Download Homework Assignment 2 Solution Set

Homework Assignment 2 Solution Set
PHYCS 4420
30 January, 2004
Problem 1 (Griffiths 1.35)
a From product rule (v) we know
~ × A)
~ =A
~ × (∇f
~ )+∇
~ × (f A).
~
f (∇
Integrating both sides over any surface (closed or not) gives
Z
Z
Z
~
~
~
~
~ × (f A))
~ · d~a.
f (∇ × A) · d~a = (A × (∇f )) · d~a + (∇
S
S
S
The second integral on the right may be changed according to the fundamental
theorem for curls, Stokes’ Theorem. Thus,
Z
Z
I
~ × A)
~ · d~a = (A
~ × (∇f
~ )) · d~a + (f A)
~ · d~l.
f (∇
S
S
P
b From product rule (iv) we know
~ · (∇
~ × A)
~ =A
~ · (∇
~ × B)
~ +∇
~ · (A
~ × B).
~
B
We just need to integrate both sides over any volume and transform the second
integral on the right using the fundamental theorem for divergence, Gauss’ Law,
to obtained the desired result:
Z
Z
I
~ · (∇
~ × A)dV
~
~ · (∇
~ × B)dV
~
~ × B)
~ · d~a.
B
=
A
+ (A
V
V
S
Problem 2 (Griffiths 1.39)
~v = (r cos θ)r̂ + (r sin θ)θ̂ + (r sin θ cos φ)φ̂
∂
1
∂
~ · ~v = 1 ∂ (r2 vr ) + 1
∇
(sin θvθ ) +
(vφ )
2
r ∂r
r sin θ ∂θ
r sin θ ∂φ
1 ∂
1
∂
1
∂
= 2 (r3 cos θ) +
(r sin2 θ) +
(r sin θ cos φ)
r ∂r
r sin θ ∂θ
r sin θ ∂φ
= 3 cos θ + 2 cos θ − sin φ = 5 cos θ − sin φ
1
We’d like to verify the divergence law
Z
I
~ · ~v dV =
∇
~v · d~a
V
S
for a hemisphere of radius R whose base is centered at the origin and lies in the
xy-plane. For the left hand side we integrate over all three variables using the
volume element dV = r2 sin θdrdθdφ.
Z
Z R Z π2
Z 2π
~ · ~v dV =
∇
dr
dθ
dφ(5 cos θ − sin φ)r2 sin θ
V
0
0
R3
=
3
Z
π
2
0
Z
2π
dφ(5 cos θ sin θ − sin φ sin θ)
θ= π2 R3 2π
5
2
=
dφ
sin θ + cos θ sin φ
3 0
2
θ=0
Z
R3 2π
5
=
dφ( − sin φ)
3 0
2
R3 5
5πR3
=
2π =
3 2
3
For the right hand side we need two integrals. The surface element along the
hemisphere is d~a = 4πr2 sin θdθdφr̂, with r = R everywhere along the surface.
We also need to include the base, though. There we should hold θ = π2 , and our
surface element is then d~a = rdrdφθ̂. Thus,
I
Z π2
Z 2π
Z R Z 2π
~v · d~a =
dθ
dφR2 sin θvr +
dr
dφrvθ
dθ
0
0
Z
S
0
Z
0
π
2
Z
dθ
=
0
= πR3 +
0
2π
0
Z
3
dφR sin θ cos θ +
0
R
Z
2π
dr
0
dφr2 sin θ
0
2πR3
5πR3
=
.
3
3
Problem 3 (Griffiths 1.42(c))
~v = s(2 + sin2 φ)ŝ + s sin φ cos φφ̂ + 3z ẑ
1 ∂vz
∂vφ
∂vs
∂vz
1 ∂svφ
∂s
~
∇ × ~v = ŝ
−
+ φ̂
−
+ ẑ (
−
)
s ∂φ
∂z
∂z
∂s
s ∂s
∂φ
1 ∂
∂
∂
∂
2
= ŝ
(3z) −
(s sin φ cos φ) + φ̂
(s(2 + sin φ)) −
(3z)
s ∂φ
∂z
∂z
∂s
1 ∂
∂
(s(2 + sin2 φ)))
+ẑ ( (s2 sin φ cos φ) −
s ∂s
∂φ
1
= ŝ [0 − 0] + φ̂ [0 − 0] + ẑ (2s sin φ cos φ − 2s sin φ cos φ
s
=0
2
Problem 4 (Griffiths 2.4)
From example 2.1 in the text we see that the electric field a distance r from
a line of uniformly distributed charge of length 2L is
~ r) =
E(~
1
2λL
√
r̂
4π0 r r2 + L2
where ~r points directly away from the center of the line and perpindicular to it.
So, to find the field a distance z from the center of the square loop shown in the
figure we need to just sum up four such electric fields.
The p
distance from the point P to any of the four sides of the square will be
r = z 2 + ( a2 )2 . Each contribution to the electric field will have a component
in the z direction as well as a component parallel to the plane of the square
loop. However, these parallel components sum to zero because of the symmetry
of the loop. Therefore, we just need to add up the z components to find the
~
total E:
1
2λL
~
√
E(z) = 4
sin θr ẑ
4π0 r r2 + L2
2λ a2
1
z
p
ẑ
=
π0 r r2 + ( a2 )2 r
=
1
λaz
q
π0 (z 2 + a2 ) z 2 +
4
ẑ.
a2
2
Problem 5 (Griffiths 2.12)
Gauss’ Law for electric fields tells us
I
Z
Z
1
qencl
~
~
~
~
E · dA =
∇ · EdV =
ρdV =
0
0 V
S
V
regardless of the bounding surface. We should pick a surface (our ”Gaussian
surface”) that uses the symmetry of the physical problem to our advantage.
For a sphere of uniformly distributed charge the electric field on the surface of
any concentric sphere will be radially symmetric. So, the surface integral of
3
the electric field over any such sphere will be the constant value of the field
times the area of the sphere. The amount of charge enclosed by such a sphere
is qencl = ρ 43 πr3 , where r is the radius of our Gaussian sphere (r < R).
I
~ · dA
~ = qencl
E
0
S
4
ρ
πr3
E(r)4πr2 = 3
.
0
So,
ρr
~
r̂.
E(r)
=
30
Problem 6 (Griffiths 2.16)
For each region we should choose a cylinder as our Guassian surface since
the charge distribution is cylindrically symmetric. There will be no component
of electric field in the axial direction, so our surface integral will only include
the curved surface of whatever cylinder we pick (the ends are perpindicular to
the field lines, so the surface integral is zero for them). For all three regions we
will get
I
~ · dA
~ = qencl
E
0
S
qencl
E(s)2πsl =
0
where l is the length (arbitrary) of whatever Gaussian cylinder we construct.
i
s<a
4
Inside the inner cylinder the charge enclosed depends on the radius s of our
Gaussian cylinder as qencl = ρπs2 l. Gauss’ Law becomes
E(s)2πsl =
so
ρπs2 l
0
ρs
~
E(s)
=
ŝ.
20
ii
a<s<b
Between the inner cylinder and outer shell the charge enclosed by the coaxial
Gaussian cylinder is the total charge of a length l of the inner cylinder.
E(s)2πsl =
so
ρπa2 l
0
ρa2
~
ŝ.
E(s)
=
20 s
iii s > b Everywhere outside the outer shell the total charge enclosed by a
~
coaxial Gaussian cylinder is always zero. Therefore, E(s)
= 0.
~
Here is the plot (in arbitrary units) of |E(s)|
with a = 1 and b = 3:
1.4
1.2
1
0.8
0.6
0.4
0.2
1
2
5
3
4
Problem 7 (Griffiths 2.17)
Since charge is distributed infinitely in the xz-plane we know the electric
field will only be a function of y. We can choose a variety of Gaussian surfaces,
but whatever surface we choose should have boundaries that are parallel and
perpindicular to the xz-plane. Let’s use a ”pillbox” (as it is called in the text)
centered at the origin with area A parallel to the xz-plane. In the y direction our
surface should have dimension 2y. This insures that the electric field strength is
uniform over two surfaces (the two perpindicular to the y-axis), and the surface
integral over the other four surfaces is zero. Therefore, Gauss’ Law will give us
E(y)2A =
qencl
.
0
i y<d
For a Gaussian pillbox inside the slab the charge enclosed is qencl = ρA2y.
We have
ρA2y
E(y)2A =
0
so
ρy
~
E(y)
=
ŷ
0
ii y > d
Outside the slab our Gaussian pillbox encloses an amount of charge limited
by the thicness of the slab: qencl = ρA2d. Therefore,
E(y)2A =
so
ρA2d
0
ρd
~
E(y)
=
ŷ
0
Here is a plot of the electric field due to the slab. The verticle axis is the
electric field strength in units of ρd
0 and the horizontal axis is the distance in
6
the y direction in units of d.
E
3
2
1
-3
-2
-1
1
2
3
y
-1
-2
-3
Note the similarities among the results of Problems 5, 6 and 7. In each case
the electric field inside a uniform charge distribution is linear in some coordinate. Outside the charge distribution the field weakens differently depending on
the geometry of the situation.
7