it at time t = 0. First we
need the electric field at +q, due to q when it was at the retarded point P (Eq. 10.72). From the figure,
r̂ = cos ✓ x̂ sin ✓ ŷ, r = 2R cos ✓. The velocity of q (at P ) was v = !R (sin 2✓ x̂ + cos 2✓ ŷ), and its
acceleration was a = ! 2 R (cos 2✓ x̂ sin 2✓ ŷ) . Quantities we will need in Eq. 10.72 are:
r
u = c r̂
v = (c cos ✓ + !R sin 2✓) x̂ (c sin ✓ !R cos 2✓) ŷ,
· u = 2R cos ✓(c + !R sin ✓), r · a = 2(!R cos ✓)2 .
y
P
r
-q
E=
q
R
q
x
n
q
2R cos ✓
[c2
3
3
4⇡✏0 (2R cos ✓) (c + !R sin ✓)
(!R)2 + 2(!R cos ✓)2 ][(c cos ✓ + !R sin 2✓) x̂
o
2R cos ✓(c + !R sin ✓)! 2 R (cos 2✓ x̂ sin 2✓ ŷ)
(c sin ✓
!R cos 2✓) ŷ]
The total torque (about the origin) is
N = 2(R x̂) ⇥ (qE) =
=
=
n
2q 2 R
ẑ
2
3
4⇡✏0 (2R cos ✓) (c + !R sin ✓)
⇥ 2
⇤
c
(!R)2 + 2(!R cos ✓)2 (c sin ✓
o
+ 2(!R)2 cos ✓(c + !R sin ✓) sin 2✓
⇥ 3
q2
ẑ
c sin ✓ + c2 !R(2 cos2 ✓
2
3
4⇡✏0 2R cos ✓(c + !R sin ✓)
⇥
q2
1
sin ✓ + (2 cos2 ✓ 1) +
2
3
4⇡✏0 2R cos ✓(1 + sin ✓)
!R cos 2✓)
1) + c(!R)2 (2 cos2 ✓ + 1) sin ✓ + (!R)3
2
(2 cos2 ✓ + 1) sin ✓ +
3
⇤
ẑ,
⇤
where ⌘ !R/c. [Since E and r both lie in the xy plane, B = (1/c) r̂ ⇥ E is along the z direction, v ⇥ B is
radial, and hence the magnetic contribution to the torque is zero.]
The angle ✓ is determined by the retarded time condition, r = ctr (note that tr is negative, here), and
2✓ is the angle through which the dipole rotates in time tr , so 2R cos ✓ = ctr = c(2✓/!), or ✓ = cos ✓. We
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244
can use this to eliminate the trig functions:
N=
q2
4⇡✏0
⇣
p
2R✓2 1 +
2
✓2
Meanwhile, expanding in powers of ✓:
⌘3
h
(
4
+ 2✓2 ) +
2
p
✓2 (
2
2
i
1) ẑ.
+ 2✓2
1
5
61 7
= ✓ sec ✓ = ✓ + ✓3 + ✓5 +
✓ + ....
2
24
720
This can be “solved” (for ✓ as a function of ) by reverting the series:
1
2
✓=
Then
✓2 =
h
(
✓
4
3
4
1
1
= 2 1+
✓2
2
2
4
1
2
2
+
+ 2✓ ) +
2
To leading order in , then,
p
N=
3
+
13
24
◆ p
+ ... ,
+ ... , ⇣
✓2 =
2
1+
✓2 ( 2
2
541
720
5
1
p
+ 2✓
2
2
q2
1 8 4
ẑ =
4⇡✏0 2R 2 3
(b) The radiation reaction force on +q is (Eq. 11.80) F =
net torque (counting both ends) is
N=
2R
µ0 q 2 3
! R ẑ =
6⇡c
7
2
+ ....
✓
2
3
1
⌘3 = 1
✓2
i 8
1) =
3
4
✓
2
+
3
2
1
4
5
4
◆
+ ... ,
+ ...,
4
5
2
◆
+ ... .
q2 4 3
ẑ.
4⇡✏0 3R
µ0 q 2
ȧ. In this case ȧ =
6⇡c
!2 v =
! 3 R ŷ, so the
q2 4 3
ẑ.
4⇡✏0 3R
Adding this to the interaction torque from (a), the total is
N=
(c) p̈ = 2qr̈ = 2q( ! 2 )r =
q2 8 3
ẑ =
4⇡✏0 3R
µ0 p2 ! 3
ẑ.
6⇡c
! 2 p, so Eq. 11.60 says the power radiated is P =
associated with the torque in (b) is N ! =
µ0 p2 ! 3
!, so they are in agreement. X
6⇡c
µ0 4 2
! p . The power
6⇡c
Problem 11.22
p
(a) This is an oscillating electric dipole, with amplitude p0 = qd and frequency ! = k/m. The (averaged)
✓
◆
µ0 p20 ! 4 sin2 ✓
Poynting vector is given by Eq. 11.21: hSi =
r̂, so the power per unit area of floor is
32⇡ 2 c
r2
✓
◆
µ0 p20 ! 4 sin2 ✓ cos ✓
R
h
If = hSi · ẑ =
. But sin ✓ = , cos ✓ = , and r2 = R2 + h2 .
32⇡ 2 c
r2
r
r
✓
◆
µ0 q 2 d2 ! 4
R2 h
=
.
2
2
32⇡ c
(R + h2 )5/2
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245

dIf
d
R2
2R
5
R2
=0)
=0)
2R = 0 )
2
2
5/2
2
2
5/2
2
dR
dR (R + h )
2 (R + h2 )7/2
(R + h )
p
5 2
3
(R2 + h2 )
R = 0 ) h2 = R2 ) R = 2/3h, for maximum intensity.
2
2
(b)
✓
◆ Z 1
Z
Z
µ0 (qd)2 ! 4
R3
P =
If (R) da = If (R) 2⇡R dR = 2⇡
h
dR.
32⇡ 2 c
(R2 + h2 )5/2
0
Z 1
Z
R3
1 1
x
1 (2) (1/2)
2
dR
=
dx =
=
.
2
2
5/2
2
5/2
2 0 (x + h )
2h
(5/2)
3h
(R + h )
0
✓
◆
µ0 q 2 d2 ! 4
2
µ0 q 2 d2 ! 4
= 2⇡
h
=
,
32⇡ 2 c
3h
24⇡c
Let x ⌘ R2 :
which should be (and is) half the total radiated power (Eq. 11.22)—the rest hits the ceiling, of course.
(c) The amplitude is x0 (t), so U = 12 kx20 is the energy, at time t, and dU/dt = 2P is the power radiated:
1 d 2
µ0 ! 4 2 2
d
µ0 ! 4 q 2 2
k (x0 ) =
q x0 ) (x20 ) =
(x0 ) = x20 ) x20 = d2 e t or x0 (t) = de t/2 .
2 dt
12⇡c
dt
6⇡kc
2
12⇡kc 2
12⇡cm2
⌧= =
m
=
.

µ0 q 2 k 2
µ0 q 2 k
✓
◆
Problem 11.23
µ0 m20 ! 4 sin2 ✓
(a) From Eq. 11.39, hSi =
r̂. Here sin ✓ =
32⇡ 2 c3
r2
p
R/r, r = R2 + h2 , and the total radiated
is
✓ power
◆ (Eq. 11.40)
µ0 m20 ! 4
12P
R2
P =
. So the intensity is I(R) =
=
12⇡c3
32⇡ (R2 + h2 )2
3P
R2
.
8⇡ (R2 + h2 )2
(b) The intensity directly below the antenna (R = 0) would (ideally) have been zero. The engineer should
have measured it at the position of maximum intensity:

dI
3P
2R
2R2
3P
2R
=
2R =
R2 + h2 2R2 = 0 ) R = h.
dR
8⇡ (R2 + h2 )2
(R2 + h2 )3
8⇡ (R2 + h2 )3
At this location the intensity is I(h) =
(c) Imax =
3P h2
3P
=
.
2
2
8⇡ (2h )
32⇡h2
3(35 ⇥ 103 )
= 0.026 W/m2 = 2.6 µW/cm2 .
32⇡(200)2
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246
Problem 11.24
(a)
✓
◆
p0 !
cos ✓±
V± = ⌥
sin[!(t r± /c)]. Vtot = V+ + V .
4⇡✏0 c
r±
✓
◆
p
p
d
2
2
⇠
⇠
r± = r + (d/2) ⌥ 2r(d/2) cos ✓ = r 1 ⌥ (d/r) cos ✓ = r 1 ⌥
cos ✓ .
2r
✓
◆
1 ⇠ 1
d
1±
cos ✓ .
=
r±
r
2r
✓
◆ ✓
◆
r cos ✓ ⌥ (d/2)
d 1
d
d
d
cos ✓± =
= r cos ✓ ⌥
1±
cos ✓ = cos ✓ ±
cos2 ✓ ⌥
r±
2r r
2r
2r
2r
d
d
= cos ✓ ⌥ (1 cos2 ✓) = cos ✓ ⌥
sin2 ✓.
2r
2r
⇢ 
✓
◆
✓
◆
r
d
!d
sin[!(t r± /c)] = sin ! t
1⌥
cos ✓
= sin !t0 ±
cos ✓ , where t0 ⌘ t r/c.
c
2r
2c
✓
◆
✓
◆
!d
!d
!d
= sin(!t0 ) cos
cos ✓ ± cos(!t0 ) sin
cos ✓ ⇠
cos ✓ cos(!t0 ).
= sin(!t0 ) ±
2c
2c
2c
⇢✓
◆✓
◆
p0 !
d
d
!d
V± = ⌥
1±
cos ✓
cos ✓ ⌥
sin2 ✓ sin(!t0 ) ±
cos ✓ cos(!t0 )
4⇡✏0 cr
2r
2r
2c
⇢✓
◆
p0 !
d
d
!d
=⌥
cos ✓ ⌥
sin2 ✓ ±
cos2 ✓ sin(!t0 ) ±
cos ✓ cos(!t0 )
4⇡✏0 cr
2r
2r
2c

p0 !
!d
d
=⌥
cos ✓ sin(!t0 ) ±
cos2 ✓ cos(!t0 ) ±
cos2 ✓ sin2 ✓ sin(!t0 .
4⇡✏0 cr
2c
2r

p0 !
!d
d
Vtot =
cos2 ✓ cos(!t0 ) +
cos2 ✓ sin2 ✓ sin(!t0 )
4⇡✏0 cr c
r
i
p0 ! 2 d h 2
c
2
2
=
cos
✓
cos(!t
)
+
cos
✓
sin
✓
sin(!t
0
0 .
4⇡✏0 c2 r
!r
Meanwhile
!/c) the second term is negligible, so V =
p0 ! 2 d
cos2 ✓ cos[!(t
4⇡✏0 c2 r
r/c)].
µ0 p0 !
4⇡r±
µ0 p0 !
=⌥
4⇡r
µ0 p0 !
=⌥
4⇡r
A± = ⌥
Atot
sin[!(t r± /c)] ẑ
⇢✓
◆
d
!d
1±
cos ✓ sin(!t0 ) ±
cos ✓ cos(!t0 ) ẑ
2r
2c

!d
d
sin(!t0 ) ±
cos ✓ cos(!t0 ) ±
cos ✓ sin(!t0 ) ẑ.
2c
2r

µ0 p0 ! !d
d
= A+ + A=
cos ✓ cos(!t0 ) + cos ✓ sin(!t0 ) ẑ
4⇡r
c
r
h
i
2
µ0 p0 ! d
c
=
cos ✓ cos(!t0 ) +
sin(!t0 ) ẑ.
4⇡cr
!r
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247
µ0 p0 ! 2 d
cos ✓ cos[!(t r/c)] ẑ.
4⇡cr
µ0 p0 ! 2 d
(b) To simplify the notation, let ↵ ⌘
. Then
4⇡
In the radiation zone, A =
cos2 ✓
cos[!(t r/c)];
r
⇢
@V
1 @V ˆ
1
!
2
rV =
r̂ +
✓ = ↵ cos ✓
cos[!(t r/c)] +
sin[!(t r/c)] r̂
2
@r
r @✓
r
rc
2
2 cos ✓ sin ✓
ˆ = ↵ ! cos ✓ sin[!(t r/c)] r̂ (in the radiation zone).
+↵
cos[!(t
r/c)]
✓
r2
c r
⇣
⌘
⇣
⌘
↵ cos ✓
@A
↵! cos ✓
A =
cos[!(t r/c)] cos ✓ r̂ sin ✓ ✓ˆ .
=
sin[!(t r/c)] cos ✓ r̂ sin ✓ ✓ˆ .
c r
@t
c r
V =↵
E=
rV
@A
=
@t
↵!
sin[!(t
cr
⇣
r/c)] cos2 ✓ r̂
⌘
cos2 ✓ r̂ + sin ✓ cos ✓ ✓ˆ
↵!
ˆ
sin ✓ cos ✓ sin[!(t r/c)] ✓.
cr

1 @
@Ar ˆ
B = r⇥A=
(rA✓ )
r @r
@✓
⇢

↵
@
@ cos2 ✓
ˆ
=
(cos ✓ cos[!(t r/c)]( sin ✓))
cos[!(t r/c)]
cr @r
@✓
r
↵
!
↵!
=
( sin ✓ cos ✓) sin[!(t r/c)] ˆ (in the radiation zone) =
sin ✓ cos ✓ sin[!(t
cr
c
c2 r
=
Notice that B =
r/c)] ˆ.
1
(r̂ ⇥ E) and E · r̂ = 0.
c
⇤
1
1
1 ⇥ 2
E2
(E ⇥ B) =
E ⇥ (r̂ ⇥ E) =
E r̂ (E · r̂)E =
r̂
µ0
µ0 c
µ0 c
µ0 c
o2
⌘2
1 n ↵!
1 ⇣ ↵!
=
sin ✓ cos ✓ sin[!(t r/c)] r̂.
I=
sin ✓ cos ✓ .
µ0 c rc
2µ0 c rc
Z
Z
Z ⇡
⇣
⌘
1 ↵! 2
1 ⇣ ↵! ⌘2
P = hSi · da =
sin2 ✓ cos2 ✓ sin ✓ d✓ d =
2⇡
(1
µ0 c c
2µ0 c c
0
cos3 ✓ ⇡ cos5 ✓ ⇡
2 2
4
The integral is :
+
=
=
.
3 0
5 0
3 5
15
1 ! 2 µ20
4
µ0
=
(p0 d)2 ! 4 2⇡
=
(p0 d)2 ! 6 .
2µ0 c c2 16⇡ 2
15
60⇡c3
S=
cos2 ✓) cos2 ✓ sin ✓ d✓.
Notice that it goes like ! 6 , whereas dipole radiation goes like ! 4 .
Problem 11.25
(a) m(t) = M cos ẑ + M sin [cos(!t) x̂ + sin(!t) ŷ]. As in Prob. 11.4, the power radiated will be twice
that of an oscillating magnetic dipole with dipole moment of amplitude m0 = M sin . Therefore (quoting
Eq. 11.40): P =
µ0 M 2 ! 4 sin2
6⇡c3
. (Alternatively, you can get this from the answer to Prob. 11.11.)
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248
(b) From Eq. 5.88, with r ! R, m ! M , and ✓ = ⇡/2: B =
M=
4⇡R3
4⇡(6.4 ⇥ 106 )3 (5 ⇥ 10
B =
µ0
4⇡ ⇥ 10 7
µ0 M
, so
4⇡ R3
5
)
= 1.3 ⇥ 1023 A m2 .
✓
◆4
)(1.3 ⇥ 1023 )2 sin2 (11 )
2⇡
= 4 ⇥ 10 5 W (not much).
6⇡(3 ⇥ 108 )3
24 ⇥ 60 ⇥ 60
µ0 (4⇡R3 B/µ0 )2 ! 4 sin2
8⇡
2
=
! 2 R3 B sin
. Using the average value (1/2) for sin2 ,
(d) P =
3
3
6⇡c
3µ0 c
"✓
#2
◆2
8⇡
2⇡
1
4 3
8
P =
(10 ) (10 )
= 2 ⇥ 1036 W (a lot).
3(4⇡ ⇥ 10 7 )(3 ⇥ 108 )3
10 3
2
(c) P =
(4⇡ ⇥ 10
7
Problem 11.26
(a) Write p(t) = q(t)d ẑ, with q(t) = kt2 , where kd = (1/2)p̈0 . As in Eq. 11.5,

1
k(t r + /c)2
k(t r /c)2
V (r, t) =
r+
r
4⇡✏0
"
#
2
2
t
(2t/c) r + + r + /c2
t2 (2t/c) r + r 2+ /c2
k
=
r+
r
4⇡✏0
 ✓
◆
k
1
1
1
+ 2 (r + r ) .
=
t2
r+ r
4⇡✏0
c
From Eqs. 11.8 and 11.9,
r
±
✓
◆
d
=r 1⌥
cos ✓ ,
2r
1
r
±
=
1
r
✓
1±
◆
d
cos ✓ ,
2r
so
 2✓
◆
k
t
d
V (r, t) =
cos ✓
4⇡✏0 r r
r
c2
✓
d
cos ✓
r
◆
k
=
d cos ✓
4⇡✏0 c2
"✓
ct
r
◆2
#
µ0 p̈0
1 =
cos ✓
8⇡
"✓
ct
r
◆2
#
1 .
As in Eq. 11.15, I(t) = (dq/dt) ẑ = 2kt ẑ, so (following Eqs. 11.16, and 11.17),
µ0
A(r, t) =
ẑ
4⇡
Z
d/2
d/2
2k(t
r
r
/c)
✓
µ0 (t r/c)
µ0 p̈0
dz =
2k
d ẑ =
4⇡
r
4⇡c
(
"✓ ◆

2
@A
µ0 p̈0
(ct)2
1
ct
E = rV
=
cos ✓
2 3
r̂
sin ✓
@t
8⇡
r
r
r
"
✓ ◆2
✓ ◆2
µ0 p̈0
ct
1
ct
1
=
cos ✓
r̂ + sin ✓
✓ˆ
sin ✓ ✓ˆ (cos ✓ r̂
4⇡r
r
2
r
2
("✓ ◆
#
"✓ ◆
#
)
2
2
µ0 p̈0
ct
1
ct
=
1 cos ✓ r̂ +
+ 1 sin ✓ ✓ˆ .
4⇡r
r
2
r
# )
1 ✓ˆ
#
ct
r
◆
1 ẑ.
µ0 p̈0 ⇣ c ⌘
ẑ
4⇡c r
ˆ
sin ✓ ✓)
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249
⇢✓ ◆
µ0 p̈0
ct
ˆ
r⇥A=
r⇥
1 (cos ✓ r̂ sin ✓ ✓)
4⇡c
r
⇢
✓ ✓ ◆
◆
✓✓ ◆
◆
µ0 p̈0
@
ct
@
ct
=
r
1 sin ✓
1 cos ✓
4⇡cr
@r
r
@✓
r
⇢
✓ ◆
µ0 p̈0
ct
µ0 p̈0 t
=
sin ✓ +
1 sin ✓ ˆ =
sin ✓ ˆ.
4⇡cr
r
4⇡r2
B =
ˆ
(b) The Poynting vector is
1
µ0 p̈20 t
S=
(E ⇥ B) =
sin ✓
µ0
16⇡ 2 r3
("✓
µ0 p̈20 t
S · da =
32⇡ 2
µ0 p̈20 t
P (r, t) =
32⇡ 2 r
"✓
ct
r
◆2
"✓
ct
r
ct
r
#
#
◆2
◆2
+ 1 2⇡
ˆ +1
1 cos ✓( ✓)
2
#
+1
Z
⇡
0
"✓
ct
r
◆2
#
)
+ 1 sin ✓ r̂ .
sin2 ✓ 2
(r sin ✓ d✓ d ).
r3
µ0 p̈20 t
sin ✓ d✓ =
12⇡r
3
"✓
ct
r
◆2
#
+1 .
 ✓
◆
✓
◆"
✓ ◆2 #
µ0 p̈20 ⇣
r ⌘ c2 2
r r2
µ0 p̈20
ct0
ct0
ct0
(c) P (r, t0 + r/c) =
t0 +
t + 2t0 + 2 + 1 =
1+
2+2
+
.
12⇡r
c
r2 0
c
c
12⇡c
r
r
r
Prad (t0 ) = lim P (r, t0 + r/c) =
r!1
µ0 p̈20
,
6⇡c
in agreement with Eq. 11.60.
Problem 11.27
The momentum flux density is (minus) the Maxwell stress tensor (Section 8.2.3),
✓
◆
✓
◆
1
1
1
2
2
Tij = ✏0 Ei Ej
+
Bi Bj
,
ij E
ij B
2
µ0
2
(Eq. 8.17) so the momentum radiated per unit (retarded) time is
I
dp
1
\$
=
T · dA,
dtr
(@tr /@t)
where (Eq. 10.78)
@tr
=
@t
rc
r ·u =
✓
1
r̂
·v
c
◆
1
.
(This factor is 1, when v = 0; it can be ignored in deriving the Larmor formula, but it does contribute to the
momentum radiated.) The integration is over a large spherical surface centered on the charge:
dA =
r
2
sin ✓ d✓ d
r̂
.
As in the case of the power radiated, only the radiation fields contribute (Eq. 11.66):
r [ r ⇥ (u ⇥ a)],
q
4⇡✏0 ( r · u)3
1
c
r̂
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250
Thus

\$
T · dA = ✏0 E(E · dA)

1 2
1
E dA +
B(B · dA)
2
µ0
1 2
B dA =
2

1
1 2
✏0 E 2 +
B dA =
2
µ0
✏0 E 2 dA
(note that B · dA = 0, and, for radiation fields, E · dA = 0, (1/µ0 )B 2 = (1/µ0 c2 )( r̂ ⇥ E) · ( r̂ ⇥ E) =
✏0 r̂ · [E ⇥ ( r̂ ⇥ E)] = ✏0 r̂ · [ r̂ E 2 E( r̂ · E)] = ✏0 E 2 ). So
◆
I ✓
I
r ·u 2
r [ r ⇥ (u ⇥ a)]2 dA.
dp
q2
= ✏0
E dA =
2
rc
dtr
16⇡ ✏0 c
( r · u)5
We expand the integrand to first order in
r
⌘ v/c: u = c r̂
1
1
=
r 5 c5 [1 + 5( r̂ · )],
( r · u)5
⇥ (u ⇥ a) = u( r · a) a( r · u) = r c r̂
⇥
= r c r̂ ( r̂ · a) a + a( r̂ · )
[ r ⇥ (u ⇥ a)]2 =
r
( r · u)5
r
h
c ( r̂ · a)2
2 2
+ 2(a · )( r̂
[ r ⇥ (u ⇥ a)]2 =
=
=
=
r
r
r
1
2 c3
1
2 c3
1
2 c3
⇥ 2
a
⇥ 2
a
r
v=c
( r̂ · a)
⇤
( r̂ · a) .
r̂
a r c(1
,
r̂
r
· u = c r (1
· ),
· )
2( r̂ · a)2 + a2 + 2( r̂ · a)2 ( r̂ · ) 2( r̂ · )( r̂ · a)2 2a2 ( r̂ · )
i
⇥
⇤
· a) = r 2 c2 a2 ( r̂ · a)2 2a2 ( r̂ · ) + 2( r̂ · a)(a · ) .
1
2 c3
( r̂ · a)2
⇥
[1 + 5( r̂ · )] a2
( r̂ · a)2
⇥
( r̂ ⇥ a)2 +
r̂
⇤
2a2 ( r̂ · ) + 2( r̂ · a)(a · )
2a2 ( r̂ · ) + 2( r̂ · a)(a · ) + 5( r̂ · )a2
( r̂ · a)2 + 3a2 ( r̂ · ) + 2( r̂ · a)(a · )
· [3a2 + 2(a · )a]
5( r̂ · )( r̂ · a)2
5( r̂ · )( r̂ · a)2
⇤
5( r̂ · )( r̂ · a)2 .
To integrate the first term we set the polar axis along a, so ( r̂ ⇥ a)2 = a2 sin2 ✓, while
sin ✓ sin ŷ + cos ✓ ẑ. The integral kills the x̂ and ŷ components, leaving
I
Z ⇡
1
2
2
( r̂ ⇥ a) dA = 2⇡a ẑ
sin3 ✓ cos ✓ d✓ = 0.
2
r
r̂
⇤
r̂
⇤
= sin ✓ cos x̂ +
0
[Note that if v = 0 then dp/dtr = 0—a particle instantaneously at rest radiates no momentum. That’s why
we had to carry the expansion to first order in , whereas in deriving the Larmor formula we could a↵ord to
set v = 0.] The second term is of the form ( r̂ · g), for a constant vector g. Setting the polar axis along g, so
r̂ · g = g cos ✓, the x and y components again vanish, leaving
I
Z ⇡
1
4⇡
( r̂ · g) dA = 2⇡g ẑ
cos2 ✓ sin ✓ d✓ =
g.
2
r
3
0
The last term involves ( r̂ · )( r̂ · a)2 ; this time we orient the polar axis along a and let v lie in the xz plane:
= x x̂ + z ẑ, so ( r̂ · ) = x sin ✓ cos + z cos ✓. Then
I
Z
⇥
⇤
1
2
2
r̂
r̂
(
·
)(
·
a)
dA
=
a
( x sin ✓ cos + z cos ✓) cos2 ✓ sin ✓ cos x̂ + sin ✓ sin ŷ + cos ✓ ẑ sin ✓ d✓ d
2
r
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251
= a2
✓
x x̂
Z
sin3 ✓ cos2 ✓ cos2 d✓ d +
z ẑ
Putting all this together,
dp
q2 1
=
dtr
16⇡ 2 ✏0 c4
⇢
Z
cos4 ✓ sin ✓ d✓ d
⇤
4⇡ ⇥
2a(a · ) + 3a2
3
I
dL
=
dtr
5
◆
=
4⇡ 2
a (
15
x
x̂ + 3
⇤
4⇡ ⇥ 2
a + 2(a · )a
15
=
z
ẑ) =
4⇡ 2
a + 2(a · )a .
15
µ0 q 2 2
a v.
6⇡c3
⇣
⌘
1
r ⇥\$
T · dA.
(@tr /@t)
Because of the “extra” r in the integrand, it seems at first glance that the radiation fields alone will
⇣ produce⌘a
\$
result that grows without limit (as r ! 1); however, the coefficient of this term is precisely zero: r ⇥ T ·
✏0 E 2 ( r ⇥ dA) = 0 (for radiation fields). The finite contribution
comes from the cross terms, ◆
in which
✓
r
q
\$
one field (in T ) is a radiation field and the other a Coulomb field Ecoul ⌘
(c2 v 2 )u :
4⇡✏0 ( r · u)3
dA =
⇣
(r) (c)
(c) (r)
Tij = ✏0 Ei Ej + Ei Ej
This time
(r)
ij E
h
\$
T · dA = ✏0 E(r) (E(c) · dA)
so
⌘
1 ⇣ (r) (c)
(c) (r)
· E(c) +
Bi Bj + Bi Bj
µ0
⇣
dL
=
dtr
q2 c
16⇡ 2 ✏0
I
r
n
r
(r)
⌘
· B(c) .
⌘
i
E(r) · E(c) dA
\$
( r ⇥ T ) · dA = ✏0
Thus
ij B
⇣
r
⌘
1 ⇣ (r)
B · B(c) dA,
µ0
⌘⇣
⌘
⇥ E(r) E(c) · dA .
o
⇥ [ r ⇥ (u ⇥ a)] u · dA, where
1
⌘p
.
(v/c)2
⇥
⇤
Now, { r ⇥ [ r ⇥ (u ⇥ a)]} = r [ r ·(u⇥a)] (u⇥a) r 2 , and (u⇥a) = c( r̂
)⇥a = c ( r̂ ⇥ a) ( ⇥ a) ,
⇥
⇤
r̂ · (u ⇥ a) = c r̂ · ( ⇥ a), so { r ⇥ [ r ⇥ (u ⇥ a)]} = r 2 c ( r̂ ⇥ a) ( ⇥ a) + r̂ r̂ · ( ⇥ a) .
Meanwhile u · dA = (u · r̂ ) r 2 d⌦, where d⌦ ⌘ sin ✓ d✓ d . Expanding to first order in ,
I
r ( r 2 c) ( r̂ ⇥ a) ( ⇥ a) + r̂ ⇥ r̂ · ( ⇥ a)⇤ r ( r · u) d⌦
dL
µ0 q 2 c3
=
2
dtr
16⇡
( r · u)5
Z
⇥
⇤
⇥
⇤
µ0 q 2
=
1 + 4( r̂ · ) ( r̂ ⇥ a) ( ⇥ a) + r̂ r̂ · ( ⇥ a) d⌦
2
16⇡
Z n
⇤o
µ0 q 2
r̂
r̂
r̂
r̂
r̂
=
(
⇥
a)
(
⇥
a)
+
[
·
(
⇥
a)]
+
4(
·
)(
⇥
a)
d⌦
16⇡ 2
2
( r · u)5
The first integral is
Z
r̂ sin ✓ d✓ d =
a⇥
a⇥
Z
the second is
( ⇥ a)
1
(sin ✓ cos x̂ + sin ✓ sin ŷ + cos ✓ ẑ) sin ✓ d✓ d = 0,
Z
sin ✓ d✓ d =
4⇡( ⇥ a),
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252
for the third we set the polar axis along g ⌘
g
Z
r̂
⇥ a and get
cos ✓ sin ✓ d✓ d = 2⇡g ẑ
Z
cos2 ✓ sin ✓ d✓ =
and for the last we put the polar axis along a and let
4a
Z
= 4a
=
(
Z
x
sin ✓ cos +
(
z
=
x x̂
+
z ẑ
4⇡
4⇡
g=
( ⇥ a),
3
3
lie in the xz plane:
cos ✓) sin ✓( ˆ) sin ✓ d✓ d
sin ✓ cos + z cos ✓)(sin x̂ cos ŷ) sin2 ✓ d✓ d
Z
16⇡
16⇡
4a x ŷ cos2 sin3 ✓ d✓ d =
a x ŷ =
( ⇥ a).
3
3
x
Putting this all together, we conclude
dL
µ0 q 2
=
dtr
16⇡ 2

4⇡( ⇥ a) +
4⇡
16⇡
µ0 q 2
( ⇥ a) +
( ⇥ a) =
(v ⇥ a).
3
3
6⇡c
Problem 11.28
Z
µ0
K(tr )
(a) A(x, t) =
r da
4⇡
Z
µ0 ẑ
K(tr )
p
=
2⇡r dr
4⇡
r2 + x2
p
Z
µ0 ẑ
K(t
r2 + x2 /c)
p
=
r dr.
2
r2 + x2
p
The maximum r is given by t
r2 + x2 /c = 0;
p
rmax = c2 t2 x2 (since K(t) = 0 for t < 0).
(i)
µ0 K0 ẑ
A(x, t) =
2
E(x, t) =
Z
0
@A
=
@t
B(x, t) = r ⇥ A =
rm
p
r
µ0 K0 ẑ p 2
r + x2
dr =
2
r2 + x2
rm
0
=
µ0 K0 c
ẑ, for ct > x, and 0, for ct < x.
2
µ0 K0 ẑ ⇣p 2
rm
2
x2
⌘ µ K (ct
0 0
x =
2
x)
@Az
µ0 K0
ŷ =
ŷ, for ct > x, and 0, for ct < x.
@x
2
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ẑ.
253
(ii)
p
 Z rm
Z
Z
r2 + x2 /c
µ0 ↵ ẑ rm t
µ0 ↵ẑ
r
1 rm
p
p
A(x, t) =
r dr =
t
dr
r dr
2
2
c 0
r2 + x2
r2 + x2
0
0

µ0 ↵ ẑ
1 2 2
µ0 ↵ ẑ 2
µ0 ↵(x ct)2
=
t(ct x)
(c t
x2 ) =
(x
2ctx + c2 t2 ) =
ẑ.
2
2c
4c
4c
@A
µ0 ↵(x
=
@t
2
E(x, t) =
B(x, t) = r ⇥ A =
t
ct)
@Az
ŷ =
@x
ẑ, for ct > x, and 0, for ct < x.
µ0 ↵
(x
2c
ct) ŷ, for ct > x, and 0, for ct < x.

⌘
1 ⇣p 2
1 1
1
1
r
2
p
(b) Let u ⌘
r +x
x , so du =
2r dr = p
dr, and
2
2
2
c
c 2 r +x
c r + x2
p
Z
r2 + x2
x
µ0 c ẑ 1 ⇣
=t
u, and as r : 0 ! 1, u : 0 ! 1. Then A(x, t) =
K t
c
c
2
0
Z
@A
µ0 c ẑ 1
E(x, t) =
=
@t
2
0
Z 1
⇣
µ0 c
@
=
ẑ
K t
2
@u
0
µ0 c
=
K(t x/c) ẑ,
2
⌘
@ ⇣
x
K t
u du. But
@t
c
⌘
x
µ0 c h ⇣
u du =
ẑ K t
c
2
[if K( 1) = 0].
x
c
⌘
u du. qed
⌘
⇣
⌘
@ ⇣
x
@
x
K t
u =
K t
u .
@t
c
@u
c
⌘i
1
x
µ0 c
u
=
[K(t x/c) K( 1)] ẑ
c
2
0
Note that (i) and (ii) are consistent with this result. Meanwhile
B(x, t) =
=
=
S=
Z 1
⇣
⌘
⇣
⌘ 1 @
⇣
⌘
@Az
µ0 c
@
x
@
x
x
ŷ =
ŷ
K t
u du. But
K t
u =
K t
u .
@x
c
@x
c
@x
c
c @u
c
0
Z 1
⇣
⌘
h
⇣
⌘i
1
µ0
@
x
µ0
x
µ0
ŷ
K t
u du =
ŷ K t
u
=
[K(t x/c) K( 1)] ŷ
2
@u
c
2
c
2
0
0
µ0
K(t x/c) ŷ, [if K( 1) = 0].
2
1
1 ⇣ µ0 c ⌘ ⇣ µ0 ⌘
µ0 c
2
(E ⇥ B) =
K(t x/c) [ ẑ ⇥ ŷ] =
[K(t x/c)] x̂.
µ0
µ0
2
2
4
This is the power per unit area that reaches x at time t; it left the surface at time (t x/c). Moreover, an
µ0 c
2
equal amount of energy is radiated downward, so the total power leaving the surface at time t is
[K(t)] .
2
Problem 11.29
With ↵ = 90 , Eq. 7.68 gives E0 = cB, B0 =
1
0
E, qm
=
c
cqe . Use this to “translate” Eqs. 10.72, 10.73,
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254
and 11.70:
E =
0
B0 =
=
P =
✓
◆
1
r̂ ⇥ E = r̂ ⇥ ( cB0 ) = c( r̂ ⇥ B0 ).
c
c
r ⇥(c2 v2 )u + r ⇥ (u ⇥ a)⇤
1
1 qe
E=
c
c 4⇡✏0 ( r · u)3
0
r ⇥(c2 v2 )u + r ⇥ (u ⇥ a)⇤ = µ0 qm0
r ⇥(c2
1 ( qm
/c)
3
c 4⇡✏0 ( r · u)
4⇡ ( r · u)3
✓
◆
2
µ0 a2 2
µ0 a2
1 0
µ0 a2 0 2
qe =
qm
=
(q ) .
6⇡c
6⇡c
c
6⇡c3 m
v 2 )u +
r
⇤
⇥ (u ⇥ a) .
Or, dropping the primes,
B(r, t) =
E(r, t) =
=
P
Problem 11.30
Z
Z
(a) Wext = F dx = F
T
r ⇥(c2
µ0 qm
4⇡ ( r · u)3
v 2 )u +
c( r̂ ⇥ B).
Wext
=
2
F
m
"Z
T
v(t) dt. From Prob. 11.19, v(t) =
t dt + ⌧
0

1 2
T + ⌧T
2
Z
T
dt
⌧e
T /⌧
0
⌧ 2e
⇤
⇥ (u ⇥ a) .
2 2
µ0 qm
a
.
6⇡c3
0
F2
=
m
r
Z
T
t/⌧
e
0
T /⌧
⇣
eT /⌧
#
dt =
F h
t+⌧
m

F 2 t2
+ ⌧t
m 2
✓
⌘
F2 1 2
1 =
T + ⌧T
m 2
(b) From Prob. 11.19, the final velocity is vf = (F/m)T , so Wkin =
R
P dt. According to the Larmor formula, P =
a(t) =
8
⇥
< (F/m) 1
:
⇥
(F/m) 1
e
e(t
T /⌧
⇤
⇤
T )/⌧
i
. So
T
⌧e
T /⌧
⌧ et/⌧
0
⌧ 2 + ⌧ 2e
T /⌧
◆
.
1
1 F2
F 2T 2
mvf2 = m 2 T 2 =
.
2
2 m
2m
µ0 q 2 a2
, and (again from Prob. 11.19)
6⇡c
et/⌧ ,
T )/⌧
⌧ e(t
,
(t  0);
(0  t  T ).
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255
=
=
=
=
(
)
Z Th
⌘2 Z 0
i2
µ0 q 2 F 2 ⇣
T /⌧
2t/⌧
(t T )/⌧
1 e
e
dt +
1 e
dt
6⇡c m2
1
0
(
)
Z T
Z T
Z T
⌘2 ⇣ ⌧
⌘0
F2 ⇣
1 e T /⌧
e2t/⌧
+
dt 2e T /⌧
et/⌧ dt + e 2T /⌧
e2t/⌧ dt
⌧
m
2
0
0
0
1

⌘2
⇣
⌘T
⇣⌧
⌘T
⌧F2 ⌧ ⇣
1 e T /⌧ + T 2e T /⌧ ⌧ et/⌧
+ e 2T /⌧
e2t/⌧
m 2
2
0
0
⌘
⇣
⌘
⇣
⌘i
2 h ⇣
⌧F ⌧
⌧
1 2e T /⌧ + e 2T /⌧ + T 2⌧ e T /⌧ eT /⌧ 1 + e 2T /⌧ e2T /⌧ 1
m 2
2
2 h
⌧F ⌧
⌧
⌧
⌧ 2T /⌧ i
⌧F2 ⇣
⌧ e T /⌧ + e 2T /⌧ + T 2⌧ + 2⌧ e T /⌧ +
e
=
T ⌧ + ⌧e
m 2
2
2
2
m
T /⌧
⌘
.
Energy conservation requires that the work done by the external force equal the final kinetic energy plus
F 2T 2
⌧F2 ⇣
=
+
T
2m
m
Problem 11.31
k
(a) a = ⌧ ȧ +
(t) )
m
Z
✏
⌧ + ⌧e
a(t) dt = v(✏)
T /⌧
⌘
F2
=
m
v( ✏) = ⌧
✏
If the velocity is continuous, so v(✏) = v( ✏), then a(✏)
Z
✏
✏
✓
1 2
T + ⌧T
2
da
k
dt +
dt
m
a( ✏) =
Z
◆
= Wext . X
(t) dt = ⌧ [a(✏)
a( ✏)] +
⌧ +⌧ e
2
✏
k
, so the general solution is a(t) =
m⌧
⇢
T /⌧
✏
k
.
m
k
.
m⌧
When t < 0, a = ⌧ ȧ ) a(t) = Aet/⌧ ; when t > 0, a = ⌧ ȧ ) a(t) = Bet/⌧ ;
)B=A
2
a=B
A=
k
m⌧
Aet/⌧ ,
(t < 0);
[A (k/m⌧ )] et/⌧ , (t > 0).
To eliminate the runaway we’d need A = k/m⌧ ; to eliminate preacceleration we’d need A = 0. Obviously,
⇢
(k/m⌧ )et/⌧ , (t < 0);
you can’t do both. If you choose to eliminate the runaway, then a(t) =
0,
(t > 0).
Z t
k
k ⇣ t/⌧ ⌘ t
k
et/⌧ dt =
⌧e
= et/⌧ (for t < 0);
m⌧
m⌧
m
1
1
1
⇢
Z t
k
(k/m)et/⌧ , (t < 0);
for t > 0, v(t) = v(0) +
a(t) dt = v(0) = . So v(t) =
(k/m),
(t > 0).
m
0
⇢
Z t
k
0,
(t < 0);
For an uncharged particle we would have a(t) =
(t), v(t) =
a(t) dt =
(k/m),
(t > 0).
m
1
v(t) =
Z
t
a(t) dt =
The graphs:
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256
(b)
Wext =
Z
F dx =
Z
F v dt = k
Z
k2
.
m
(t)v(t) dt = kv(0) =
✓ ◆2
1
1
k
k2
mvf2 = m
=
.
2
2
m
2m
✓
◆2 Z
Z
Z
µ0 q 2
k
2
=
[a(t)] dt = ⌧ m
6⇡c
m⌧
Wkin =
Wext
Clearly, Wext = Wkin + Wrad . X
0
e2t/⌧ dt =
1
Problem 11.32
U0
Our task is to solve the equation a = ⌧ ȧ +
[ (x) + (x
m
(1) x continuous at x = 0 and x = L;
(2) v continuous at x = 0 and x = L;
(3) a = ±U0 /m⌧ v (plus at x = 0, minus at x = L).
k 2 ⇣ ⌧ 2t/⌧ ⌘
e
m⌧ 2
0
=
1
k2 ⌧
k2
=
.
m⌧ 2
2m
L)], subject to the boundary conditions
The third of these follows from integrating the equation of motion:
Z
Z
Z
dv
da
U0
dt = ⌧
dt +
[ (x) + (x L)] dt,
dt
dt
m
Z
U0
dt
v = ⌧ a+
[ (x) + (x L)]
dx = 0,
m
dx
Z
U0
1
U0
a==
[ (x) + (x L)] dx = ±
.
m⌧
v
m⌧ v
In each of the three regions the force is zero (it acts only at x = 0 and x = L), and the general solution is
a(t) = Aet/⌧ ;
v(t) = A⌧ et/⌧ + B;
x(t) = A⌧ 2 et/⌧ + Bt + C.
(I’ll put subscripts on the constants A, B, and C, to distinguish the three regions.)
Region iii (x > L): To avoid the runaway we pick A3 = 0; then a(t) = 0, v(t) = B3 , x(t) = B3 t + C3 . Let
the final velocity be vf (= B3 ), set the clock so that t = 0 when the particle is at x = 0, and let T be the time
it takes to traverse the barrier, so x(T ) = L = vf T + C3 , and hence C3 = L vf T . Then
a(t) = 0;
v(t) = vf ,
x(t) = L + vf (t
T ),
(t < T ).
Region ii (0 < x < L): a = A2 et/⌧ , v = A2 ⌧ et/⌧ + B2 , x = A2 ⌧ 2 et/⌧ + B2 t + C2 .
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257
U0
U0
) A2 =
e T /⌧ .
m⌧ vf
m⌧ vf
U0
U0
(2) ) vf = A2 ⌧ eT /⌧ + B2 =
+ B2 ) B2 = vf
.
mvf
mvf
U0 ⌧
U0 T
U0
(1) ) L = A2 ⌧ 2 eT /⌧ + B2 T + C2 =
+ vf T
+ C2 = vf T +
(⌧
mvf
mvf
mvf
U0
C2 = L vf T +
(T ⌧ ).
mvf
(3) ) 0
A2 eT /⌧ =
U0 (t T )/⌧
e
;
m⌧ vf
h
i
U0
v(t) = vf +
e(t T )/⌧ 1 ;
mvf
U0 h (t
x(t) = L + vf (t T ) +
⌧e
mvf
T ) + C2 )
a(t) =
(0 < t < T ).
T )/⌧
t+T
i
⌧ ;
[Note: if the barrier is sufficiently wide (or high) the particle may turn around before reaching L, but we’re
interested here in the régime where it does tunnel through.]
In particular, for t = 0 (when x = 0):
i
i
U0 h
U0 h
0 = L vf T +
⌧ e T /⌧ + T ⌧ ) L = vf T
⌧ e T /⌧ + T ⌧ . qed
mvf
mvf
Region i (x < 0): a = A1 et/⌧ , v = A1 ⌧ et/⌧ + B1 , x = A1 ⌧ 2 et/⌧ + B1 t + C1 . Let vi be the incident velocity
(at t ! 1); then B1 = vi . Condition (3) says
U0
e
m⌧ vf
A1 =
T /⌧
U0
,
m⌧ v0
where v0 is the speed of the particle as it passes x = 0. From the solution in region (ii) it follows that
⌘
U0 ⇣ T /⌧
v0 = vf +
e
1 . But we can also express it in terms of the solution in region (i): v0 = A1 ⌧ + vi .
mvf
Therefore
⌘
⌘
U0 ⇣ T /⌧
U0 ⇣ T /⌧
U0
U0
vi = vf +
e
1
A1 ⌧ = vf +
e
1 +
e T /⌧
mvf
mvf
mv0
mvf
(
)
✓
◆
U0
U0
U0
vf
U0
vf
⇥
⇤
= vf
+
= vf
1
= vf
1
mvf
mv0
mvf
v0
mvf
vf + (U0 /mvf ) e T /⌧ 1
(
)
U0
1
⇥
⇤ . qed
= vf
1
mvf
1 + (U0 /mvf2 ) e T /⌧ 1
If 12 mvf2 = 12 U0 , then
L = vf T
h
vf ⌧ e
vi = vf
T /⌧
+T

vf 1
i
h
⌧ = vf T
1
1+e
T /⌧
1
⌧e
T /⌧
⇣
= vf 1
i
⇣
T + ⌧ = ⌧ vf 1
1 + eT /⌧
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reproduced, in any form or by any means, without permission in writing from the publisher.
⌘
e
= vf eT /⌧ .
T /⌧
⌘
;
258
Putting these together,
L
=1
⌧ vf
e
T /⌧
)e
T /⌧
=1
L
) eT /⌧ =
⌧ vf
1
1
) vi =
(L/⌧ vf )
1
vf
. qed
(L/vf ⌧ )
✓ ◆2
1
2
vf
4
KEi
vi
16
2 mvi
In particular, for L = vf ⌧ /4, vi =
= vf , so
= 1
=
=
)
2
1 1/4
3
KEf
v
9
f
2 mvf
16
16 1
8
KEi =
KEf =
U0 = U0 .
9
9 2
9
Problem 11.33
r ⇥(c2 v2 )u + ( r · a)u ( r · u)a⇤. Here u = c r̂ v, r =
(q/2)
(a) From Eq. 10.72, E1 =
4⇡✏0 ( r · u)3
r · v = c r lv. We want only the x
l x̂ + d ŷ, v = v x̂, a = a x̂, so r · v = lv, r · a = la, r · u = c r
component. Noting that ux = (c/ r )l v = (cl v r )/ r , we have:

r
q
1
2
2
3
r (cl v r )(c v + la) a(c r lv)
8⇡✏0 (c r
lv)
⇥
⇤
q
1
=
(cl v r )(c2 v 2 ) + cl2 a v r la ac r 2 + alv r . But r
8⇡✏0 (c r
lv)3
⇥
⇤
q
1
=
(cl v r )(c2 v 2 ) acd2 .
3
8⇡✏0 (c r
lv)
2
⇥
⇤
q
1
=
(cl v r )(c2 v 2 ) acd2 x̂. (This generalizes Eq. 11.90.)
3
8⇡✏0 (c r
lv)
E1x =
Fself
Now x(t) x(tr ) = l = vT + 12 aT 2 + 16 ȧT 3 + · · · , where T = t
retarded time tr .
(cT )2 =
c2 T 2 (1
r
2
2
= l 2 + d2 .
tr , and v, a, and ȧ are all evaluated at the
1
1
1
1
= l2 + d2 = d2 + (vT + aT 2 + ȧT 3 )2 = d2 + v 2 T 2 + vaT 3 + v ȧT 4 + a2 T 4 ;
2
6
3
4
v 2 /c2 ) = c2 T 2 /
2
= d2 + vaT 3 +
✓
◆
1
1
v ȧ + a2 T 4 . Solve for T as a power series in d:
3
4
✓
◆ 4
3 3
d
c2 2 d2
d
v ȧ a2
2
2
2 2
2
T =
1 + Ad + Bd + · · · ) 2 2 1 + 2Ad + 2Bd + A d = d +va 3 (1+3Ad)+
+
d4 .
c
c
c
3
4 c4
Comparing like powers of d: A =
3
1
3va
va 3 ; 2B + A2 =
2 c
c3
3
A+
✓
v ȧ a2
+
3
4
◆
4
c4
.
✓
◆
3
6 2
3va 3 1
1 2 2 6
v ȧ 4
a2 4
v ȧ 4
a
1
v2
3 v 2 a2 6
2B =
va
v
a
+
+
=
+
+
2
c3 2 c3
4
c6
3 c4
4c4
3 c4
4c4
c2
2 c6

✓
◆

✓
◆
4
2 2
2
2
2
4
2 2
2
v ȧ a
v
v
v
v ȧ
a
v
= 4
+
1
+6 2
)B= 4
+
1+4 2
.
2
2
c
3
4
c
c
c
2c
3
4
c
⇢

✓
◆
4
2 2
d
va 3
v ȧ
a
v2
T =
1+
d
+
+
1
+
4
d2 + ( ) d4 + · · · (generalizing Eq. 11.93).
c
2 c3
2c4 3
4
c2
protected under all copyright laws as they currently exist. No portion of this material may be
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259
1
1
l = vT + aT 2 + ȧT 3 + · · ·
6
⇢2

✓
◆

4
2 2
3
v d
va 3
v ȧ
a
v2
1 2 d2
1 3
2
=
1+
d
+
+
1
+
4
d
+
a
1 + va 3 d + ȧ 3 d3
3
4
2
2
c
2 c
2c
3
4
c
2 c
c
6 c
✓
◆
⇢

✓
◆
⇣v ⌘
4
2
2
4
2 2
2
2
3
a
v
v
v
v ȧ
a
v
1
1 3
=
d+
1
+ 2 d2 +
+
1+4 2
+ a 2 va 3 + ȧ 3
2
2
4
c
2c
c
c
2c c
3
4
c
2 c
c
6 c
✓ 4◆
 ✓
◆
✓
◆
⇣v ⌘
3
2
4 2
2
2
a
ȧ
v
v a
1 v
v
=
d+
d2 + 3
1+ 2 2 +
+ 2 +1
d3
c
2c2
2c 3
c
c2
4
c
c2
✓ 4◆

⇣v ⌘
5
ȧ 5 v 2 a2 3
a
2
=
d+
d
+
+
d + ( ) d4 + · · ·
c
2c2
2c3 3 4 c2
d3
⇢

✓
◆
4
va 3
v ȧ
1 v2
2 2
= cT = d 1 +
d
+
+
a
+
d2 + ( ) d4 + · · ·
2 c3
2c4 3
4
c2

✓
◆

5
5
va 4 2
v ȧ
1 v2
v2
av 4 2
v ȧ 5 v 2 a2 3
2 2
3
lv = c d +
d
+
+
a
+
d
d
d
+
d + ···
2c2
2c3 3
4
c2
c
2c2
2c3 3 4 c2
✓
◆

✓
◆
5
v2
v ȧ
1 v2
v ȧ 5 v 2 2 a2 3
2 2
=c d 1
+
+
a
+
d + ···
2
3
2
c
2c
3
4
c
3
4 c2
5 2
c
a 3
= d+
d + ( ) d4 + · · ·
8c3
r
cr
cl
vr =
=
=
=
(c r
lv)
3
=
✓
◆

✓
◆
5
a 4 2
ȧ 5 v 2 a2
v 2 a 4 2 v 5 v ȧ
1 v2
3
2 2
v d+
d + 2
+
d
v d
d
+ a
+ 2
d3
2c
2c
3 4 c2
2 c3
2c4 3
4
c
✓
◆

✓
◆
5
a 4
v2
ȧ 5 v 2 a2
v 2 ȧ v 2 a2 1 v 2
2
1
d + 2
+
+ 2
d3 + ( ) d4 + · · ·
2c
c2
2c 3 4 c2
c2 3
c2
4
c
✓ 2◆

✓
◆
5
a
ȧ
v 2 a2 5 1 v 2
d2 + 2
+
d3 + ( ) d4 + · · ·
2c
2c 3 2
c2
4 4
c2
✓ 2◆
✓
◆
3
a
ȧ v 2 a2
d2 + 2
+
d3 + ( ) d4 + · · ·
2c
2c
3
c2
 ✓
◆ 3 ⇣ ⌘ ✓
◆
6 2
6 2
3
cd
a 2
a 2
1+
d
=
1 3 4 d + ···
8c4
cd
8c
✓
◆ ⇢✓ 2 ◆
✓
◆
6 2
3
q 2 ⇣ ⌘3
a
a
ȧ v 2 a2
c2
1 3 4 d2
d2 + 2
+
d3 2 acd2 x̂
2
8⇡✏0 cd
8c
2c
2c
3
c
✓
◆
✓
◆
2
3
6 2
2 2
q
3 a 2
ac
ȧ v a
=
1
d
+
+
d x̂
8⇡✏0 c3 d
8 c4
2
2 3
c2

✓
◆
3
q2
1
ȧ v 2 a2
=
ac
+
+
d + ( ) d2 + · · · x̂
8⇡✏0 c3 d 2
3
c2

✓
◆
4
q2
ȧ v 2 a2
3 a
=
+
+
+ ( ) d + · · · x̂ (generalizing Eq. 11.95).
4⇡✏0
4c2 d 4c3 3
c2
Fself =
Switching to t: v(tr ) = v(t) + v̇(t)(tr t) + · · · = v(t) a(t)T = v(t)
doesn’t matter—to this order—whether we evaluate at t or at tr .)
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reproduced, in any form or by any means, without permission in writing from the publisher.
a d/c. (When multiplied by d, it
260


✓
◆
2
v(tr )
[v(t)2 2va d/c]
v(t)2
2av 3 d
1
=1
= 1
1+
, so
c
c2
c2
c3
"
#
✓
◆2 1/2
✓
◆
v(tr )
va 3
ȧ
= 1
= (t) 1
d ; a(tr ) = a(t) T ȧ = a(t)
d.
c
c3
c
Evaluating everything now at time t:
"
#
✓
◆
4
3va 3 d/c3 (a ȧ d/c)
q2
ȧ v 2 a2
3 1
2
Fself =
+ 3
+
+ ( ) d + · · · x̂
4⇡✏0
4c2 d
4c
3
c2

✓
◆
✓
◆
3
3
4
q2
a
ȧ
va2 2
ȧ v 2 a2
=
+
+3 3
+ 3
+
+ ( ) d + · · · x̂
4⇡✏0
4c2 d 4c2
c
c
4c
3
c2

✓
◆
3
4
q2
a
ȧ
va2 2
v 2 a2
=
+
ȧ
+
+
3
+
+ ( ) d + · · · x̂
4⇡✏0
4c2 d 4c3
3
c2
c2

✓
◆
3
4
q2
a
va2 2
=
+
ȧ
+
3
+ ( ) d + · · · x̂ (generalizing Eq. 11.96).
4⇡✏0
4c2 d 3c3
c2
The first term
mass; the radiation reaction itself is the second term:
✓ is the electromagnetic
◆
2
2 2
µ
q
va
0
int
4
=
ȧ + 3 2
(generalizing Eq. 11.99), so the generalization of Eq. 11.100 is
12⇡c
c
µ0 q 2
6⇡c
4
Z
✓
ȧ +
t2
3
2 2
a v
c2
t1
◆
✓
ȧ + 3
va2
c2
, where A ⌘
µ0 q 2
. P = Aa2
6⇡c
Z
or
t2
P dt,
t1
Z
t2
t1
(except for boundary terms—see Sect. 11.2.2).
Z t2
Z t2
4
Rewrite the first term:
ȧv dt =
(
t1
4
t1
4
v)
4
✓
ȧv + 3
da
dt =
dt !
4
va
6
2
◆
.
(Eq. 11.75). What we must show is that
v 2 a2
c2
t2
t1
2
Z
◆
t2
dt =
d
(
dt
4
v) = 4
Problem 11.34
3d
dt
v+
4
a;
t2
a2
6
dt
t1
d
(
dt
4
v)a dt.
✓
◆
1
2va
va 3
= 2 . So
2
2
2
3/2
c
c
v /c )
t1
d
d
1
1
p
=
=
dt
dt
2 (1
1 v 2 /c2
✓
◆
✓
◆
3
2
2
d 4
va
v
v
v2
6
( v) = 4 3 v 2 + 4 a = 6 a 1
+
4
=
a
1
+
3
.
dt
c
c2
c2
c2
✓
◆
Z t2
Z
t2
t2
v2
4
6 2
ȧv dt = 4 va
a 1 + 3 2 dt, and hence
c
t1
◆ t1
✓
◆
Zt1t2 ✓
Z t2 
t2
3 2 a2 v 2
v2
a2 v 2
4
4
6 2
ȧv +
dt
=
va
+
a
1
+
3
+3 6 2
2
2
c
c
c
t1
t1
t1
Now
Z
dt =
4
va
t2
t1
Z
t2
6 2
a dt.
qed
t1
p
µ0 q 2 a2 6
c2 t
(Eq. 11.75). w = b2 + c2 t2 (Eq. 10.52); v = ẇ = p
;
6⇡c
b2 + c2 t2
2
2 2
2
2
2
c
c t(c t)
c
b c
a = v̇ = p
= 2
b2 + c2 t2 c2 t2 = 2
;
2
2
2
3/2
2
2
3/2
2
2
2
(b + c t )
(b + c t )
(b + c2 t2 )3/2
b +c t
(a) P =
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261
2
=
1
=
v 2 /c2
1
1
1
[c2 t2 /(b2
+
c2 t2 )]
=
b2
b2 + c2 t2
1
= 2 b2 + c2 t2 . So
+ c2 t2 c2 t2
b
µ0 q
b c
(b + c t )
q2 c
=
.
Yes, it radiates (in fact, at a constant rate).
6⇡c (b2 + c2 t2 )3
b6
6⇡✏0 b2
✓
◆
✓
◆
µ0 q 2 4
3 2 a2 v
3 b2 c2 (2c2 t)
3b2 c4 t
3 2 a2 v
ȧ +
;
ȧ
=
=
;
ȧ
+
=
6⇡c
c2
2 (b2 + c2 t2 )5/2
c2
(b2 + c2 t2 )5/2
3b2 c4 t
3 (b2 + c2 t2 )
b4 c4
c2 t
p
+
No, the radiation reaction is zero.
c2
b2
(b2 + c2 t2 )3 b2 + c2 t2
(b2 + c2 t2 )5/2
P =
2
4 4
2
2 2 3
Problem 11.35
The fields of a nonstatic ideal dipole (Problem 10.34) contain terms that go like 1/r, 1/r2 , and 1/r3 , for
fixed t0 ⌘ t r/c; radiation comes from the first of these:
Er =
µ0
[p̈
4⇡r
r̂(r̂ · p̈)],
Br =
µ0
(r̂ ⇥ p̈)
4⇡rc
where the dipole moments are evaluated at time t0 . The 1/r2 term in the Poynting vector is
1
µ0
µ0
(Er ⇥ Br ) =
[p̈ r̂(r̂ · p̈)] ⇥ (r̂ ⇥ p̈) =
{p̈ ⇥ (r̂ ⇥ p̈) (r̂ · p̈)[r̂ ⇥ (r̂ ⇥ p̈)]}
2
2
µ0
16⇡ cr
16⇡ 2 cr2
⇥ 2
⇤
µ0
µ0
=
r̂ p̈2 p̈(r̂ · p̈) (r̂ · p̈)[r̂(r̂ · p̈) p̈] =
p̈
(r̂ · p̈)2 r̂.
16⇡ 2 cr2
16⇡ 2 cr2
⇥
⇤
Setting the z axis along p̈, p̈2 (r̂ · p̈)2 = p̈2 sin2 ✓, and the power radiated is
Sr =
P =
I
Sr · da =
µ0 2
p̈
16⇡ 2 c
I
1
µ0 2
sin2 ✓r2 sin ✓ d✓ d =
p̈
2
r
8⇡c
Z
⇡
0
sin3 ✓ d✓ =
µ0 2
p̈ .
6⇡c
For the sinusoidal case, p = p0 cos !t, we have p̈ = ! 2 p0 cos !t, p̈2 = ! 4 p20 cos2 !t, and its time average
is (1/2)p20 ! 4 , so
µ0 2 4
hP i =
p̈ ! ,
12⇡c 0
in agreement with Eq. 11.22. And in the case of quadratic time dependence our result agrees with the answer
to Problem 11.26.
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262
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Chapter 12
Electrodynamics and Relativity
Problem 12.1
Let u be the velocity of a particle in S, ū its velocity in S̄, and v the velocity of S̄ with respect to S.
Galileo’s velocity addition rule says that u = ū + v. For a free particle, u is constant (that’s Newton’s first
law in S.
(a) If v is constant, then ū = u v is also constant, so Newton’s first law holds in S̄, and hence S̄ is inertial.
(b) If S̄ is inertial, then ū is also constant, so v = u ū is constant.
Problem 12.2
(a) mA uA + mB uB = mC uC + mD uD ;
ui = ūi + v.
mA (ūA + v) + mB (ūB + v) = mC (ūC + v) + mD (ūD + v),
mA ūA + mB ūB + (mA + mB )v = mC ūC + mD ūD + (mC + mD )v.
Assuming mass is conserved, (mA + mB ) = (mC + mD ), it follows that
mA ūA + mB ūB = mC ūC + mD ūD , so momentum is conserved in S̄.
(b) 12 mA u2A + 12 mB u2B = 12 mC u2C + 12 mD u2D )
1
1
1
1
2
2
2
2
2
2
2
2 mA (ūA + 2ūA · v + v ) + 2 mB (ūB + 2ūB · v + v ) = 2 mC (ūC + 2ūC · v + v ) + 2 mD (ūD
1
1
1 2
2
2
2 mA ūA + 2 mB ūB + v · (mA ūA + mB ūB ) + 2 v (mA + mB )
1
1
2
2
= 2 mC ūC + 2 mD ūD + v · (mC ūC + mD ūD ) + 12 v 2 (mC + mD ).
+ 2ūD · v + v 2 )
But the middle terms are equal by conservation of momentum, and the last terms are equal by conservation
of mass, so 12 mA ū2A + 12 mB ū2B = 12 mC ū2C + 12 mD ū2D . qed
Problem 12.3
vAB +vBC
vAB vBC
(a) vG = vAB + vBC ; vE = 1+v
; ) vGvGvE =
2 ⇡ vG 1
c2
AB vBC /c
8
In mi/h, c = (186, 000 mi/s) ⇥ (3600 sec/hr) = 6.7 ⇥ 10 mi/hr.
)
(b)
vG vE
vG
1
2c
=
(5)(60)
(6.7⇥108 )2
+ 34 c / 1 +
1
2
·
3
4
= 6.7 ⇥ 10
=
5
4c
(c) To simplify notation, let
2
=
2
1
+2 1
(1 + 2
+2 1 2+
2 2
1 2 + 1 2)
2
/
16
. ) 6.7 ⇥ 10
11
8
= vAC /c,
2
2
=
1+2
(1 + 2
=
1
% error (pretty small!)
10
c (still less than c)
11
= vAB /c,
+
2+
1 2
1
14
vAB vBC
.
c2
2
1
2
1
2
2
2
2)
2
= vBC /c. Then Eq. 12.3 says:
(1 + 12
(1 + 2
2
2
1 2
2
1
+
2
2)
2 2
1 2)
=1
(1
=
1+ 2
1+
2
1 )(1
(1 +
1
2
2
1 2)
, or:
2
2)
=1
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,
263
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
where
⌘ (1
2
1 )(1
2
2 )/(1
+
2
1 2)
is clearly a positive number. So
2
Problem 12.4
< 1, and hence |vAC | < c. qed.
(a) Velocity of bullet relative to ground is 12 c + 13 c = 56 c = 10
12 c.
9
Velocity of getaway car is 34 c = 12
c. Since vb > vg , bullet does reach target .
(b) Velocity of bullet relative to ground is
Velocity of getaway car is 34 c =
21
28 c.
1
1
2 c+ 3 c
1+ 12 · 13
=
5
6c
7
6
= 57 c =
20
28 c.
Since vg > vb , bullet does not reach target .
Problem 12.5
(a) Light from 90th clock took
90⇥109 m
3⇥108 m
= 300 sec = 5 min to reach me, so the time I see on the clock is
11:55 am .
(b) I observe 12 noon .
Problem 12.6
(
light signal leaves a at time t0a ; arrives at earth at time ta = t0a + dca
light signal leaves b at time t0b ; arrives at earth at time tb = t0b + dcb
h
( v t0 cos ✓)
= t0 1
c
c
(Here da is the distance from a to earth, and db is the distance from b to earth.)
)
t = tb
s = v t0 sin ✓ =
⇥
v (1
du
=
d✓
✓max
v
c
ta = t0b
t0a +
v sin ✓ t
(1
(1t
(db
da )
v
cos ✓)
c
=
t0 +
) u=
(1
i
v
cos ✓
c
v sin ✓
is the the apparent velocity.
v
c cos ✓)
⇤
cos ✓)(cos ✓) sin ✓( vc sin ✓)
= 0 ) (1
(1 vc cos ✓)2
v
v
cos ✓) cos ✓ = sin2 ✓
c
c
v
v
) cos ✓ = (sin2 ✓ + cos2 ✓) =
c
c
p 2 2
v 1 v /c
= cos 1 (v/c) At this maximal angle, u = 1 v2 /c2 = p v 2 2
1 v /c
As v ! c, u ! 1 , because the denominator ! 0 — even though v < c.
Problem 12.7
The student has not taken into account time dilation of the muon’s “internal clock”. In the laboratory, the
muon lasts ⌧ = p ⌧ 2 2 , where ⌧ is the “proper” lifetime, 2 ⇥ 10 6 sec. Thus
1 v /c
v=
d
p
t/ 1
⇣ ⌧ ⌘2
v2 = 1
d
v2
;
c2
v2
1
=
2;
c2
1 + (⌧ c/d)
r
v2
, where d = 800 meters.
c2
⇣⇣ ⌧ ⌘2
1⌘
1
v2
+ 2 = 1; v 2 =
.
2
d
c
(⌧ /d) + (1/c)2
d
=
2
2
⌧
v /c
1
⌧c
(2 ⇥ 10 6 )(3 ⇥ 108 )
6
3
=
= = ;
d
800
8
4
v2
1
16
=
=
;
c2
1 + 9/16
25
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v=
4
c.
5
264
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.8
(a) Rocket clock runs slow; so earth clock reads t = p
1
1 v 2 /c2
· 1 hr. Here
=p
=p
1
1 v 2 /c2
1
1 9/25
= 54 .
) According to earth clocks signal was sent 1 hr, 15 min after take-o↵.
(b) By earth observer, rocket is now a distance 35 c 54 (1 hr) = 34 c hr (three-quarters of a light hour) away.
Light signal will therefore take 34 hr to return to earth. Since it left 1 hr and 15 min after departure, light
signal reaches earth 2 hrs after takeo↵
(c) Earth clocks run slow: trocket =
Problem 12.9
Lc = 2Lv ;
Lc
c
=
Lv
v
; so
2
c
=
1
v
· (2 hrs) =
q
= 1
1 2
2
5
4
· (2 hrs) = 2.5 hrs
=
q
3 1
4 ; c2
v2
c2
=1
=
v2
c2
=
3
v2
16 . c2
=1
3
16
=
13
16 ;
v=
p
13
c
4
Problem 12.10
¯ horizontal projection
Say length of mast (at rest) is ¯l. To an observer on the boat, height of mast is ¯l sin ✓,
¯ To observer on dock, the former is una↵ected, but the latter is Lorentz contracted to 1 ¯l cos ✓.
¯
is ¯l cos ✓.
Therefore:
¯l sin ✓¯
tan ✓¯
¯
tan ✓ = 1 ¯
= tan ✓,
or
tan ✓ = p
¯
l cos ✓
1 v 2 /c2
Problem 12.11
p
Naively, circumference/diameter = 1 (2⇡R)/(2R) = ⇡/ = ⇡ 1 (!R/c)2 — but this is nonsense. Point
is: an accelerating object cannot remain rigid, in relativity. To decide what actually happens here, you need a
specific model for the internal forces holding the disc together.
Problem 12.12
(iv) ) t = t̄ + vx
c2 . Put this into (i), and solve for x:
x̄ = x
x
Similarly, (i) ) x =
v
x̄
t̄ = t
⇣ t̄
+
⇣
vx ⌘
=
x
1
c2
v2 ⌘
c2
v t̄ = x
1
v t̄ =
2
x
v t̄; x = (x̄ + v t̄) X
+ vt. Put this into (iv) and solve for t:
v ⇣ x̄
c2
⌘
⇣
+ vt = t 1
v2 ⌘
c2
v t
x̄
c2
v
x̄; t =
c2
⇣
t̄ +
v ⌘
x̄ X
c2
Problem 12.13
Let brother’s accident occur at origin, time zero, in both frames. In system S (Sophie’s), the coordinates
v
of Sophie’s cry are x = 5 ⇥ 105 m, t = 0. In system S̄ (scientist’s), t̄ = (t
vx/c2 . Since
c2 x) =
this is negative, Sophie’s cry occurred before the accident, in S̄.
= p 1
= p16913 144 = 13
5 . So
2
1 (12/13)
t̄ =
13
5
12
13 c
(5 ⇥ 10 )/c =
5
2
12 ⇥ 10 /3 ⇥ 10 = 10
5
8
3
. 4 ⇥ 10
3
seconds earlier
Problem 12.14
(a) In S it moves a distance dy in time dt. In S̄, meanwhile, it moves a distance dȳ = dy in time dt̄ =
(dt cv2 dx).
)
dȳ
dy
=
=
dt̄
(dt cv2 dx)
(dy/dt)
; or ūy =
1 cv2 dx
dt
uy
1
vux
c2
; ūz =
uz
1
vux
c2
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265
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
(b) S = dock frame; S 0 = boat frame; we need reverse transformations (v ! v):
ūy / 1 + vcū2x
uy
1
ūy
¯ ūy = c sin ✓,
¯ so
tan ✓ =
=
. In this case ūx = c cos ✓;
v ūx =
ux
(ūx + v)
(ūx + v)/ 1 + c2
✓
◆
1
sin ✓¯
1
c sin ✓¯
tan ✓ =
=
¯
( c cos ✓+v)
cos ✓¯ v/c
¯
sin ✓
[Contrast tan ✓ = cos
in Prob. 12.10. The point is that velocities are sensitive not only to the transfor✓¯
mation of distances, but also of times. That’s why there is no universal rule for translating angles — you have
to know whether it’s an angle made by a velocity vector or a position vector.]
That’s how the velocity vector of an individual photon transforms. But the beam as a whole is a snapshot
of many di↵erent photons at one instant of time, and it transforms the same way the mast does.
Problem 12.15
3
1
c 1c
c
5
2
c. Outlaws relative to police: 4 3 2 1 = 45 = c.
7
5
1 4·2
8
5
3
1
c
c
c
1
28
Bullet relative to outlaws: 7 5 4 3 = 13
=
c . [Velocity of A relative to B is minus the velocity of
13
1 7·4
28
B relative to A, so all entries below the diagonal are trivial. Note that in every case vbullet < voutlaws , so no
matter how you look at it, the bad guys get away.]
Bullet relative to ground:
speed of !
Ground
Police
Outlaws
0
1
2c
3
4c
2
5c
Ground
relative
to #
Police
Outlaws
Bullet
0
1
2
3
4c
5
7c
Bullet
5
7c
1
3c
1
13 c
0
2
5c
1
3c
Do they escape?
0
1
13 c
Yes
Yes
Yes
Yes
Problem 12.16
(a) Moving clock runs slow, by a factor
5
3
= p
1
1 (4/5)2
=
5
3.
Since 18 years elapsed on the moving clock,
⇥ 18 = 30 years elapsed on the stationary clock. 51 years old
(b) By earth clock, it took 15 years to get there, at 45 c, so d = 45 c ⇥ 15 years = 12c years (12 light years)
(c) t = 15 yrs, x = 12c yrs
(d) t̄ = 9 yrs, x̄ = 0. [She got on at the origin in S̄, and rode along on S̄, so she’s still at the origin. If you
doubt these values, use the Lorentz Transformations, with x and t in (c).]
⇢
(e) Lorentz Transformations: x̃ = (x + vt)
[note that v is negative, since S̃ us going to the left]
t̃ = (t + cv2 x)
) x̃ = 53 (12c yrs + 45 c · 15 yrs) = 53 · 24c yrs = 40c years.
t̃ = 53 (15 yrs +
4 c
5 c2
· 12c yrs) =
5
3
15 +
48
5
yrs = (25 + 16)yrs = 41 years.
(f) Set her clock ahead 32 years, from 9 to 41 (t̄ ! t̃). Return trip takes 9 years (moving time), so her clock
will now read 50 years at her arrival. Note that this is 53 · 30 years—precisely what she would calculate if the
stay-at-home had been the traveler, for 30 years of his own time.
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266
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
(g) (i) t̄ = 9 yrs, x = 0. What is t? t =
v
c2 x
+
=
t̄
3
5
· 9 yrs =
= 5.4 years, and he started at age 21, so he’s
27
5
26.4 years old (Younger than traveler (!) because to the traveller it’s the stay-at-home who’s moving.)
(ii) t̃ = 41 yrs, x = 0. What is t? t =
=
t̃
3
5
45.6 years old.
· 41 yrs, or
yrs, or 24.6 yrs, and he started at 21, so he’s
123
5
(h) It will take another 5.4 years of earth time for the return, so when she gets back, she will say her twin’s
age is 45.6 + 5.4 = 51 years—which is what we found in (a). But note that to make it work from traveler’s
point of view you must take into account the jump in perceived age of the stay-at-home when she changes
coordinates from S̄ to S̃.)
Problem 12.17
ā0 b̄0 + ā1 b̄1 + ā2 b̄2 + ā3 b̄3 =
=
2
=
2 0 0
=
Problem 12.18
0 1 0
ct̄
1
B x̄ C B
C B
(a) B
@ ȳ A = @ 0
z̄
0
0
0
1
0
0
0
1
0
0
B 0
(b) ⇤ = B
@
0
0
0
1
0
2
(a0
(a0 b0
a1 )(b0
a1 b0 +
a0 b1
a b (1
2
b1 ) +
)+
2
2 1 1
a b
a b (1
2 1 1
2
(a1
a0 )(b1
a0 ) + a2 b2 + a3 b3
a1 b1 + a1 b0 + a0 b1
a b ) + a2 b2 + a3 b3
2 0 0
) + a2 b2 + a3 b3
a0 b0 + a1 b1 + a2 b2 + a3 b3 . qed [Note:
2
(1
2
) = 1.]
10 1
0
ct
BxC
0C
C B C (using the notation of Eq. 12.24, for best comparison)
0A @ y A
1
z
0
0
1
0
0C
C
0A
1
0
¯
B 0
(c) Multiply the matrices: ⇤ = B
@ ¯¯
0
0
1
0
0
¯¯
0
¯
0
10
0
B
0C
CB
0A @ 0
1
0
0
0
0
0
1
0
1
0
0C
C=
0A
1
0
¯
B
B
@ ¯ ¯
0
¯
¯ ¯
0
¯¯
0
¯
0
1
0
0C
C
0A
1
Yes, the order does matter. In the other order “bars” and “no-bars” would be switched, and this would yield
a di↵erent matrix.
Problem 12.19
2
sinh ✓
(a) Since tanh ✓ = cosh
sinh2 ✓ = 1, we have:
✓ , and cosh ✓
1
=p
1
0
v 2 /c2
=p
1
cosh ✓
sinh ✓
B sinh ✓ cosh ✓
) ⇤=B
@ 0
0
0
0
0
0
1
0
1
2
tanh ✓
cosh ✓
=p
cosh2 ✓
1
0
0
cos
C
0C
@ sin
Compare:
R
=
0A
0
1
sinh2 ✓
sin
cos
0
= cosh ✓ ;
= cosh ✓ tanh ✓ = sinh ✓.
1
0
0A
1
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267
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
u
v
u v
ū
tanh
tanh ✓
c
c
=
) tanh ¯ =
, where tanh = u/c, tanh ✓ = v/c;
uv )
u
v
1 c2
c
1
tanh
tanh ✓
1
c
c
tanh ¯ = ū/c. But a “trig” formula for hyperbolic functions (CRC Handbook, 18th Ed., p. 204) says:
(b) ū =
tanh
tanh ✓
= tanh(
1 tanh tanh ✓
Problem 12.20
(a) (i) I = c2 t2 + x2 + y 2 + z 2 =
(ii) No. (In such a system
y
(iii) Yes.
✻
8
) tanh ¯ = tanh(
✓).
(5
15)2 + (10
B
1
100 + 25 + 25 =
50
5
)v=
5x̂ 5ŷ
=
10/c
c
x̂
2
c
ŷ
2
!
❂
A!
2
(3
0)2 =
✓
S̄ travels in the direction from B toward A,
making the trip in time 10/c.
\$
"#
4
(b) (i) I =
3)2 + (0
t̄ = 0, so I would have to be positive, which it isn’t.)
6
2
1)2 + (5
5)2 + (8
✓), or: ¯ =
"#
5
4
6
2)2 + 0 + 0 =
\$
2
Note that vc2 =
less than c.
✲x
8
10
4+9= 5
1
4
+
1
4
= 12 , so v =
p1 c,
2
safely
(ii) Yes. By Lorentz Transformation: (ct̄) =
(ct)
( x) . We want t̄ = 0, so (ct) = ( x); or
✒
v
(ct)
(3 1)ct 2
2
=
=
=
✻ . So v = c in the +x direction.
c
( x)
(5 2)
3
3
■
(iii) No. (In such a system x = y = z = 0 so I would be negative, which it isn’t.)
Problem 12.21
Using Eq. 12.18 (iv): t̄ = ( t
world line
of player 1
#
✒
✮
✛
■
10 ft
✲
v
c2
x) = 0 )
t=
v
c2
x, or v =
t 2
xc
=
tB
xB
tA 2
c
xA
world line of
player 2
✲x
world line of
the ball
✠
ct
✻
✒
✐
✶
A
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reproduced, in any form or by any means, without permission in writing from the publisher.
B
✲x
B
\$
8
"#
6
4
2
2
!
❂
A!
"#
5
268
6
4
5
\$
8
✲x
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
10
Problem 12.22
(a)
✒
ct
✻
■
world line
of player 1
Truth is, you never do communicate with
the other person right now —you communicate
with the person he/she will be when the message gets there; and the response comes back
to an older and wiser you.
#
✛
■
world line of
player 2
✒
✮
10 ft
✲
✲x
world line of
the ball
(b) No way It is true that a moving observer
might say she arrived at B before she left A,
but for the round trip everyone must agree
that she arrives back after she set out.
✠
ct
✻
✒
✐
✶
A
B
✲x
c
⃝2005
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reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 12.24
(a) 1
u2
c2
⌘ 2 = u2 ; u2 1 +
⌘2
c2
1
= ⌘2 ; u = p
⌘.
1 + ⌘ 2 /c2
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269
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
(b) p
=p
1
1 u2 /c2
=p
1
1 tanh2 ✓
cosh ✓
cosh2 ✓ sinh2 ✓
Problem 12.25
(a) ux = uy = u cos 45 =
(b) p
=p
1
1 u2 /c2
(c) ⌘0 = c =
(f) p
1
1 ū2 /c2
1 4/5
=
p
p 5
5 4
r
=
=
p
= cosh ✓ c tanh ✓ = c sinh ✓.
1
u
1 u2 /c2
2
c.
5
5. ) ⌘ = p
u
1 u2 /c2
) ⌘x = ⌘y =
p
2
p
(d) Eq. 12.45)
(e) ⌘¯x = (⌘x
1
p1 p2 c
2 5
)⌘= p
= cosh ✓
5 c.
8
>
>
< ūx =
>
>
: ūy =
ux v
1 ucx2v
1
⇣
=
uy
1
ux v
c2
p2
5
⌘
c
1
=
p2
5
2
5
q
1
c
= 0.
p2
2
5 1
5
c
2
5
r
p
2
2/5
p
=
c=
c.
3/5
3
q
q p
p
p
2
1 25 ( 2 c
⌘¯y = ⌘y = 2 c.
5 5 c) = 0.
p
⇢
p
p
⌘
¯
=
0X
x
1
p3 ūx = p
=p
= 3; ) ⌘
¯ = 3 ū )
1 2/3
⌘¯y = 3 ūy = 2 c X
⌘0 ) =
Problem 12.26
⌘ µ ⌘µ = (⌘ 0 )2 + ⌘ 2 =
1
(1 u2 /c2 ) (
c2 + u2 ) =
Problem 12.27
Use the result of Problem 12.24(a): u = p
c2 (1
(1
u2 /c2 )
u2 /c2 )
1
=
⌘. Here
c2 .
Timelike.
⌘
4
1
3
= , so p
= , and hence
c
3
5
1 + 16/9
1+
3
8
8
u = (4 ⇥ 10 ) = 2.4 ⇥ 10 m/s.
Innocent.
5
Problem 12.28
p
p
R
R
(a) From Prob. 11.34 we have = 1b b2 + c2 t2 . ) ⌧ = 1 dt = b pb2dt
= cb ln(ct + b2 + c2 t2 ) + k; at
+c2 t2

p
b
1
b
b
t = 0 we want ⌧ = 0: 0 = c ln b + k, so k = c ln b; ⌧ = ln (ct + b2 + c2 t2 )
c
b
p
p
(b) x2 b2 + x = bec⌧ /b ; x2 b2 = bec⌧ /b x; x2 b2 = b2 e2c⌧ /b 2xbec⌧ /b + x2 ; 2xbec⌧ /b = b2 (1 + e2c⌧ /b );
p
c⌧ /b
c⌧ /b
x = b e +e
= b cosh(c⌧ /b) . Also from Prob. 11.34: v = c2 t/ b2 + c2 t2 .
2
p
q
⇣ c⌧ ⌘
p
2
c
2 cosh2 (c⌧ /b)
2 = c cosh (c⌧ /b) 1 = c sinh(c⌧ /b) = c tanh
v = xc x2 b2 = b cosh(c⌧
b
b
.
/b)
cosh(c⌧ /b)
cosh(c⌧ /b)
b
⇣
⌘
c⌧
c⌧
(c) ⌘ µ = (c, v, 0, 0). = xb = cosh c⌧b , so ⌘ µ = cosh c⌧b c, c tanh c⌧b , 0, 0 = c cosh , sinh , 0, 0 .
b
b
Problem 12.29
(a) mA uA + mB uB = mC uC + mD uD ;
⌘ 2 /c2
ūi + v
.
1 + (ūi v/c2 )
ūA + v
ūB + v
ūC + v
ūD + v
mA
+ mB
= mC
+ mD
.
2
2
2
1 + (ūA v/c )
1 + (ūB v/c )
1 + (ūC v/c )
1 + (ūD v/c2 )
This time, because the denominators are all di↵erent, we cannot conclude that
mA ūA + mB ūB = mC ūC + mD ūD .
As an explicit counterexample, suppose all the masses are equal, and uA = uB = v, uC = uD = 0. This is
a symmetric “completely inelastic” collision in S, and momentum is clearly conserved (0=0). But the Einstein
ui =
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270
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
velocity addition rule gives ūA = 0, ūB =
momentum is not conserved:
2v/(1 + v 2 /c2 ), ūC = ūD =
✓
◆
2v
m
6= 2mv.
1 + v 2 /c2
v, so in S̄ the (incorrectly defined)
(b) mA ⌘A + mB ⌘B = mC ⌘C + mD ⌘D ; ⌘i = (¯
⌘i + ⌘¯i0 ). (The inverse Lorentz transformation.)
0
0
0
0
mA (¯
⌘A + ⌘¯A ) + mB (¯
⌘B + ⌘¯B ) = mC (¯
⌘C + ⌘¯C
) + mD (¯
⌘D + ⌘¯D
). The gamma’s cancel:
0
0
0
0
mA ⌘¯A + mB ⌘¯B + (mA ⌘¯A + mB ⌘¯B ) = mC ⌘¯C + mD ⌘¯D + (mC ⌘¯C + mD ⌘¯D
).
0
0
But mi ⌘i = pi = Ei /c, so if energy is conserved in S̄ (ĒA + ĒB = ĒC + ĒD ), then so too is the momentum
(correctly defined):
mA ⌘¯A + mB ⌘¯B = mC ⌘¯C + mD ⌘¯D . qed
Problem 12.30
2
1
mc2 mc2 = nmc2 ) = n + 1 = p1 2 2 ) 1 uc2 = (n+1)
2
1
u /c
p
2
n(n + 2)
1
n2 +2n+1 1
) uc2 = 1 (n+1)
= n(n+2)
c
2 =
(n+1)2
(n+1)2 ; u =
n+1
Problem 12.31
ET = E1 + E2 + · · · ; pT = p1 + p2 + · · · ; p̄T = (pT
v = c pT /ET = c (p1 + p2 + · · · )/(E1 + E2 + · · · )
2
2
ET /c) = 0 )
= v/c = pT c/ET
Problem 12.32
Eµ =
(m2⇡ + m2µ ) 2
c = mµ c2 )
2m⇡
v2
=1
c2
1
2
=1
=
(m2⇡ + m2µ )
1
=p
;
2m⇡ mµ
1 v 2 /c2
1
v2
1
= 2;
2
c
4m2⇡ m2µ
m4⇡ + 2m2⇡ m2µ + m4µ 4m2⇡ m2µ
(m2⇡ m2µ )2
=
=
;v =
2
2
2
2
2
2
(m⇡ + mµ )
(m⇡ + mµ )
(m2⇡ + m2µ )2
(m2⇡ m2µ )
(m2⇡ + m2µ )
!
Problem 12.33
p
Initial momentum: E 2 p2 c2 = m2 c4 ) p2 c2 = (2mc2 )2 m2 c4 = 3m2 c4 ) p = 3 mc.
Initial energy: 2mc2 + mc2 = 3mc2 .
p
Each is conserved, so final energy is 3mc2 , final momentum is 3 mc.
E2
p2 c2 = (3mc2 )2
p
( 3 mc)2 c2 = 6m2 c4 = M 2 c4 .
) M=
p
6 m ⇡ 2.5m
(In this process some kinetic energy was converted into rest energy, so M > 2m.)
p
pc2
3 mc c2
c
v=
=
= p = v.
E
3mc2
3
Problem 12.34
9
5
2 4
2
First calculate pion’s energy: E 2 = p2 c2 + m2 c4 = 16
m2 c4 + m2 c4 = 25
16 m c ) E = 4 mc .
5
2
Conservation of energy:
4 mc = EA + EB
2EA = 2mc2
EB
3
3
2
Conservation of momentum: 4 mc = pA + pB = EcA
)
mc
=
E
E
A
B
c
4
1 2
2
) EA = mc ;
EB = mc .
4
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reproduced, in any form or by any means, without permission in writing from the publisher.
c.
✒✻
−1t̄ = 2
=
t̄
c −c2 1
=
ct̄ = −c3t̄ 0
=
ct̄ = ct̄ −1
ct̄ = −2
✲x
ct̄ = −3
ct̄ =
✒✻
9.2RELATIVITY
CHAPTER 12. ELECTRODYNAMICS AND
271
✲x
✛
✲❄
8.7
Problem 12.35
9.2
Classically, E = 12 mv 2 . In a colliding beam experiment, the relative velocity (classically) is twice the
velocity of either one, so the relative energy is 4E.
✛
✲❄
8.7
v
✻
✻
=⇒
Let S̄ be the system in which 1 is at rest. Its
E E
Ē
⃝
1 ✲ ✛⃝
⃝
1 ✛ ⃝
2
2
speed v, relative to S, is just the speed of 1
✲S
✻
in S.
v ✲ S̄
✻
=⇒
Ē
⃝
1 ✛ ⃝
2
E E
⃝
1 ✲ ✛⃝
2
E
p̄0 = (p0
p1 ) ✲
) SĒc =
p , where✲p S̄
is the momentum of 2 in S.
c
E
E
2
E = M c , so = M c2 ; p =
MV =
M c. ) Ē =
M c c = (E + M c2
c +
h
i
2
2
1
1
E
E
2
2
= 11 2 )1
= 12 ) 2 = 1
=
.
)
Ē
=
E
+
1
M c2
2
2
M c2
M c2
Ē =
E2
M c2
+
E2
M c2
2E 2
M c2 ; x
Ē ✻
=
M c2
)
2
M c✒
.
EA
(2)(900)
1
For E = 30 GeV and M c = ✒
1 GeV, we have Ē =
A
x✻
Problem 12.36
✒
✲ ct
One photon is impossible,
B because✒in the “center of momentum” frame (Prob. 12.31) A
we’d be left with a photon
at rest, whereas photons have to travel at speed
✲c.ct
2
2
✒
1 = 1800
=◦ 60E.
✲ 1 = 1799 GeV 60
θ
m
m
EB⑦
EA
(before)
(after) ✒◦
60
✲
θ
m
m
EB⑦
(before)
(after)
B
p
8
< Cons. of energy: ⇢ p0 c2 + m2 c4 + mc2 = EA + EB
horizontal: p0 = EcA cos 60 + EcB cos ✓ ) EB cos ✓ =pp0 c
: Cons. of mom.:
EB
3
vertical:
0 = EcA sin 60
c sin ✓ ) EB sin ✓ = 2 EA
1 2
3 2
p0 cEA + EA
+ EA
4
4
q
2
p0 cEAc + EA
=
p20 c2 + m2 c4 + mc2
1
2 EA
2
EB
(cos2 ✓ + sin2 ✓) = p0 c2
2
) EB
= p0 c2
= p0 c2 +
2
EA
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q
reproduced, in any form or by any means, without permission in writing from the publisher.
2 4
2
m c + 2 p20 c2 + m2 c4 (mc2 EA ) + m2 c4 2EA mc2 + EA
. Or:
p0 cEA = 2m2 c4 + 2mc2
q
) EA (mc2 + p20 c2 + m2 c4
q ⃝2005
q
c
2
4 currently
as2 c
they
p20protected
c2 + m2under
c4 all
p20 c2laws
+m
2EA mcexist.
; No portion of this material may be
A
reproduced, in any form or by any means, without permission in writing from the publisher.
p
p0 c/2) = m2 c4 + mc2 p0 c2 + m2 c4 ;
p
p
(mc2 + p20 c2 + m2 c4 )
(mc2
p2 c2 + m2 c4 p0 c/2)
2
p
p 0
EA = mc
·
(mc2 + p20 c2 + m2 c4 p0 c/2) (mc2
p20 c2 + m2 c4 p0 c/2)
p
p
2 4
p20 c2 m2 c4 12 p0 mc3 p20 c p20 c2 + m2 c4 )
mc2 (mc + 2p0 + p20 + m2 c2 )
2 (m c
= mc
=
.
2
2
(mc + 34 p0 )
(m2 c4 p0 mc3 + p04c
p0 c2 m2 c4 )
protected under all copyright laws as they currently exist. No portion of this material may be
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9.2
✛
v
CHAPTER
✻ 12. ELECTRODYNAMICS
✻
=⇒ AND RELATIVITY
E
E
Ē
⃝
1 ✲ ✛⃝
⃝
1 ✛ ⃝
2
2
272
Problem 12.37
(
du
dt
dp
d
mu
p
=
=m p
+u
dt
dt 1 u2 /c2
1 u2 /c2
⇢
m
u(u·a)
=p
a+ 2
. qed
2
2
(c
u2 )
1 u /c
F=
✲❄
8.7
✓
1
2
◆
✲S
2u· du
dt
u2 /c2 )3/2
1
c2
(1
✲ S̄
)
Problem 12.38
At constant force you go in “hyperbolic” motion. Photon A, which left the origin at t < 0,
catches up with you, but photon B, which
passes the origin at t > 0, never does.
x✻
A
✒
✒
✲ ct
B
"
!#
Problem 12.39
d⌘
d⌘
dt
d
c
1
0
0
0
(a)
p
p
↵ =
=
=
2
2
d⌧
dt d⌧
dt
1 u /c
1 u2 /c2
✓
◆
1
c
1
1
u·a
c2 2u·a
=p
=
.
2 (1 u2 /c2 )3/2
c (1 u2 /c2 )2
1 u2 /c2
d⌘
dt d⌘
1
d
↵=
=
=p
2
2
d⌧
d⌧ dt
dt
1 u /c

1
u(u·a)
=
a+ 2
.
(1 u2 /c2 )
(c
u2 )
(b)
p
1
u
u2 /c2
!
=p
1
1
u2 /c2
(
p
a
1
u2 /c2
+ u( t)
(1
1
c2 2u·a
u2 /c2 )3/2
c
⃝2005
protected under all copyright laws as they cur
reproduced, in any form or by any means, wit

2
1
(u·a)2
1
u2
1
↵
↵
↵µ ↵ = (↵ ) + ·↵ =
+
a(1
)
+
u(u·a)
c2 (1 u2 /c2 )4
(1 u2 /c2 )4
c2
c2
(
)
✓
◆
✓
◆
2
1
u2
u2
1 2
1
2
2
2
2
2
=
(u·a) + a 1
+ 2 1
(u·a) + 4 u (u·a)
(1 u2 /c2 )4
c2
c2
c
c2
c
⇢ ✓
◆
2
1
u2
(u·a)2
u2
u2
2
=
a
1
+
+
1
+
2
2
2
(1 u2 /c2 )4
c2
c2
c2}
|
{z c
µ
0 2
(1
=
(c) ⌘ µ ⌘µ =
(d) K µ =
(1
c2 , so
d⇢µ
d⌧
=

1
(u·a)2
2
a
+
.
u2 /c2 )2
(c2 u2 )
d
µ
d⌧ (⌘ ⌘µ )
d
µ
d⌧ (m⌘ )
u2
c2 )
= ↵µ ⌘µ + ⌘ µ ↵µ = 2↵µ ⌘µ = 0, so ↵µ ⌘µ = 0.
= m↵µ .
)
K µ ⌘µ = m↵µ ⌘µ = 0.
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273
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.40
Kµ K µ = (K 0 )2 + K·K. From Eq. 12.69, K·K =
1
d
1 dE
K =
= p
c d⌧
c 1 u2 /c2 dt
0
p
1
But Eq. 12.74: u·F = uF cos ✓ = p
mc2
u2 /c2
m
1
uF cos ✓
K = p
.
c 1 u2 /c2
0
u2 /c2
!
F2
(1 u2 /c2 ) .
=p
1

(u·a) +
F2
) Kµ K =
(1 u2 /c2 )
µ
mc
u2 /c2
From Eq. 12.70:

1
( 1/c2 )
(u·a)
m
2u·a =
2 (1 u2 /c2 )3/2
c (1 u2 /c2 )2
u2 (u·a)
m(u·a)
=
,
c2 (1 u2 /c2 )
(1 u2 /c2 )3/2
so:

u2 F 2 cos2 ✓
1 (u2 /c2 ) cos2 ✓
=
F 2 . qed
c2 (1 u2 /c2 )
(1 u2 /c2 )
Problem 12.41

u(u·a)
u(u·a)
qp
F= p
a+ 2
=
q(E
+
u⇥B)
)
a
+
=
1 u2 /c2 (E + u⇥B).
c
u2
(c2 u2 )
m
1 u2 /c2
⇥
⇤
u2 (u·a)
u·a
qp
=
=
Dot in u: (u·a) + 2
1 u2 /c2 u·E + u·(u⇥B) ;
2
2
2
2
| {z }
c (1 u /c )
(1 u /c )
m
=0
r
r
u(u·a)
q
u2 u(u·E)
q
u2
1
) 2
=
1
. So a =
1
E + u⇥B
u(u·E) . qed
(c
u2 )
m
c2 c2
m
c2
c2
m
Problem 12.42
One way to see it is to look back at the general formula for E (Eq. 10.36). For a uniform infinite plane of
charge, moving at constant velocity in the plane, J̇ = 0 and ⇢˙ = 0, while ⇢ (or rather, ) is independent of t
(so retardation does nothing). Therefore the field is exactly the same as it would be for a plane at rest (except
that itself is altered by Lorentz contraction).
A more elegant argument exploits the fact that E is a vector (whereas B is a pseudovector). This means that
any given component changes sign if the configuration is reflected in a plane perpendicular to that direction.
But in Fig. 12.35(b), if we reflect in the x y plane the configuration is unaltered, so the z component of E would
have to stay the same. Therefore it must in fact be zero. (By contrast, if you reflect in a plane perpendicular
to the y direction the charges trade places, so it is perfectly appropriate that the y component of E should
reverse its sign.)
Problem 12.43
(a) Field is 0 /✏0 , and it points perpendicular to the positive plate, so:
E0 =
(b) From Eq. 12.109, Ex = Ex0 =
0
✏0
p
(cos 45 x̂ + sin 45 ŷ) = p
0
2 ✏0
; Ey = E y0 =
(c) From Prob. 12.10: tan ✓ = , so ✓ = tan
1
p
0
2 E0
0
2 ✏0
( x̂ + ŷ).
. So E = p
.
(d) Let n̂ be a unit vector perpendicular
p to the plates in S — evidently
n̂ = sin ✓ x̂ + cos ✓ ŷ; |E| = p20✏
1 + 2.
0
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0
2 ✏0
( x̂ + ŷ).
y
B
A
274
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
y✻
3
❨n̂
θ
So the angle
between n̂ and E is:
E·n̂
= cos
|E|
But
= tan ✓ =
✲x
sin ✓
cos ✓
=
p
1
=p
1+
1 cos2 ✓
cos ✓
=
2
q
cos ✓
(sin ✓ + cos ✓) = p
1+
1
cos2 ✓
1)
1 2
cos ✓
=
2
2
2
(tan ✓ + ) = p
1+
+ 1 ) cos ✓ = p
Evidently the field is not perpendicular to the plates in S.
1
1+
2
2
cos ✓
. So cos
=
✓
2
1+
2
◆
.
y 12.44
Problem
y
✻v
ŝ
x
x̂
+
y✻
0
0 ŷ
✲
(a) E =
=
.
2
qB 2⇡✏0 s
2⇡✏0 (x0 +qBy02 )
d
d
x0
y0
(x0 x̂ + y0 ŷ
✲x
✛, Ēy = ✲
(b)qĒ
=
E
=
Eyx̄=
, Ēz = Ez = 0, Ē =
.
x
x
2
2
2
2
A
q
2⇡✏0 (x0 + y0 v) A
2⇡✏0 (x0 + y0 )
2⇡✏0 (x20 + y02 )
Using the inverse Lorentz transformations (Eq. 12.19), x0 = (x + vt), y0 = y,
(x + vt) x̂ + y ŷ
1 y (x
✻ + vt) x̂ + y ŷ .
=
2
2
2
2 + y2 / 2 ]
2⇡✏0 [ (x + vt)3 + y ]
2⇡✏0 [(x + vt)
❨n̂
θ
Now S = (x+vt) x̂+y ŷ, and y = S sin ✓, so [(x+vt)2 +y 2 / 2 ] = [(x+vt)2 +y 2 (1✲ vx2 /c2 ] = S 2 (v/c)2 S 2 sin2 ✓ =
2
2
2
S [1 (v/c) sin ✓], so
p
1 (v/c)2
Ŝ
Ē =
.
2
2
2
2⇡✏0 1 v sin ✓/c S
ȳ
✻
This is reminiscent of Eq. 10.75. Yes, the field does point away fromȲthe present location of the wire.
❑
y✻
Problem 12.45
✯ X̄
Y❑
y
✻v
1 qA qB
1 qA
✲
(a) Fields of A at B: E = 4⇡✏0 d2 ŷ; B = 0. So force on qB is F =
ŷ. ✯ X
2
φ
qB✲
4⇡✏0 d
❄
✯
d x̄
✲x
vt
y
qA
φ
✻
✲x
qA qB ❑
(b) (i) From Eq. 12.67: F̄ =
ŷ.
(Note:
here
the
particle
is at rest in S̄.)
qB
4⇡✏0 d2
d
✛
✲ x̄
1 qA (1 v 2 /c2 ) 1
qA
(ii) From Eq. 12.93, with ✓ = 90 : Ē =
ŷ =
ŷ
v qA
4⇡✏0 (1 v 2 /c2 )3/2 d2
4⇡✏0 d2
(this also follows from Eq. 12.109).
qA qB
B̄ 6= 0, but since vB = 0 in S̄, there is no magnetic force anyway, and F̄ =
ŷ (as before).
4⇡✏0 d2
Ē =
✻v
✲
d
✲x
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c
⃝2005
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✻ the publisher.
reproduced, in any form or by any means, without permission in writing from
Y❑
y✻
Ȳ❑
✯ X̄
✯X
φ
q
✛
v
275
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.46
System A: Use Eqs. 12.93 and 12.112, with ✓ = 90 , R = d ŷ, and ˆ = ẑ:
q
ŷ ;
4⇡✏0 d2
E=
q v
ẑ ;
4⇡✏0 c2 d2
B=
2
2
where
2
=p
1
1
v 2 /c2
.
2
[Note that (E 2 B 2 c2 ) = 4⇡✏q0 d2
1 vc2 = 4⇡✏q0 d2
is invariant, since it doesn’t depend on v (see
Prob. 12.47b for the general proof). We’ll use this as a check.]
F = q E + ( vx̂)⇥B =
System B : The speed of
B
1
=q
1
⇥
Check : (E 2
F = qE =
q is
4v 2 /c2
(1+v 2 /c2 )2
)E=
B 2 c2 ) =
vB =
q2
ŷ
4⇡✏0 d2
v2
(x̂⇥ẑ) =
c2
v+v
2v
=
1 + v 2 /c2
(1 + v 2 /c2 )
(1 + v 2 /c2 )
(1 + v 2 /c2 )
=q
=
=
2
4
(1 v 2 /c2 )
1 2 vc2 + vc4
q 1
4⇡✏0 d2
2
2 4
q
4⇡✏0 d2
q2
v2
1
+
ŷ.
4⇡✏0 d2
c2
1+
1+
v2
ŷ ;
c2
2v 2
c2
+
B=
v4
c4
4v 2
c2
2
1+
v2
; vB
c2
B
= 2v
2
.
q 2v 2
ẑ.
4⇡✏0 c2 d2
=
2 4 1
q
4
4⇡✏0 d2
=
2
q
4⇡✏0 d2
X
⇤
q2 2
v2
1 + 2 ŷ (+q at rest ) no magnetic force). [Check : Eq. 12.67 ) FA = 1 FB . X]
2
4⇡✏0 d
c
q 1
q2 1
ŷ;
B
=
0.
F
=
qE
=
ŷ.
4⇡✏0 d2
4⇡✏0 d2
[The relative velocity of B and C is 2v/(1 + v 2 /c2 ), and corresponding is 2 (1 + v 2 /c2 ). So Eq. 12.67
1
) FC = 2 (1+v
X]
2 /c2 ) FB .
Summary:
✓
◆
✓
◆
✓
◆
q
q
q
2
2 2
ŷ
1 + v /c ŷ
ŷ
2
2
4⇡✏
4⇡✏0 d2
✓ 4⇡✏0 d ◆
✓ 0d
◆
q
v
q
2v 2
ẑ
ẑ
0
2
2
2
4⇡✏
d
c
4⇡✏
d
c2
0
0
✓
◆
✓
◆
✓
◆
q2
q2
q2
2 2
2
2 2
1 + v /c ŷ
1 + v /c ŷ
ŷ
4⇡✏0 d2
4⇡✏0 d2
4⇡✏0 d2
System C :
vC = 0.
E=
Problem 12.47
(a) From Eq. 12.109:
Ē·B̄ = Ēx B̄x + Ēy B̄y + Ēz B̄z = Ex Bx +
2
(Ey
v
{Ey By + 2 Ey Ez vBy Bz
c
✓
⇣
⇣
v2 ⌘
2
= Ex Bx +
Ey By 1
+
E
B
1
z
z
c2
= Ex Bx +
2
v
Ez ) + (Ez + vBy )(Bz
c2
v2
v
Ez Bz + Ez Bz
Ey Ez + vBy Bz
c2
c2
◆
v2 ⌘
= Ex Bx + Ey By + Ez Bz = E·B.
c2
vBz )(By +
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v
Ey )
c2
v2
Ey By }
c2
qed
276
(b)
Ē 2
⇥
c2 B̄ 2 = Ex2 +
2
= Ex2 +
c2
=
2
(Ey
Ex2
2
c
vBz )2 +
c2 Bz2 + c2 2
Bx2
2
(Ez + vBy )2
⇤
⇥
c2 Bx2 +
By +
2
2Ey vBz + v 2 Bz2 + Ez2 + 2Ez vBy + v 2 By2
Ey2
v2 2
E
c4 z
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
+
2
✓
Ey2
= (Ex2 + Ey2 + Ez2 )
⇣
v
Bz Ey
c2
1
c2
v2 2
E
c4 y
⇣
v2 ⌘
2
+
E
1
z
c2
v
Ez
c2
c2 By2
2
+
c2 2
2
⇤
v
Ey
2
c
Bz
v
By Ez
c2
c2 Bx2
v2 ⌘
c2
c2 (Bx2 + By2 + Bz2 ) = E 2
2
c
⇣
(By2 )
1
v2 ⌘
c2
2
c
Bz2
⇣
v2 ⌘
c2
1
◆
B 2 c2 . qed
(c) No. For if B = 0 in one system, then (E 2 c2 B 2 ) is positive. Since it is invariant, it must be positive in
any system. ) E 6= 0 in all systems.
Problem 12.48
(a) Making the appropriate modifications in Eq. 9.48 (and picking = 0 for convenience),
E(x, y, z, t) = E0 cos(kx
!t) ŷ,
B(x, y, z, t) =
E0
cos(kx
c
!t) ẑ,
where k ⌘
!
.
c
(b) Using Eq. 12.109 to transform the fields:
Ēx = Ēz = 0,
B̄x = B̄y = 0,
where
↵⌘
⇣
1
Ēy = (Ey
B̄z = (Bz
v⌘
=
c
s
h
vBz ) = E0 cos(kx

v
1
Ey ) = E 0
cos(kx
c2
c
!t)
!t)
v
cos(kx
c
v
cos(kx
c2
i
!t) = ↵E0 cos(kx
!t) = ↵
1 v/c
.
1 + v/c
Now the inverse Lorentz transformations (Eq. 12.19) ) x = (x̄ + v t̄) and t =
t̄ +
!t),
v ⌘
x̄ , so
c2
⇣
h⇣
i
v ⌘i
!v ⌘
! t̄ + 2 x̄ =
k
x̄
(!
kv)
t̄
= k̄x̄ !
¯ t̄,
c
c2
⇣
!v ⌘
where, recalling that k = !/c): k̄ ⌘
k
= k(1 v/c) = ↵k and !
¯ ⌘ !(1 v/c) = ↵!.
c2
Ē0
Ē(x̄, ȳ, z̄, t̄) = Ē0 cos(k̄x̄ !
¯ t̄) ŷ, B̄(x̄, ȳ, z̄, t̄) =
cos(k̄x̄ !
¯ t̄) ẑ,
c
s
Conclusion:
1 v/c
where Ē0 = ↵E0 , k̄ = ↵k, !
¯ = ↵!, and ↵ ⌘
.
1 + v/c
s
1 v/c
¯ = 2⇡ = 2⇡ = . The velocity of the wave
(c) !
¯=!
. This is the Doppler shift for light.
1 + v/c
↵k
↵
k̄
!
¯ ¯
!
in S̄ is v̄ =
=
= c.
Yup, this is exactly what I expected (the velocity of a light wave is the same
2⇡
2⇡
in any inertial system).
kx
!t =
h
⇣
E0
cos(kx
c
!t),
k(x̄ + v t̄)
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277
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
I¯ Ē02
1 v/c
= 2 = ↵2 =
.
I
E0
1 + v/c
(d) Since the intensity goes like E 2 , the ratio is
Dear Al,
The amplitude, frequency, and intensity of the light will all decrease to zero as you run faster
and faster. It’ll get so faint you won’t be able to see it, and so red-shifted even your night-vision
goggles won’t help. But it’ll still be going 3 ⇥ 108 m/s relative to you. Sorry about that.
Sincerely,
David
Problem 12.49
t̄02 = ⇤0 ⇤2 t
t̄03 = ⇤0 ⇤3 t
t̄23 = ⇤2 ⇤3 t
t̄31 = ⇤3 ⇤1 t
t̄12 = ⇤1 ⇤2 t
= ⇤00 ⇤22 t02 + ⇤01 ⇤22 t12
= ⇤00 ⇤33 t03 + ⇤01 ⇤33 t13
= ⇤22 ⇤33 t23 = t23 .
= ⇤33 ⇤10 t30 + ⇤33 ⇤11 t31
= ⇤10 ⇤22 t02 + ⇤11 ⇤22 t12
= t02 + (
= t03 + (
=(
=(
Problem 12.50
Suppose t⌫µ = ±tµ⌫ (+ for symmetric,
)t12 = (t02
)t13 = (t03
t12 ).
t13 ) = (t03 + t31 ).
)t30 + t31 = (t31
)t02 + t12 = (t12
t03 ).
t02 ).
for antisymmetric).
t̄ = ⇤µ ⇤⌫ tµ⌫
t̄

= ⇤µ ⇤⌫ tµ⌫ = ⇤⌫ ⇤µ t⌫µ
[Because µ and ⌫ are both summed from 0 ! 3,
it doesn’t matter which we call µ and and which call ⌫.]
=
⇤µ ⇤µ (±tµ⌫ )

= ±t̄ .
[Using symmetry of tµ⌫ , and writing the ⇤’s in the other order.]
qed
Problem 12.51
F µ⌫ Fµ⌫ = F 00 F 00
F 01 F 01
F 02 F 02
F 03 F 03
F 10 F 10
F 20 F 20
F 30 F 30
+ F 11 F 11 + F 12 F 12 + F 13 F 13 + F 21 F 21 + F 22 F 22 + F 23 F 23 + F 31 F 31 + F 32 F 32 + F 33 F 33
=
(Ex /c)2
= 2B 2
(Ey /c)2
⇣
2E 2 /c2 = 2 B 2
which, apart from the constant factor
Gµ⌫ Gµ⌫ = 2(E 2 /c2
F µ⌫ Gµ⌫ =
=
=
(Ez /c)2
(Ex /c)2
(Ey /c)2
(Ez /c)2 + Bz2 + By2 + Bz2 + Bx2 + By2 + Bx2
E2 ⌘
,
c2
2/c2 , is the invariant we found in Prob. 12.47(b).
B 2 ) (the same invariant).
2(F 01 G01 + F 02 G02 + F 03 G03 ) + 2(F 12 G12 + F 13 G13 + F 23 G23 )
✓
◆
1
1
1
2
Ex Bx + Ey By + Ez Bz + 2[Bz ( Ez /c) + ( By )(Ey /c) + Bx ( Ex /c)]
c
c
c
2
(E · B)
c
2
(E · B) =
c
4
(E · B),
c
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278
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
which, apart from the factor 4/c, is the invariant we found of Prob. 12.47(a). [These are, incidentally, the
only fundamental invariants you can construct from E and B.]
Problem 12.52
E=
1 2
4⇡✏0 x
x̂ =
B=
µ0 2 v
4⇡ x ŷ
=
9
µ0 c2 =
2⇡ x x̂
µ0 v
2⇡ x ŷ
;
F µ⌫
0
0
µ0 B
B c
=
2⇡x @ 0
0
c
0
0
v
1
0 0
0 vC
C
0 0A
0 0
Gµ⌫
0
0
µ0 B
B0
=
2⇡x @ v
0
0
0
0
0
1
v 0
0 0C
C
0 cA
c 0
Problem 12.53
@⌫ F µ⌫ = µ0 J µ . Di↵erentiate: @µ @⌫ F µ⌫ = µ0 @µ J µ .
But @µ @⌫ = @⌫ @µ (the combination is symmetric) while F ⌫µ = F µ⌫ (anti symmetric).
) @µ @⌫ F µ⌫ = 0. [Why? Well, these indices are both summed from 0 ! 3, so it doesn’t matter which we
call µ, which ⌫: @µ @⌫ F µ⌫ = @⌫ @µ F ⌫µ = @µ @⌫ ( F µ⌫ ) = @µ @⌫ F µ⌫ . But if a quantity is equal to minus itself,
it must be zero.] Conclusion: @µ J µ = 0. qed
Problem 12.54
We know that @⌫ Gµ⌫ = 0 is equivalent to the two homogeneous Maxwell equations, r·B = 0 and r⇥E =
@B
@t . All we have to show, then, is that @ Fµ⌫ + @µ F⌫ + @⌫ F µ = 0 is also equivalent to them. Now this
equation stands for 64 separate equations (µ = 0 ! 3, ⌫ = 0 ! 3, = 0 ! 3, and 4 ⇥ 4 ⇥ 4 = 64). But many
of them are redundant, or trivial.
Suppose two indices are the same (say, µ = ⌫). Then @ Fµµ + @µ Fµ = @µ F µ = 0. But Fµµ = 0 and
Fµ = F µ , so this is trivial: 0 = 0. To get anything significant, then, µ, ⌫, must all be di↵erent. They
could beall spatial (µ, ⌫, = 1, 2, 3 = x, y, z — or some permutation thereof), or one temporal and two spatial
(µ = 0, ⌫, = 1, 2 or 2, 3, or 1, 3 — or some permutation). Let’s examine these two cases separately.
All spatial : say, µ = 1, ⌫ = 2,
= 3 (other permutations yield the same equation, or minus it.)
@3 F12 + @1 F23 + @2 F31 = 0 )
One temporal : say, µ = 0, ⌫ = 1,
@
@
@
(Bz ) +
(Bx ) +
(By ) = 0 ) r·B = 0.
@z
@x
@y
= 2 (other permutations of these indices yield same result, or minus it).
@2 F01 + @0 F12 + @1 F31 = 0 )
@E
@
@y
Ex
@
@
Ey
=
(Bz ) +
+
= 0.
c
@(ct)
@x
c
@Ex
y
z
or: @B
= 0, which is the z-component of
@t +
@y
@x
y component; for ⌫ = 2, = 3 we get the x component.)
Conclusion: @ Fµ⌫ + @µ F⌫ + @⌫ F
@⌫ Gµ⌫ = 0. qed
µ
@B
@t
= r⇥E. (If µ = 0, ⌫ = 1,
= 0 is equivalent to r·B = 0 and
@B
@t
=
= 2, we get the
r⇥E, and hence to
Problem 12.55
q
u·E. Now from Eq. 12.70 we know that
c
1
K 0 = 1c dW
d⌧ , where W is the energy of the particle. Since d⌧ = dt, we have:
K 0 = q⌘⌫ F 0⌫
q(⌘1 F 01 + ⌘2 F 02 + ⌘3 F 03 ) = q(⌘·E)/c =
1 dW
q
dW
=
(u·E) )
= q(u·E)
c dt
c
dt
This says the power delivered to the particle is force (qE) times velocity (u) — which is as it should be.
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279
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.56
@
@0 =
=
@ x̄0
1 @
1
=
c @ t̄
c
@t
From Eq. 12.19, we have:
=
@ t̄
1 @
@
So @ 0 =
+v
or
c @t
@x
@ @t @ @x @ @y @ @z
+
+
+
@t @ t̄ @x @ t̄
@y @ t̄
@z @ t̄
,
@x
@ t̄
= v,
@y
@ t̄
=
(since ct = x0 =
@z
@ t̄
= 0.
x0 ): @ 0 =
@
@x0
@1 =
@
@ @t
@ @x @ @y @ @z
v @
@
=
+
+
+
= 2
+
=
1
@ x̄
@t @x
@x @ x̄ @y @ x̄ @z @ x̄
c @t
@x
@2 =
@
@ @t
@ @x @ @y @ @z
@
=
+
+
+
=
= @2 .
@ ȳ
@t @ ȳ
@x @ ȳ
@y @ ȳ
@z @ ȳ
@y
@3 =
@
@ @t
@ @x @ @y @ @z
@
=
+
+
+
=
= @3 .
@ z̄
@t @ z̄
@x @ z̄
@y @ z̄
@z @ z̄
@z
v @
=
c @x1
@
@x1
⇥ 0
(@ )
v @
=
c @x0
⇤
(@ 1 ) .
⇥ 1
(@ )
⇤
(@ 0 ) .
Conclusion: @ µ transforms in the same way as aµ (Eq. 12.27)—and hence is a contravariant 4-vector. qed
Problem 12.57
According to Prob. 12.54,
of Prob. 12.56):
@Gµ⌫
@x⌫
= 0 is equivalent to Eq. 12.130. Using Eq. 12.133, we find (in the notation
@Fµ⌫
@F⌫
@F µ
+
+
= @ Fµ⌫ + @µ F⌫ + @⌫ F µ
@x
@xµ
@x⌫
= @ (@µ A⌫ @⌫ Aµ ) + @µ (@⌫ A
@ A⌫ ) + @⌫ (@ Aµ @µ A )
= (@ @µ A⌫ @µ @ A⌫ ) + (@µ @⌫ A
@⌫ @µ A ) + (@⌫ @ Aµ @ @⌫ Aµ ) = 0. qed
[Note that @ @µA⌫ =
@ 2 A⌫
@x @x⌫
=
@ 2 A⌫
@x⌫ @x
Problem 12.58
From Eqs. 12.40 and 12.42, ⌘ µ =
cr + v · r =
( r c r · v).
= @⌫ @ A⌫ , by equality of cross-derivatives.]
(c, v), while
r
µ
= (ct
ctr , r
w(tr )) = ( r ,
q
⌘0
q
c
1
qc
1
=
=
= V
⌫
4⇡✏0 c (⌘ r ⌫ )
4⇡✏0 c ( r c r · v)
4⇡✏0 c ( r c r · v)
c
(Eq. 10.46),
q
⌘
q
v
1
qv
=
=
=A
⌫
4⇡✏0 c (⌘ r ⌫ )
4⇡✏0 c ( r c r · v)
4⇡✏0 c ( r c r · v)
(Eq. 10.47), so (Eq. 12.132)
q
⌘µ
= Aµ . X
⌫
4⇡✏0 c (⌘ r ⌫ )
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r
), so ⌘ ⌫ r
⌫
=
✛
v qA
qA
280
✲ x̄
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.59
ȳ
Step 1 : rotate from xy to XY , using Eq. 1.29:
y✻
Y❑
X = cos x + sin y
Y = sin x + cos y
✻
Ȳ❑
✯ X̄
✯X
φ
❄
✯
✲ x̄
vt
Step 2 : Lorentz-transform from XY to X̄ Ȳ , using
Eq. 12.18:
φ
❑
✲x
X̄ = (X vt) = [cos x + sin y
ct]
Ȳ = Y = sin x + cos y
Z̄ = Z = z
⇥
⇤
ct̄ = (ct
X) = ct
(cos x + sin y)
Step 3 : Rotate from X̄ Ȳ to x̄ȳ, using Eq. 1.29 with negative :
x̄ = cos X̄
sin Ȳ =
cos [cos x + sin y
sin [ sin x + cos y]1
cos (ct)
ct) + cos ( sin c + cos y)
ct]
= ( cos + sin )x + (
1) sin cos y
ȳ = sin X̄ + cos Ȳ = sin (cos x + sin y
2
=(
2
1) sin cos x + ( sin2
+ cos2 )y
sin (ct)
sin
40
(
1) sin cos
0
( sin2
Problem 12.60
In center-of-momentum system, threshold occurs when incident energy is just sufficient to cover the rest energy of the resulting particles,
with none “wasted” as kinetic energy. Thus, in lab system, we want
the outgoing K and ⌃ to have the same velocity, at threshold:
✲
π
p
✲
0Inc.,
1Upper
c
⃝2005
Pearson Education,
cos of this material may
sin be
ct̄laws as they currently exist. No portion
reproduced, in any form or
any means,
writing
Bby
B without
x̄ C
cos permission
( cos2 in+
sin2 from
) ( the publisher.
1) sin cos
C
B
In matrix form: B
@ ȳ A = @
z̄
π
+ cos2
0
1
0
0C
C
) 0A
1
before (CM)
✛
0 1
ct K Σ
BxC
B C
@yA
z
π
✲
✛
p
K Σ
after (CM)
before (CM)
after (CM)
✲
K Σ
p
After
Before
Initial momentum: p⇡ ; Initial energy of ⇡: E
p2 c2 = m2 c4 ) E⇡2 = m2⇡ c4 + p2⇡ c2 .
p
Total initial energy: mp c2 = m2⇡ c4 + p2⇡ c2 . These are also the final energy and momentum: E 2
(mK + m⌃ )2 c4 .
2
⇣
mp c2 +
p
m2⇡ c4 + p2⇡ c2
⌘2
p2 c2 =
p2⇡ c2 = (mK + m⌃ )2 c4
p
m2p c4 + 2mp c2 m2⇡ c2 + p2⇡ c + m2⇡ c4 + p2⇡ c2 p2⇡ c2 = (mK + m⌃ )2 c4
c4
2mp p 2 2
m⇡ c + p2⇡ = (mK + m⌃ )2 m2p m2⇡
c
✲
✲
π
p
Σ
c 2012 PearsonK
Education,
This material is
protected under all copyright laws as they currently exist. No portion of this material may be
After
Before
reproduced, in any
form or by any means, without permission in writing from the publisher.
✲
✲
π
p
CHAPTER 12. ELECTRODYNAMICS
AND
K
Σ RELATIVITY
After
Before
(m2⇡ c2 + p2⇡ )
4m2p
= (mK + m⌃ )4
c2
4m2p 2
p = (mK + m⌃ )4
c2 ⇡
p⇡ =
281
c q
(mK + m⌃ )4
2mp
2(m2p + m2⇡ )(mK + m⌃ )2 + m4p + m4⇡ + 2m2p m2⇡
2(m2p + m2⇡ )(mK + m⌃ )2 + (m2p
2(m2p + m2⇡ )(mK + m⌃ )2 + (m2p
m2⇡ )2
m2⇡ )2
q
(mK c2 + m⌃ c2 )4 2 (mp c2 )2 + (m⇡ c2 )2 (mK c2 + m⌃ c2 )2 + (mp c2 )2 (m⇡ c2 )2
(2mp
q
2
1
= 2c(900)
(1700)4 2 (900)2 + (150)2 (1700)2 + (900)2 (150)2
p
1
1
(8.35 ⇥ 1012 ) (4.81 ⇥ 1012 ) + (0.62 ⇥ 1012 ) = 1800c
(2.04 ⇥ 106 ) = 1133 MeV/c
= 1800c
=
1
2
c2 )c
Problem 12.61
p
✲
In CM :
rµ
❘ p
✒
φ
p
✛
y
✻
✲x
(p = magnitude of 3-momentum
in CM, φ = CM scattering angle)
p✠
■
sµ
After
Before
5
Outgoing 4-momentua: rµ =
✲
In Lab:
E
E
µ
c , p cos , p sin , 0 ; s = c , p cos , p sin , 0 .
c
⃝2005
Pearson Education, r̄Inc.,
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form ✒
or by any means, without permission in writing from the publisher.
θ
❘
Before
Lorentz transformation: r̄x = (rx
Now E = mc2 ; p =
) r̄x =
p cos +
cos ✓ =
r̄·s̄
= q⇥
r̄s̄
s̄µ
r0 ); r̄y = ry ; s̄x = (sx
mv (v here is to the left; E 2
pc E
E c
Problem: calculate θ, in terms of p, φ.
s0 ); s̄y = sy .
p2 c2 = m2 c4 , so
= p(1 + cos ); r̄y = p sin ; s̄x = p(1
pc
E.
=
cos ); s̄y =
p sin .
p (1
cos2 ) p2 sin2
⇤⇥
⇤
2 p2 (1 + cos )2 + p2 sin2
2 p2 (1
cos )2 + p2 sin2
2 2
2
( 2 1) sin2
φ)
= q⇥ / sin
⇤⇥
2 (1
(τ2
τ / sin
)2 φ+ sin2
√ 1+ (1 + cos
θ
( 2 1)
= rh 1
ih
2
2 1+cos
2 1 cos
+
1
sin
sin
cos )2 + sin2
i=q
2
+1
2
⇤
cot
(
2
2
2
1)
+ 1 tan2
2
+1
protected under all copyright laws as they currently exist. No portion of this material may be
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d/2
y
✻
l
+q!"#\$
✕
282
cos ✓ = q
=q
=q
sin ✓ =
⌧
sin
Or, since (
=q
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
1 + cot2
2
+ ! cot2
1 + tan2
2
2
+ ! tan2
!
csc2
2
+ ! cot
sec2
2
2
+ ! tan
2
2
2
1
2 ! sin
1+
! 12 (1
+ cos ) 1 +
sin
2
!
+1
.
⌧2 =
2
1) =
2
cos2
4
! 2 (1
2
⇣
1
+ !) =
1
2
⌘
=q
(
=
2
4
! 12 (1
sin
4
!2
1)2
2 v2
c2 ,
+
2
4
!
+ sin2
, so tan ✓ =
tan ✓ =
! sin
1+
= q⇥
2
! cos2
2
!
=q
1+
(
Before
2
=q
cos )
✲
In Lab:
(where ! ⌘ 2 1)
!
cos
2
2
1 + ! sin2
sin
⇤⇥
+ 1 + cos
1
2
1) sin
2
⌧2
sin2
2
!
2
+1
, where ⌧ 2 =
cos
4
4
+ .
2
!
!
.
2c2
2
v sin
⇤
2
√
1+
(τ
θ
in
/s
φ)
τ / sin φ
1
Problem 12.62
dp
dp dt
dt
p 1 2 2 ; p = p mu2 2 .
d⌧ = K (a constant) ) dt d⌧ = K. But d⌧ =
1 u /c
1 u /c
⇣
⌘
p
d p u
K
2 /c2 . Multiply by dt = 1 :
) dt
=
1
u
m
dx
u
2
2
1 u /c
p
✓
◆
✓
◆
dt d
u
d
u
K 1 u2 /c2
u
p
p
=
=
. Let w = p
.
dx dt
dx
m
u
1 u2 /c2
1 u2 /c2
1 u2 /c2
dw
K 1
=
;
dx
mw
) w2 =
Let
2K
m x+
w
dw
1 d 2
k
=
w =
;
dx
2 dx
m
d(w2 )
2K
2K
=
) d(w2 ) =
(dx).
dx
m
m
constant. But at t = 0, x = 0 and u = 0 (so w = 0), and hence the constant is 0.
w2 =
2K
x=
m
1
u2 =
2Kx/m
c2
=
;
mc2
1 + 2Kx
1 + 2Kx
mc2
u2
;
u2 /c2
u2 =
2Kx
m
2Kx 2
u ;
mc2
dx
c
=q
dt
1+
mc2
2Kx
2Kx
2Kx
=
.
2
mc
m
Z r
mc2
ct =
1+
dx
2Kx
u2 1 +
;
R px+a2
p
p
= a2 ; ct =
dx. Let x = y 2 ; dx = 2y dy; x = y.
x
Z p 2
Z
h p
i
p
p
y + a2
ct =
2y dy = 2
y 2 + a2 dy = y y 2 + a2 + a2 ln(y + y 2 + a2 ) + constant.
y
At t = 0, x = 0 ) y = 0. ) 0 = a2 ln a+ constant, so constant = a2 ln a.
r⇣ ⌘
⇣ ⌘ r⇣ ⌘
✓
◆
p
p
2
y
y
y
y 2
2
2
) ct = y y 2 + a2 = a ln y/a + (y/a)2 + 1 = a
+ 1 + ln
+
+1
a
a
a
a
mc2
2K
Let: z =
y
a
=
q
p
p
p q 2K
2Kt
2 + ln(z +
x mc2 = 2Kx
=
z.
Then
=
z
1
+
z
1 + z 2 ).
2
mc
mc
c
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Pearson Education, Inc., Upper Saddle River, NJ
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reproduced,
any formThis
or by
any means,
without permi
c 2012 Pearson Education, Inc., Upper Saddle River, NJ.
All rightsinreserved.
material
is
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
283
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.63
hp
i
F
(a) x(t) = ↵c
1 + (↵t)2 1 , where ↵ = mc
. The force of +q on
q will be the mirror image of the force of q on +q (in the x-axis),
so the net force is in the x direction (the net magnetic force is zero).
So all we need is the x-component of E.
The field at +q due to q is: (Eq. 10.72)
r ⇥u(c2 v2 ) + u( r ·a)
q
E=
4⇡✏0 ( r ·u)3
u = cr
v ) ux = c rl
Ex =
=
q
4⇡✏0 (c r
v=
r
vl)3
1
r
⇥ 1
r
⇥
q
1
(cl
3
4⇡✏0 (c r
vl)
(cl
(cl
v r );
r
v r )(c2
v r )(c2
v2 )
·u = c r
v2 ) =
1
r
d/2
y
✻
l
+q!"#\$
−d/2
⇤
a( r ·u) .
·v = (c r
lv);
r
✲x
−q
·a = la. So:
⇤
(cl v r )la a(c r
lv)
|r
{z
}
1
2
2
2
r ca(l r ) = cad / r
⇤
The force on +q is qEx , and there is an equal force on
q, so the net force on the dipole is:
c
⃝2005
2
protected under all2q
laws
No portion of⇤this material
may beto determine
⇥ currently exist.
1 as they
It remains
reproduced,
F in
= any form or by any means,
(clwithout
v permission
)(c2 v 2 )in writing
x̂ the publisher.
3
v,
and
a, and plug these
4⇡✏0 (c
lv)
r
r
✕r
d
r
, l,
in.
p
dx
c1
1
c↵t
c↵tr
p
=
2↵2 t = p
; v = v(tr ) =
, where T = 1 + (↵tr )2 .
2
2
dt
↵ 2 1 + (↵t)
T
1 (↵t)
✓
◆ 2
dv
c↵
1 2↵ tr
c↵
c↵
a(tr ) =
=
+ c↵tr
= 3 1 + (↵tr )2 (↵tr )2 = 3
dtr
T
2
T3
T
T
p
⇥p
⇤
Now calculate tr : c2 (t tr )2 = r 2 = l2 + d2 ; l = x(t) x(tr ) = ↵c
1 + (↵t)2
1 + (↵tr )2 , so
p
p
⇥
⇤
t2 2ttr + t2r = ↵12 1 + (↵t)2 + 1 + (↵tr )2 2 1 + (↵t)2 1 + (↵tr )2 + (d/c)2
p
p
2
(F) 1 + (↵t)2 1 + (↵tr )2 = 1 + ↵2 ttr + 12 ↵d
. Square both sides:
c
v(t) =
1 + (↵t)2 + (↵tr )2 + ↵4 t2 t2r = 1 + ↵4 t2 t2r +
t2 + t2r
2ttr
ttr
⇣ ↵d ⌘2
c
⇣ d ⌘2
c
⇣ ↵d ⌘2
⇣ ↵d ⌘2
1 ⇣ ↵d ⌘4
+ 2↵2 ttr +
+ ↵2 ttr
4 c
c
c
↵ 2 ⇣ d ⌘4
=0
4 c
At this point we could solve for tr (in terms of t), but since v and a are already expressed in terms of tr , it is
simpler to solve for t (in terms of tr ), and express everything in terms of tr :


⇣ ↵d ⌘2
⇣ d ⌘2 ↵ 2 ⇣ d ⌘4
t2 ttr 2 +
+ t2r
= 0 =)
c
c
4 c
s 
( 
)
⇣ ↵d ⌘2
⇣ ↵d ⌘2 ⇣ ↵d ⌘4
⇣ d ⌘2
⇣ d ⌘4
1
t=
tr 2 +
± t2r 4 + 4
+
4t2r + 4
+ ↵2
2
c
c
c
c
c
s


⇣ ↵d ⌘
⇥
⇤⇣ d ⌘2
1 ⇣ ↵d i 2
= tr 1 +
) ±
1 + (↵tr )2
1+
2 c
c
2c
protected under all copyright laws as they currently exist. No portion of this material may be
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284
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Which sign? For small ↵ we want t ⇡ tr + d/c, so we need the + sign:
r

⇣ ↵d ⌘2
1 ⇣ ↵d ⌘2
d
t = tr 1 +
+ T D, where D = 1 +
2 c
c
2c
So
r
p
2
= ct2r ↵d
+ dT D. Now go back to Eq. (F) and solve for 1 + (↵t)2 :
c
⇢
 ⇣
p
1
1 ⇣ ↵d ⌘2
1 ⇣ ↵d ⌘2
d
2
2
1 + (↵t) =
1+
+ ↵ tr t r 1 +
+ TD
T
2 c
2 c
c
⇢


⇣
⌘
2
⇤
1 ⇥
1 ↵d 2
↵ tr d
1 ⇣ ↵d ⌘2
↵ 2 tr d
2
=
1 + (↵tr ) 1 +
+
TD = 1 +
T+
D
T | {z }
2 c
c
2 c
c
T2
⇢
i
⇣d
⌘
p
c hp
c
1 ⇣ ↵d ⌘2
↵ 2 tr d
l=
1 + (↵t2 )
1 + (↵tr )2 =
1+
T+
D T = ↵d
T + tr D
↵
↵
2 c
c
2c
= c(t
tr ) )
r
Putting all this in, the numerator in square brackets in F becomes:
[
n
⇣d
⌘ c↵t h ct ⇣ ↵d ⌘2
ioh
c2 ↵2 t2r i
c↵
r
r
] = c↵d
T + tr D
+ dT D
c2
c d2
2
2c
T
2
c
T
T3
hd
i c2 ⇥
2
2
⇤
d(atr )2
c
↵d
= c↵d
T + tr D
tr D
1 + (↵tr )2 (↵tr )2
2c
2cT
T2
T3
i
i
2
2h
2
2h
c ↵d 1 2 1
c ↵d
c2 ↵d2
2
2
2
=
T
(↵t
)
1
=
1
+
(↵t
)
(↵t
)
2
=
r
r
r
T3 2
2
2T 3
2T 3
)F=
(c r
q2
c2 ↵d2
⇥
⇤3 x̂.
4⇡✏0 (c r
lv)T
It remains to compute the denominator:
⇢  ⇣ ⌘
⇣d
⌘ c↵t
ctr ↵d 2
r
c
+ dT D
↵d
T + tr D
T
2
c
2c
T
h1
⇥
⇤
1 2 2 cd(↵tr )2 i
= ↵2 tr d2 + cdT D
↵ tr d
D T = cdD T 2 (↵tr )2 = dcD
|
{z
}
2
2
T
lv)T =
1+(↵tr )2
)F=
q 2 c2 d2 ↵
q2
↵
x̂
=
4⇡✏0 c3 d3 D3
4⇡✏0 cd⇥1 + ↵d
2c
x̂
2 ⇤3/2
⇣
↵=
(↵tr )2
F ⌘
mc
Energy must come from the “reservoir” of energy stored in the electromagnetic fields.
h
⇣ ↵d ⌘2 i3/2
⇣ µ q2 ⌘
1 q2
↵
q2
0
(b) F = mc↵ =
) 1+
=
=
.
⇥
⇤
2
2
3/2
2 4⇡✏0 cd 1 + ↵d
2c
8⇡✏0 mc d
8⇡md
2c
"
(force on one end only)
r
2c ⇣ µ0 q 2 ⌘2/3
)↵=
1,
d
8⇡md
so
2mc2
F =
d
r
⇣ µ q 2 ⌘2/3
0
8⇡md
1.
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285
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.64
(a) Aµ = (V /c, Ax , Ay , Az ) is a 4-vector (like xµ = (ct, x, y, z)), so (using Eq. 12.19): V = (V̄ + v Āx ). But
V̄ = 0, and
µ0 (m⇥r̄)x
Āx =
4⇡
r̄3
Now (m⇥r̄)x = my z̄ mz ȳ = my z mz y. So
V = v
µ0 (my z mz y)
4⇡
r̄3
Now x̄ = (x vt) = Rx , ȳ = y = Ry , z̄ = z = Rz , where R is the vector (in S) from the (instantaneous)
location of the dipole to the point of observation. Thus
r̄2 =
2
Rx2 + Ry2 + Rz2 =
2
(Rx2 + Ry2 + Rz2 ) + (1
2
)(Ry2 + Rz2 ) =
2
R2
v2 2 2
R sin ✓
c2
(where ✓ is the angle between R and the x-axis, so that Ry2 + Rz2 = R2 sin2 ✓).
)V =
µ0 v (my Rz mz Ry )
; v·(m⇥R) = v(m⇥R)x = v(my Rz
4⇡ 3 R3 1 v22 sin2 ✓ 3/2
c
2
V =
or, using µ0 =
1
✏0 c2
µ0 v·(m⇥R) 1 vc2
,
4⇡ R3 1 v22 sin2 ✓ 3/2
c
2
b
R·(v⇥m)
1 vc2
1
and v·(m⇥R) = R·(v⇥m): V =
4⇡✏0 c2 R2 1 v22 sin2 ✓ 3/2
c
(b) In the nonrelativistic limit (v 2 ⌧ c2 ):
V =
b
b
1 R·(v⇥m)
1 R·p
=
,
2
2
4⇡✏0 c R
4⇡✏0 R2
which is the potential of an electric dipole.
Problem 12.65
(a) B = µ20 Kŷ (Eq. 5.58); N = m⇥B (Eq. 6.1), so N =
µ0
N=
mKx̂ = µ20 ( vl2 )( v)x̂ = µ20 v 2 l2 x̂.
2
with p =
µ0
2 mK(ẑ⇥ŷ).
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v⇥m
,
c2
mz Ry ),
so
286
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
(b)
Charge density in the front side: 0 ( = 0 );
Charge density on the back side: ¯ = ¯ 0 , where v̄ =
1
so ¯ = q
4v 2 /c2
(1+v 2 /c2 )2
1
=q
(1 + v 2 /c2 )
2
1 + 2 vc2 +
v4
c4
2
4 vc2
1 + v 2 /c2
=q
2
1 2 vc2 +
v4
c4
=
(1 + v 2 /c2 )
=
(1 v 2 /c2 )
2v
1+v 2 /c2 ,
2
⇣
1+
v2 ⌘
c2
Length of front and back sides in this frame: l/ . So net charge on back side is:
l
q+ = ¯ =
2
Net charge on front side is:
⇣
q =
v2 ⌘
c2
1+
l
0
l
l
=
=
⇣
v2 ⌘
= 1+ 2 l
c
1
2
l
So dipole moment (note: charges on sides are equal):
⇣
l
v2 ⌘ l
(q ) ŷ = 1 + 2 l
2
c
2
l
p = (q+ ) ŷ
2
1
2
l
l
l2 ⇣
v2
ŷ =
1+ 2
2
2
c
1+
v2 ⌘
ŷ =
c2
l2 v 2
ŷ.
c2
l2 v 2
1 µ0
(ŷ⇥ẑ) =
l2 v 2 x̂.
c2 2✏0
2
So apart from the relativistic factor of the torque is the same in both systems — but in S it is the torque
exerted by a magnetic field on a magnetic dipole, whereas in S̄ it is the torque exerted by an electric field on
an electric dipole.
Problem 12.66
Choose axes so that E points in the z direction and B in the yz plane: E = (0, 0, E); B = (0, B cos , B sin ).
Go to a frame moving at speed v in the x direction:
E=
2✏0 ẑ,
0
where
=
0,
Ē = 0,
so N = p⇥E =
vB sin , (E + vB cos ) ; B̄ 0, (B cos +
(I used Eq. 12.109.) Parallel provided
vB sin
(E + vB cos )
=
,
v
(B cos + c2 E)
B sin
v
E (E + vB cos ) = EB cos + vB 2 cos2
c2
⇣
v
v2 ⌘
v
EB cos
=
0 = vB 2 + 2 E 2 + EB cos 1 + 2 ;
2
2
c
c
1 + v /c
B 2 + E 2 /c2
vB 2 sin2
Now E⇥B =
= B cos +
x̂
ŷ
0
0
0 B cos
ẑ
E
B sin
=
v
E), B sin
c2
EB cos . So
+
E⇥B
v
= 2
.
1 + v 2 /c2
B + E 2 /c2
.
or
v 2 v2
E + 2 EB cos
c2
c
qed
No, there can be no frame in which E ? B, for (E·B) is invariant, and since it is not zero in S it can’t be
zero in S̄.
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v☛ v☛
✕v ✕v
287
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.67
❖ ✻ ✗✻ ✗
❑ ❑ ❖
✕ ✕
✎☛
❯❲ ❄
❯✎☛❲ ❄
✲✲
−q−q
✕✗ ✻
✕❖❑✗ ✻
❖❑
✛✛
+q+q
Just before:
Field lines emanate
from present position
of particle.
✲✲
x x
☛ ☛
❯
✎ ✎❄ ❲❄ ❲ ❯
❖ ✻✻
▼ ▼❖ ✗ ✗
✍ ◆✍❲ ◆❄
❲✎ ❄
✌ ✎✌
!!
E E
t
c ✶ct✶
−−
++
E
✲✲
x x
E
!!
❖▼✻❖▼
◆ ✍◆✗ ✍✻
✗
✌ ✌ ❲ ❲
✎ ❄
✎ ❄
Just after : Field lines outside sphere of radius ct emanate from
position particle would have reached, had it kept going on its
original “flight plan”. Inside the sphere E = 0. On the surface the lines connect up (since they cannot simply terminate
in empty space), as suggested in the figure.
This produces a dense cluster of tangentially-directed field
lines, which expand with the spherical shell. This is a pictorial way of understanding the generation of electromagnetic
Problem 12.68
Equation 12.67 assumes the particle is (instantaneously) at rest in S. Here the particle is at rest in S̄. So
F? = 1 F̄? , Fk = F̄k . Using F̄ = q Ē, then,
Fx = F̄x = q Ēx ,
Fy =
1
F̄y =
1
Fz =
q Ēy ,
1
F̄z =
1
q Ēz .
Invoking Eq. 12.109:
Fx = qEx ,
But
v⇥B=
1
q (Ey
vBz x̂ + vBy ẑ,
Problem 12.69
z
✻
z̄
✻
=⇒v
✻
E
✰ ✰B
Fy =
✲y
✰
x̄
✲ ȳ
so
vBz ) = q(Ey
vBz ) Fz =
F = q(E + v ⇥ B).
1
q (Ez + vBy ) = q(Ez + vBy ).
qed
1
c ⃝2005
⃝2005
Pearson
Education,
Inc.,
Upper
NJ.zNJ.
All
rights
reserved.
This
material
is is
c Rewrite
Pearson
Education,
Inc.,
Upper
River,
reserved.
This
material
Eq.
12.109
with
y, River,
y!
z,
!All
x:rights
protected
under
lawslaws
as they
currently
exist.
No No
portion
of this
material
maymay
be be
protected
under
as they
currently
exist.
portion
of this
material
reproduced,
in any
form
or by
means,
without
permission
in writing
from
the the
publisher.
reproduced,
in any
form
or any
by any
means,
without
permission
in writing
from
publisher.
Ēy = Ey
Ēz = (Ez vBx )
Ēx = (Ex + vBz )
B̄y = By
B̄z =
⇣
Bz +
⌘
v
E
x
c2
B̄x =
⇣
Bx
⌘
v
E
z
c2
This gives the fields in system S̄ moving in the y direction at speed v.
x
Now E = (0, 0, E0 ); B = (B0 , 0, 0), so Ēy = 0, Ēz = (E0
vB0 ), Ēx = 0.
If we want Ē = 0, we must pick v so that E0 vB0 = 0; i.e. v = E0 /B0
(The condition Ez̄0 /B0 < c guarantees that there is no problem getting to such a system.)
✻
Rlaws
✶ as they currently exist. No portion of this material may be
✒
reproduced, in any form or by any means, without permission in writing from the publisher.
ωt
q
✰
x̄
✲ ȳ
z
z̄
✻
=⇒v
✻
✻
E
✲
✲ y ȳ
✰ AND RELATIVITY
CHAPTER 12. ELECTRODYNAMICS
✰ ✰B x̄
x
288
With this, B̄y = 0, B̄z = 0, B̄x = (B0
v
c2 E0 )
= B0 1
v2
c2
= B0
1
2
The trajectory in S̄: Since the particle started out at rest at the origin
in S, it started out with velocity vŷ in S̄. According to Eq. 12.71
it will move in a circle of radius R, given by
p = qBR, or mv = q
⇣1
B0
⌘
1
= 1 B0 ; B̄ =
z̄
B0 x̂.
✻
R✶
✒
m 2v
R) R=
.
qB0
ωt
✲ ȳ
q
✰
x̄
v
.
R
The trajectory in S: The Lorentz transformations Eqs. 12.18 and 12.19, for the case of relative motion in
The actual trajectory is given by x̄ = 0 ; ȳ =
R sin ! t̄ ; z̄ = R(1
cos ! t̄); where ! =
x̄ = x
ȳ = (y
z̄ = z
⇣
t̄ = t
x = x̄
vt)
y = (ȳ + v t̄)
z = z̄
⇣
v ⌘
v ⌘
y
t
=
t̄
+
ȳ
c2
c2
So the trajectory in S is given by:
⇢
h ⇣
x = 0; y = ( R sin ! t̄ + v t̄) =
R sin ! t
⇣
v2 ⌘
y 1+ 2 2
|
{z c }
2 y(1
v2
c2
z = R(1
2
+ vc2
)=
=
2
vt
2y
h
cos2 ! t̄) = R 1
cos !
⇣
t
R
v
c2 y
v ⌘i
y .
c2
h ⇣
v ⌘i
v ⌘i
y
;
z
=
R
R
cos
!
t
.
c2
c2
We can get rid of the trigonometric terms by the usual trick:
⇥
⇤
v
(y vt) = R sin
⇥ ! (t v c2⇤ y)
) 2 (y vt)2 + (z
z R = R cos ! (t c2 y)
So: x = 0; y = vt
h ⇣
sin ! t
h ⇣
R sin ! t
⇣
v ⌘
v ⌘i
y
+
v
t
y , or
c2
c2
v ⌘i)
h ⇣
y
c2
(y vt) = R sin ! t
⌘i
;
R)2 = R2 .
Absent the 2 , this would be the cycloid we found back in Ch. 5 (Eq. 5.9). The 2 makes it, as it were, an
elliptical cycloid — same picture as Fig. 5.7, but with the horizontal axis stretched out.
Problem 12.70
(a) D = ✏0 E + P suggests E ! ✏10 D
but it’s a little cleaner if we divide by µ0 while we’re at it, so that
H = µ10 B M suggests B ! µ0 H
c
⃝2005
Pearson Education, Inc., Upper Saddle River, NJ. Al
protected
under all copyright laws as they currently exist. N
8
9
in any form or by any means, without permissio
cDx cDy cDz reproduced,
>
>
E ! µ01✏0 D = c2 D, B ! H. Then:
> 0
>
<
=
cDx 0
Hz
Hy
µ⌫
D =
cDy Hz 0
Hx >
>
>
>
:
;
cDz Hy
Hx 0
protected under all copyright laws as they currently exist. No portion of this material may be
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289
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Then (following the derivation in Sect. 12.3.4):
@
@
1 @
@Dµ⌫
D0⌫ = cr·D = c⇢f = Jf0 ;
D1⌫ =
( cDx ) + (r⇥H)x = (Jf )x ; so
= Jfµ ,
⌫
⌫
@x
@x
c @t
@x⌫
where Jfµ = (c⇢f , Jf ). Meanwhile, the homogeneous Maxwell equations r·B = 0, E =
and hence
(b)
@Gµ⌫
= 0.
@x⌫
@B
@t
are unchanged,
8
9
0
Hx
Hy
Hz >
>
>
>
<
=
Hx 0
cDz cDy
=
Hy cDz
0
cDx >
>
>
>
:
;
Hz cDy cDx
0
H µ⌫
(c) If the material is at rest, ⌘⌫ = ( c, 0, 0, 0), and the sum over ⌫ collapses to a single term:
Dµ0 ⌘0 = c2 ✏F µ0 ⌘0 ) Dµ0 = c2 ✏F µ0 )
H µ0 ⌘0 =
1 µ0
1
G ⌘0 ) H µ0 = Gµ0 )
µ
µ
cD =
H=
c2 ✏
E
) D = ✏E (Eq. 4.32), X
c
1
1
B ) H = B (Eq. 6.31). X
µ
µ
(d) In general, ⌘⌫ = ( c, u), so, for µ = 0:
D0⌫ ⌘⌫ = D01 ⌘1 + D02 ⌘2 + D03 ⌘3 = cDx ( ux ) + cDy ( uy ) + cDz ( uz ) = c(D · u),
Ex
Ey
Ez
( ux ) +
( uy ) +
( uz ) = (E · u), so
c
c
c
c
⇣ ⌘
0⌫
2
0⌫
2
D ⌘⌫ = c ✏F ⌘⌫ ) c(D · u) = c ✏
(E · u) ) D · u = ✏(E · u).
c
F 0⌫ ⌘⌫ = F 01 ⌘1 + F 02 ⌘2 + F 03 ⌘3 =

H 0⌫ ⌘⌫ = H 01 ⌘1 + H 02 ⌘2 + H 03 ⌘3 = Hx ( ux ) + Hy ( uy ) + Hz ( uz ) = (H · u),
G0⌫ ⌘⌫ = G01 ⌘1 + G02 ⌘2 + G03 ⌘3 = Bx ( ux ) + By ( uy ) + Bz ( uz ) = (B · u), so
H 0⌫ ⌘⌫ =
1 0⌫
1
1
G ⌘⌫ ) (H · u) =
(B · u) ) H · u = (B · u).
µ
µ
µ

Similarly, for µ = 1:
D1⌫ ⌘⌫ = D10 ⌘0 + D12 ⌘2 + D13 ⌘3 = ( cDx )(
⇥ 2
⇤
=
c D + (u ⇥ H) x ,
c) + Hz ( uy ) + ( Hy )( uz ) = (c2 Dx + uy Hz
Ex
( c) + Bz ( uy ) + ( By )( uz ) = (Ex + uy Bz
c
D1⌫ ⌘⌫ = c2 ✏F 1⌫ ⌘⌫ )
F 1⌫ ⌘⌫ = F 10 ⌘0 + F 12 ⌘2 + F 13 ⌘3 =
=
[E + (u ⇥ B)]x ,
so
⇥ 2
⇤
1
c D + (u ⇥ H) x = c2 ✏ [E + (u ⇥ B)]x ) D + 2 (u ⇥ H) = ✏ [E + (u ⇥ B)] .
c
H 1⌫ ⌘⌫ = H 10 ⌘0 + H 12 ⌘2 + H 13 ⌘3 = ( Hx )( c) + ( cDz )( uy ) + (cDy )( uz )
= c(Hx uy Dz + uz Dy ) = c [H (u ⇥ D)]x ,
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u z Hy )
uz By )

290
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
◆
✓ ◆
Ez
Ey
G ⌘⌫ = G ⌘0 + G ⌘2 + G ⌘3 = ( Bx )( c) +
( uy ) +
( uz )
c
c
⇥ 2
⇤
1
= (c2 Bx uy Ez + uz Ey ) =
c B (u ⇥ E) x , so H 1⌫ ⌘⌫ = G1⌫ ⌘⌫ )
c
c
µ

⇤
1 ⇥ 2
1
1
c [H (u ⇥ D)]x =
c B (u ⇥ E) x ) H (u ⇥ D) =
B
(u ⇥ E) .
µc
µ
c2
1⌫
10
12
✓
13
Use Eq.  as an expression for H, plug this into Eq. , and solve for D:
⇢

1
1
1
D + 2 u ⇥ (u ⇥ D) +
B
(u ⇥ E)
= ✏ [E + (u ⇥ B)] ;
c
µ
c2
D+
1 ⇥
(u · D)u
c2
1
1
(u ⇥ B) + 4 [u ⇥ (u ⇥ E)] .
2
µc
µc
⇤
u2 D = ✏ [E + (u ⇥ B)]
Using Eq.  to rewrite (u · D):
✓
◆
⇤
u2
✏
1
1 ⇥
D 1
=
(E · u)u + ✏[E + (u ⇥ B)]
(u ⇥ B) + 4 (E · u)u u2 E
2
2
2
c
c
µc
µc
⇢


2
u
1
1
1
=✏ 1
E
1
(E · u)u + (u ⇥ B) 1
.
✏µc4
c2
✏µc2
✏µc2
Let
⌘p
1
1
u2 /c2
,
1
v⌘p .
✏µ
D=
2
✏
⇢✓
Then
u2 v 2
c4
1
◆
✓
E+ 1
v2
c2
◆
(u ⇥ B)
1
(E · u)u
c2
.
Now use Eq.  as an expression for D, plug this into Eq. , and solve for H:
⇢

1
1
1
H u⇥
(u
⇥
H)
+
✏[E
+
(u
⇥
B)]
=
B
(u ⇥ E) ;
2
c
µ
c2

⇤
1 ⇥
1
1
2
H + 2 (u · H)u u H =
B
(u ⇥ E) + ✏(u ⇥ E) + ✏[u ⇥ (u ⇥ B)].
c
µ
c2
Using Eq.  to rewrite (u · H):
✓
◆

⇥
u2
1
1
1
H 1
=
(B
·
u)u
+
B
(u ⇥ E) + ✏(u ⇥ E) + ✏ (B · u)u
2
2
2
c
µc
µ
c
⇢

⇥
⇤
1
1
=
1 µ✏u2 B + ✏µ
[(u ⇥ E) + (B · u)u] .
µ
c2
H=
2
µ
⇢✓
1
u2
v2
◆
B+
✓
1
v2
1
c2
◆
u2 B
⇤
[(u ⇥ E) + (B · u)u] .
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291
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.71
We know that (proper) power transforms as the zeroth component of a 4-vector K 0 =
2
1 dW
c d⌧
. The Larmor
1 2q
2
4⇡✏0 3 c3 a
formula says that for v = 0,
=
(Eq. 11.70). Can we think of a 4-vector whose zeroth component
reduces to this when the velocity is zero?
Well, a2 smells like (↵⌫ ↵⌫ ), but how do we get a 4-vector in here? How about ⌘ µ , whose zeroth component
is just c, when v = 0? Try, then:
1 2 q2 ⌫
Kµ =
(↵ ↵⌫ )⌘µ
4⇡✏0 3 c5
dW
d⌧
This has the right transformation properties, but we must check that it does reduce to the Larmor formula
when v = 0:
dW
1 dW
1
1 µ0 q 2 ⌫
dW
µ0 q 2 ⌫
0
0
=
= cK 0 = c
(↵
↵
)⌘
,
but
⌘
=
c
,
so
=
(↵ ↵⌫ ) . [Incidentally, this tells
⌫
dt
d⌧
6⇡c3
dt
6⇡c
us that the power itself (as opposed to proper power) is a scalar. If this had been obvious from the start, we
could simply have looked for a Lorentz scalar that generalizes the Larmor formula.]
In Prob. 12.39(b) we calculated (↵⌫ ↵⌫ ) in terms of ordinary velocity and acceleration:
h
i
(v·a)2 i
1
6
2
2
2
=
(v·a)
a
+
(c2 v 2 )
c2
h ⇣
⌘
i
n
v2
1
1⇥ 2 2
= 6 a2 1
+ 2 (v·a)2 = 6 a2
v a
2
c
c
c2
↵⌫ ↵⌫ =
4
h
a2 +
(v·a)2
Now v·a = va cos ✓, where ✓ is the angle between v and a, so:
⇤o
.
(v·a)2 = v 2 a2 (1 cos2 ✓) = v 2 a2 sin2 ✓ = |v⇥a|2 .
⇣
v⇥a 2 ⌘
↵⌫ ↵⌫ = 6 a2
.
c
dW
µ0 q 2 6 ⇣ 2
v⇥a 2 ⌘
=
a
, which is Liénard’s formula (Eq. 11.73).
dt
6⇡c
c
v 2 a2
Problem 12.72
(a) It’s inconsistent with the constraint ⌘µ K µ = 0 (Prob. 12.39(d)).
µ
⌫
µ
µ
(b) We want to find a 4-vector bµ with the property that d↵
d⌧ +b ⌘µ = 0. How
d⌧ ⌘⌫ ⌘ ? Then
d↵⌫
d↵µ
d↵⌫
d↵µ
d↵⌫
µ
µ
µ
2
2
c , so this becomes d⌧ ⌘µ
c  d⌧ ⌘⌫ , which is zero,
d⌧ + b ⌘µ = d⌧ ⌘µ +  d⌧ ⌘⌫ (⌘ ⌘µ ). But ⌘ ⌘µ =
⌘
2⇣
µ
⌫
µ0 q d↵
1 d↵
µ
if we pick  = 1/c2 . This suggests Krad
=
+ 2
⌘⌫ ⌘ µ . Note that ⌘ µ = (c, v) , so the spatial
6⇡c d⌧
c d⌧
components of bµ vanish in the nonrelativistic limit v ⌧ c, and hence this still reduces to the Abraham-Lorentz
⌫
d
d↵⌫
⌫ d⌘⌫
formula. [Incidentally, ↵⌫ ⌘⌫ = 0 ) d⌧
(↵⌫ ⌘⌫ ) = 0 ) d↵
↵⌫ ↵⌫ , and hence bµ can
d⌧ ⌘⌫ + ↵ d⌧ = 0, so d⌧ ⌘⌫ =
1
⌫
µ
just as well be written c2 (↵ ↵⌫ )⌘ .]
Problem 12.73
Define the electric current 4-vector as before: Jeµ = (c⇢e , Je ), and the magnetic current analogously:
µ
Jm = (c⇢m , Jm ). The fundamental laws are then
@⌫ F µ⌫ = µ0 Jeµ ,
@⌫ Gµ⌫ =
µ0 µ
J ,
c m
⇣
qm µ⌫ ⌘
K µ = qe F µ⌫ +
G
⌘⌫ .
c
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292
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
The first of these reproduces r·E = (1/✏0 )⇢e and r⇥B = µ0 Je + µ0 ✏0 @E/@t, just as before; the second
yields r·B = (µ0 /c)(c⇢m ) = µ0 ⇢m and (1/c)[@B/@t + r⇥E] = (µ0 /c)Jm , or r⇥E = µ0 Jm @B/@t
(generalizing Sec. 12.3.4). These are Maxwell’s equations with magnetic charge (Eq. 7.44). The third says
"
✓
◆
✓ ◆#
qe
qm
c
uy
Ez
uz
Ey
1
p
K = p
[E + (u ⇥ B)]x +
( Bx ) + p
+p
,
c
c
c
1 u2 /c2
1 u2 /c2
1 u2 /c2
1 u2 /c2
⇢

1
1
K = p
qe [E + (u ⇥ B)] + qm B
(u ⇥ E) , or
2
2
c2
1 u /c

1
F = qe [E + (u ⇥ B] + qm B
(u ⇥ E) ,
c2
which is the generalized Lorentz force law (Eq. 7.69).
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293
CONCORDANCE
4th ed
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.20
1.21
1.22
1.23
1.24
1.25
1.26
1.27
1.28
1.29
1.30
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.40
1.41
1.42
1.43
1.44
1.45
Same?
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X
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3rd ed
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
–
1.19
1.20
1.21
1.22
1.23
1.24
1.25
1.26
1.27
1.28
1.29
1.30
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.40
1.41
1.42
1.43
1.44
4th ed
1.46
1.47
1.48
1.49
1.50
1.51
1.52
1.53
1.54
1.55
1.56
1.57
1.58
1.59
1.60
1.61
1.62
1.63
1.64
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
2.21
2.22
2.23
2.24
2.25
2.26
Same?
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mod
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mod
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X
3rd ed
1.45
1.46
1.47
1.48
1.49
1.50
1.51
1.52
1.53
1.54
1.55
1.56
1.57
1.58
1.59
1.60
1.61
1.62
–
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
2.21
2.22
2.23
2.24
2.25
2.26
4th ed
2.27
2.28
2.29
2.30
2.31
2.32
2.33
2.34
2.35
2.36
2.37
2.38
2.39
2.40
2.41
2.42
2.43
2.44
2.45
2.46
2.47
2.48
2.49
2.50
2.51
2.52
2.53
2.54
2.55
2.56
2.57
2.58
2.59
2.60
2.61
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
Same?
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X
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reproduced, in any form or by any means, without permission in writing from the publisher.
3rd ed
2.27
2.28
2.29
2.30
2.31
–
–
2.32
2.33
2.34
–
2.35
2.36
–
2.37
2.38
2.39
2.40
2.41
–
2.43
2.44
2.45
2.46
–
2.47
2.48
2.49
2.50
2.51
2.52
–
–
–
–
3.1
3.2
3.3
–
3.4
3.5
3.6
3.7
3.8
3.9
294
4th ed
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.25
3.26
3.27
3.28
3.29
3.30
3.31
3.32
3.33
3.34
3.35
3.36
3.37
3.38
3.39
3.40
3.41
3.42
3.43
3.44
3.45
3.46
3.47
3.48
3.49
3.50
3.51
3.52
3.53
3.54
3.55
Same?
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3rd ed
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.25
3.26
–
3.27
3.28
3.29
3.30
3.31
3.32
–
3.33
–
–
3.35
3.36
–
–
3.37
3.38
3.39
3.40
3.41
3.42
–
3.43
3.44
3.45
3.46
3.47
3.48
CONCORDANCE
4th ed
3.56
3.57
3.58
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
4.14
4.15
4.16
4.17
4.18
4.19
4.20
4.21
4.22
4.23
4.24
4.25
4.26
4.27
4.28
4.29
4.30
4.31
4.32
4.33
4.34
4.35
4.36
4.37
4.38
4.39
4.40
4.41
4.42
Same?
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3rd ed
3.49
–
–
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
4.14
4.15
4.16
4.17
4.18
4.19
4.20
4.21
4.22
4.23
4.24
4.25
4.26
4.27
4.28
4.29
4.30
–
–
4.31
–
4.32
4.33
4.34
4.35
4.36
4.37
4.38
4.39
4th ed
4.43
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
5.12
5.13
5.14
5.15
5.16
5.17
5.18
5.19
5.20
5.21
5.22
5.23
5.24
5.25
5.26
5.27
5.28
5.29
5.30
5.31
5.32
5.33
5.34
5.35
5.36
5.37
5.38
5.39
5.40
5.41
5.42
5.43
5.44
Same?
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X
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X
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3rd ed
4.40
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
–
5.12
5.13
5.14
5.15
5.16
5.17
5.18
5.19
5.20
5.21
5.22
5.23
5.24
5.25
5.26
5.27
5.28
5.29
5.30
5.31
5.32
5.33
5.34
5.37
5.35, 5.36
5.60
–
5.38
5.39
5.40
5.41
5.42
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295
CONCORDANCE
4th ed
5.45
5.46
5.47
5.48
5.49
5.50
5.51
5.52
5.53
5.54
5.55
5.56
5.57
5.58
5.59
5.60
5.61
5.62
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
6.15
6.16
6.17
6.18
6.19
6.20
6.21
6.22
6.23
6.24
6.25
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3rd ed
5.43
5.44
5.46
5.47
5.48
5.49
–
5.50
5.51
5.52
5.53
5.54
5.55
5.56
5.57
5.58
5.59
5.61
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
6.15
6.16
6.17
6.18
6.19
6.20
6.21
6.22
6.25
–
6.23
4th ed
6.26
6.27
6.28
6.29
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
7.10
7.11
7.12
7.13
7.14
7.15
7.16
7.17
7.18
7.19
7.20
7.21
7.22
7.23
7.24
7.25
7.26
7.27
7.28
7.29
7.30
7.31
7.32
7.33
7.34
7.35
7.36
7.37
7.38
7.39
Same?
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3rd ed
6.24
6.26
6.27
6.28
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
7.10
7.11
7.12
7.13
7.14
7.15
7.16
7.17
7.18
7.19
–
–
7.20
7.21
7.22
7.23
7.24
7.25
7.26
7.27
7.28
7.29
7.30
–
7.31
7.32
7.33
7.34
7.35
7.36
4th ed
7.40
7.41
7.42
7.43
7.44
7.45
7.46
7.47
7.48
7.49
7.50
7.51
7.52
7.53
7.54
7.55
7.56
7.57
7.58
7.59
7.60
7.61
7.62
7.63
7.64
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
8.10
8.11
8.12
8.13
8.14
8.15
8.16
8.17
8.18
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protected under all copyright laws as they currently exist. No portion of this material may be
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3rd ed
7.37
7.39
7.41
7.57
7.42
7.43
7.44
7.45
7.46
7.47
7.48
–
7.49
7.50
–
7.51
7.52
7.53
7.54
–
7.55
7.56
7.58
7.59
7.60
8.1
8.2
8.3
8.4
–
8.6
8.5
8.7
–
8.8
–
–
8.9
–
–
8.10
8.11
–
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4th ed
8.19
8.20
8.21
8.22
8.23
8.24
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
9.11
9.12
9.13
9.14
9.15
9.16
9.17
9.18
9.19
9.20
9.21
9.22
9.23
9.24
9.25
9.26
9.27
9.28
9.29
9.30
9.31
9.32
9.33
9.34
9.35
9.36
9.37
9.38
9.39
9.40
Same?
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3rd ed
8.12
–
8.13
8.14
8.15
–
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
–
9.11
9.12
9.13
9.14
9.15
9.16
9.17
9.18
9.19
9.20
9.21
9.22
9.23
9.24
–
9.26
9.27
9.28
9.29
9.30
9.31
9.32
–
9.33
9.34
9.35
9.36
9.37
9.38
CONCORDANCE
4th ed
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9
10.10
10.11
10.12
10.13
10.14
10.15
10.16
10.17
10.18
10.19
10.20
10.21
10.22
10.23
10.24
10.25
10.26
10.27
10.28
10.29
10.30
10.31
10.32
10.33
10.34
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
11.10
11.11
11.12
Same?
X
X
mod
X
X
X
new
new
new
X
X
X
X
X
X
X
X
X
X
X
mod
X
X
X
X
new
X
X
new
new
X
X
new
new
X
X
X
X
X
X
X
new
X
X
X
X
3rd ed
10.1
10.2
10.3, 10.5
10.4
10.6
10.7
–
–
–
10.8
10.9
10.10
10.11
10.12
10.13
10.14
10.15
10.16
10.17
10.18
12.43a
10.19
10.20
10.21
10.22
–
10.23
10.24
–
–
10.25
10.26
–
–
11.1
11.2
11.3
11.4
11.5
11.6
11.7
–
11.8
11.9
11.12
11.10
4th ed
11.13
11.14
11.15
11.16
11.17
11.18
11.19
11.20
11.21
11.22
11.23
11.24
11.25
11.26
11.27
11.28
11.29
11.30
11.31
11.32
11.33
11.34
11.35
12.1
12.2
12.3
12.4
12.5
12.6
12.7
12.8
12.9
12.10
12.11
12.12
12.13
12.14
12.15
12.16
12.17
12.18
12.19
12.20
12.21
12.22
12.23
Same?
new
X
X
X
X
new
X
mod
new
X
X
X
X
new
new
X
X
X
X
X
X
X
new
X
X
X
X
X
X
X
X
X
X
X
X
X
mod
X
X
X
X
X
X
X
X
X
3rd ed
–
11.14
11.15
11.16
11.17
–
11.19
11.20
–
11.21
11.22
11.11
11.23
–
–
11.24
11.26
11.27
11.28
11.29
11.30
11.31
–
12.1
12.2
12.3
12.4
12.5
12.6
12.7
12.8
12.9
12.10
12.11
12.12
12.13
12.14
12.15
12.16
12.17
12.18
11.19
12.20
12.21
12.22
12.23
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
4th ed
12.24
12.25
12.26
12.27
12.28
12.29
12.30
12.31
12.32
12.33
12.34
12.35
12.36
12.37
12.38
12.39
12.40
12.41
12.42
12.43
12.44
12.45
12.46
12.47
12.48
Same?
X
X
mod
new
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
new
X
X
X
X
3rd ed
12.24
12.25
12.26
–
12.27
12.28
12.29
12.30
12.31
12.32
12.33
12.34
12.35
12.36
12.37
12.38
12.39
12.40
12.41
12.42
–
12.44
12.45
12.46
12.47
4th ed
12.49
12.50
12.51
12.52
12.53
12.54
12.55
12.56
12.57
12.58
12.59
12.60
12.61
12.62
12.63
12.64
12.65
12.66
12.67
12.68
12.69
12.70
12.71
12.72
12.73
Same?
X
X
X
X
X
X
X
X
X
new
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
3rd ed
12.48
12.49
12.50
12.51
12.52
12.53
12.54
12.55
12.56
–
12.57
12.58
12.59
12.60
12.61
12.62
12.63
12.64
12.65
12.66
12.67
12.68
12.69
12.70
12.71