Download Homework Assignment 6 Solution Set

Homework Assignment 6 Solution Set
PHYCS 4420
2 March, 2004
Problem 1 (Griffiths 4.26)
~ and D
~ fields everywhere and then
One way to find the energy is to find the E
~ depends only on
integrate the energy density for those fields. We know that D
the free charge and is therefore continuous across the dielectric boundary with
vacuum. Thus, the energy is
Z
1
~ ·E
~
D
W =
2 allspace
Z
Z
1 b
Q2
1 ∞
Q2
=
dτ
+
dτ
2 a (4πr2 )2
2 b 0 (4πr2 )2
Q2
1
1
1 1
=
−
+
8π0 1 + χe a b
b
2
Q
1
1 χe
=
+
.
8π0 1 + χe a
b
Problem 2 (Griffiths 4.32)
There is only one free charge: q. So inside and outside the dielectric we have
~ =
D
q
r̂.
4πr2
Inside the dielectric, then, we know
~ =
E
q
r̂
4π0 (1 + χe )r2
and
~ =
P~ = 0 χe E
χe q
r̂.
4π(1 + χe )r2
So the bound charges are
σb = P~ · r̂|r=R =
χe q
4π(1 + χe )R2
~ · P~ = 0 (except at r = 0).
ρb = −∇
χe
The total bound charge on the surface is just q 1+χ
, and it is compensated by
e
the volume charge that is localized immediately around the implanted charge.
1
Note also that the field outside the dielectric is the same as if the dielectric were
not present. The field is screened inside the dielectric but is otherwise unaffected
once you are outside the dielectric again.
Problem 3 (Griffiths 4.33 - optional)
~ is known at the interface from the boundary conditions
The behavior of E
~ D
~ and the free and bound charges.
that arise due to the relationships between E,
~
We know that the parallel component of E should be continous across the
boundary, but the perpendicular component has a discontinuity due to the sur~ only depends on free charges, so
face charges there. However, we know that D
~
the perpendicular component of D is continuous across the interface. Thus,
tan θ1
=
tan θ2
E1,k
E1,⊥
E2,k
E2,⊥
=
D2,⊥
2
D1,⊥
1
=
1
.
2
In the region 0 < θ < π2 , tan θ is everywhere positive and grows with increasing θ. Thus, electric field lines behave opposite to the analogous light waves in
media - they bend toward the normal when going from regions of large to
small . Therefore, a convex lens would be a diverging lens for electric field lines
(i.e., they would “defocus” the field).
Problem 4 (Griffiths 4.24 - optional)
This is a boundary value problem that is very similar to several others that
we have done. This time, however, we add another boundary and a few more
conditions based on the properties of dielectrics. The general solution is
X
B li
Vi (r, θ) =
Ali rli + li +1
Pli (cos θ)
r
li
where i denotes the region of interest (i = 1, 2, 3 for inside the conductor, inside
the dielectric, and outside respectively) with the following boundary conditions:
V3 (r → ∞)
V1
V1 (r = a)
V2 (r = b)
∂V2
|r=b
∂r
→
=
=
=
−E0 r cos θ
0 (for simplicity)
V2 (r = a) = 0
V3 (r = b)
∂V3
= 0
|r=b
∂r
2
(1)
(2)
(3)
(4)
(5)
From (1) we see that all Al3 vanish except for l3 = 1, for which A13 = −E0 .
Condition (4) then tells us that
B l3
1
bl3 +1
− B l2
1
bl2 +1
− Al2 bl2 = 0 (for l 6= 0)
and
l3 = l2 .
Thus,
(Bl3 − Bl2 )
1
− Al2 bl = 0.
bl+1
(6)
Now, condition (5) gives
r
X
l2
(l2 Al2 bl2 −1 − (l2 + 1)Bl2
!
1
b
)Pl2 (cos θ)
l2 +2
= −E0 cos θ +
X
−(l3 + 1)Bl3
l3
and so,
l+1
(r Bl2 − Bl3 ) − r lbl−1 Al2 = 0 (for l 6= 0).
bl+2
Now, take condition (3) which gives
Al2 = −Bl2
1
a2l+1
and combine it with results (6) & (7) above to get
b2l+1
Bl3 = Bl2 1 + 2l+1
a
lbl−1
l+1
Bl3 = Bl2 r
+
bl+2
a2l+1
=⇒ Bl3 = Bl2 = Al2 = 0 (for l 6= 0)
(7)
(8)
(9)
(10)
(11)
So, all that’s left is to find A12 and B12 (we’ll get B11 in the process). Go
back to conditions (4) and (5) and write them again, this time with l = 1
(substituting for A via (8)). We have
1
1
1
− 3
(12)
−E0 + B13 3 = B12
b
b3
a
2
2
1
E0 + B13 3 = B12 r
+ 3
(13)
3
b
b
a
−3E0 a3 b3
=⇒ B12 =
(14)
r (2a3 + b3 ) + 2b3 − 2a3 )
3E0 b3
1
(15)
=⇒ A12 = −B12 3 =
a
r (2a3 + b3 ) + 2b3 − 2a3 )
3
1
Pl (cos θ)
bl3 +2 3
THEREFORE....
3E0 b3
V2 (r, θ) =
r (2a3 + b3 ) + 2b3 − 2a3 )
a3
r− 2
r
cos θ
and, so,
~ r) = −∇V
~ (~r)
E(~
=−
3E0 b3
3
r (2a + b3 ) + 2b3 − 2a3 )
1+
2a3
r3
a3
cos θr̂ − 1 − 3 sin θθ̂ .
r
Apparently the screening by the dielectric makes it so that the electric field in
the dielectric has some θ dependence as you move away from r = a. Thus, the
bending of E0 is weaker than if we had just placed the conducting sphere there
without any dielectric.
Problem 5 (Griffiths 5.2)
We can just start from the result of Example 5.2
E
t + C3
B
z(t) = C2 cos ωt − C1 sin ωt + C4
y(t) = C1 cos ωt + C2 sin ωt +
and apply the given initial conditions to solve for the unknown constants. In
each case we have x(0) = y(0) = z(0) = 0 which tells us
C1 = −C3
C2 = −C4 .
a
dy
|t=0
dt
dz
|t=0
dt
=⇒
C2
C1
=⇒
=
E
B
=0
=0
=0
E
t
B
z(t) = 0.
y(t) =
In this case the magnetic and electric forces exactly cancel and the trajectory
is a straight line in the ŷ direction.
4
b
dy
E
|t=0 =
dt
2B
dz
|t=0 = 0
dt
=⇒
C2 = −
E
ω2B
C1 = 0
=⇒
E
E
sin ωt + t
ω2B
B
E
E
cos ωt +
.
z(t) = −
ω2B
ω2B
y(t) = −
Now the velocity in the ŷ direction is not enough to cause the magnetic force to
cancel with the electric force, so there is some “rolling” like in Example 5.2. The
trajectory is sketched below, with ~z as the verticle axis and ~y as the horizontal
Q
E
axis, both in units of B
with m
such that ω = 2πsec1 .
0.15
0.125
0.1
0.075
0.05
0.025
1
2
3
4
c
E
dy
|t=0 =
dt
B
dz
E
|t=0 =
dt
B
=⇒
C2 = 0
E
C1 = −
ωB
=⇒
E
E
E
y(t) = −
cos ωt + t +
ωB
B
ωB
E
z(t) =
sin ωt.
ωB
5
This is another cycloid, but centered along the y axis. The sketch is below in
the same units as the previous plot.
0.15
0.1
0.05
1
2
3
4
-0.05
-0.1
-0.15
Problem 6 (Griffiths 5.6)
a
~
K(r)
= σ(r)~v (r) = σωrθ̂
b
~ θ) = ρ~v (r, θ) = ρωr sin θφ̂
J(r,
Problem 7 (Griffiths 5.8)
a The field due to a finite straight segment of current is given in eq. 5.35
as
µ0 I
(sin θ2 − sin θ1 )
2πs
where s is the perpendicular distance from the current to the point and θ2 andθ1
are the angles to the ends of the current segment relative to the perpendicular
s. The direction is determined by the right hand rule. At the center of a square
loop the contribution to the magnetic field from each of the four sides points in
the same direction (normal to the plane of the loop), so the total field at the
center of a square of side 2R is just
~ = µ0 I sin π − sin − π
B
πR
4
4
√
2µ0 I
=
.
πR
~
|B(s)|
=
b Generalizing to more sides just gives
~ = N µ0 I 2 sin π
B
4πR
N
since sin θ − sin(−θ) = 2 sin θ.
6
c
As N → ∞ we get
lim
N →∞
µ0 I N
µ0 I sin x
µ0 I
sin θ = lim
=
.
x→0
2πR π
2R x
2R
This agrees with eq. 5.38, the field anywhere on the axis of a circular loop,
when z = 0.
Problem 8 (Griffiths 5.13)
a From Ampere’s law we get quite easily
~ inside = 0
B
~ outside = µ0 I θ̂
B
2πs
b
~ inside = µ0 Iencl = µ0
B
2πs
2πs
2
Cµ0 s
=
θ̂,
3
Z
s
Cs0 2πs0 ds0
0
but how do we find C? We know that the total current is I, so
Z a
Cs0 2πs0 ds0 = I
0
=⇒ C =
3
I.
2πa3
This gives
2
2
~ inside = µ0 s 3 I = µ0 Is θ̂
B
3 2πa3
2πa3
and, of course,
~ outside = µ0 I θ̂.
B
2πs
7