Homework Assignment 5 Solution Set
PHYCS 4420
23 February, 2004
Problem 1 (Griffiths 3.8)
The first image charge guarantees a potential of zero on the surface. The
second image charge won’t change the contribution to the potential from the
first, so we should just put a charge at the center of the sphere to make the
potential some uniform constant over the sphere’s surface. For a potential of V0
we should use a charge of
qimage2 = 4π0 V0 R
where R is the radius of the sphere.
The force between the real charge q and the conducting sphere will be the
same as the force felt by the real charge as due to the two image charges. If the
sphere is neutral, then all the image charges inside should add to zero (see the
text for a convincing argument of this almost obvious fact). Thus,
!
q
q
q
image2
image1
F~q (~r) =
−
r̂
2 2
4π0
r2
r − Rr
!
q
qR
qR
=
r̂
−
2 2
4π0 r3
r r − Rr
3 2
R − 2r2
q
R
=
r̂
4π0 r
(R2 − r2 )2
where r is the distance from the center of the sphere to the physical charge
outside. Since r > R the force of attraction is
3 2
R
r − 2R2
q
~
|F | =
4π0 r
(r2 − R2 )2
Problem 2 (Griffiths 3.13 - optional)
The charge density determines the field just outside the strip. Thus,
σ(y, x = 0) = −0
∂V
∂V
= −0
∂n
∂x
1
!
nπy 4V0 X 1 − nπx
e a sin
π
n
a
x=0
oddn
!
40 V0 X −nπ − nπx
nπy =−
e a sin
π
na
a
x=0
oddn
!
40 V0 X
nπy
=
sin
a
a
∂
= −0
∂x
oddn
Problem 3 (Griffiths 3.17(b))
Fortunately we know the answer before we start, since a uniform surface
charge distribution is exactly what we get from a charged conducting sphere.
2
Outside we should have V (~r) = σR
0 r and inside we should get a constant V =
σR
0 . However, the book wants us to use the example and solve it as a boundary
value problem.
We know from Example 3.9 that the general solution for a fixed, spherical
surface charge density is
X
V (r, θ)inside =
Al rl Pl (cos θ)
l
V (r, θ)outside =
X
Bl r−(l+1) Pl (cos θ)
l
with
Al =
1
20 Rl−1
Z
π
σPl (cos θ) sin θdθ
0
Bl = Al R2l+1 .
In this case, though, σ is a constant. Thus, the orthogonality of the Legendre
polynomials insures us that
σR
A0 =
0
and all other Al vanish. Thus,
B0 =
σR2
0
and all other Bl vanish. Therefore, our potential is
σR
;r < R
0
σR2
V (r, θ) =
;r > R
0 r
V (r, θ) =
which is just as we supposed to begin with.
2
Problem 4 (Griffiths 3.18)
Again, the general solution is
V (r, θ) =
∞ X
Al rl + Bl r−(l+1) Pl (cos θ)
l=0
with the following boundary conditions:
V (r → ∞) → 0
V (r → 0) = f inite
V (R, θ)outside = V (R, θ)inside = k cos 3θ.
(1)
(2)
(3)
From conditions (1) and (2) we have
V (r, θ)outside =
∞
X
Bl r−(l+1) Pl (cos θ)
l=0
and
V (r, θ)inside =
∞
X
Al rl Pl (cos θ).
l=0
From condition (3) and the results above we see that
Al Rl = Bl R−(l+1)
⇒ Al = Bl R−(2l+1)
and
V (R, θ) =
∞
X
Bl R−(2l+1) Rl Pl (cos θ) = k cos 3θ.
l=0
However, we can write cos 3θ in terms of the Legendre polynomials as
cos 3θ = 4 cos3 θ − 3 cos θ
8
3
= P3 (cos θ) − P1 (cos θ).
5
5
Therefore, all coefficients of orders other than 1 and 3 must vanish and we have
B1
1
1
3k
8k
P3 (cos θ).
P1 (cos θ) + B3 4 P3 (cos θ) = − P1 (cos θ) +
R2
R
5
5
from which we can get all of our coefficients (because of the relationship between
the Al and Bl we found before). Therefore our potential is
r3
12 r3
3r
3
cos
θ
−
k
cos θ − k
cos θ
3
3
R
5 R
5R
R4
12 R4
3 R2
= 4k 4 cos3 θ − k
cos θ − k
cos θ.
4
r
5 r
5 r2
V (r, θ)inside = 4k
V (r, θ)outside
3
To find the surface charge density we just need to find the discontinuity in
the gradient of the potential. We know that
σ
∂Vout
∂Vin = .
−
+
∂r
∂r r=R
0
So, from eq. 3.83 in the book we get
X
σ(θ) = 0
(2l + 1)Al Rl−1 Pl (cos θ)
l
9k0
28k0
840
=−
cos θ +
cos3 θ −
5R
R
5R
k0
=
140 cos3 θ − 93 cos θ .
5R
Problem 5 (Griffiths 3.20 - optional)
With an arbitrary charge on the sphere the problem differs very little from
Example 3.8 in the text. There is just one more boundary condition. The
integral of the surface charge density over the total area of the sphere should
return Q. You may be able to convince yourself that this amounts to changing
the surface charge density in the example (σ = 30 E0 cos θ) by a constant, since
the distribution of the extra charge should be the same as the distribution in
the neutral case. After all, Q=0 is just a particular case of the general solution.
So, we can only add a boundary condition that affects the l = 0 term in the
potential outside the sphere. The only such addition that integrates to give us
1
Q is V ∗ = 4π
Q. Therefore, the new potential outside the sphere is
0
V (r, θ) = −E0
R3
r− 2
r
cos θ +
1 Q
.
4π0 r
It’s as though we added the charge as a point charge at the center of the sphere!
This should make sense if you did the image charge problem correctly above
(Griffiths 3.8).
Problem 6 (Griffiths 4.13)
We could try to do this one by solving Laplace’s equation in cylindrical
coordinates using separation of variables, but that seems a little out of the
scope of what you should be expected to know at this point. However, we can
take a hint from Example 4.3 in the text and treat the uniformly polarized
cylinder as two superimposed cylinders of equal and opposite uniform charge
distribution separated by some vector d~ whose size is small. Each cylinder gives
us a field inside of
~ si )in = ρi s~i
E(~
20
and a field outside of
2
~ si )out = ρi a sˆi
E(~
20 si
4
where i denotes which of the two cylinders we are considering. Let the origin be
directly between the centers of the two overlapping cylinders. Thus, the total
field inside is
~ s) = ρs~1 − ρs~2
E(~
20
20
ρ
=
(s~1 − s~2 )
20
ρ ~
d.
=
20
Outside the sphere we get
2
2
~ s) = ρa sˆ1 − ρa sˆ2
E(~
20 s1
20 s2
2
sˆ1
sˆ2
ρa
−
=
20 s1
s2
We’d like to put this in terms of the vector ~s from the origin. We have
s~1 = ~s +
and
d~
2
d~
s~2 = ~s − .
2
Thus,
2
~ s) = ρa
E(~
20
ρa2
=
20
s~1
s~2
− 2
s21
s2
~s −
|~s −
d~
2
d~ 2
2|
−
~s +
|~s +
d~
2
d~ 2
2|
!
.
Here we need some trickery (I got stuck a little myself). We may write
d2
d~
d~
+ 2~s · )−1
|~s + |−2 = (s2 +
2
4
2
~
1
~s · d
≈ 2 (1 + 2 )−1
s
s
1
~s · d~
≈ 2 (1 − 2 )
s
s
and similarly
d~
1
~s · d~
|~s − |−2 ≈ 2 (1 + 2 )
2
s
s
5
Here we have ignored nonlinear terms of d because it is so small, and we have
~
s·d~
also expanded a binomial expansion in 2s
2 and truncated the expansion for the
same reason. Finally, combining everything together we have
!
2
~
~
~
~
d
d
~
s
·
d
~
s
·
d
ρa
1
~ s) =
E(~
(~s − )(1 − 2 ) − (~s + )(1 + 2 )
20 s2
2
s
2
s
!
~
ρa2
2~s(~s · d)
=
+ d~
−
20 s2
s2
a2 =
2ŝ(ŝ · P~ ) − P~
2
20 s
because d~ (as I chose it, for whatever reason) points from the positively charged
axis to the negatively charged axis.
Problem 7 (Griffiths 4.16(b) & (c))
b The ends of such a long cavity will not contribute very much to the field
inside. Therefore, the field inside the needle-shaped cavity will be dominated
by the boundary conditions along the long boundary, which is parallel to P~ .
The electric field at this boundary is continuous (because it is parallel to the
~ 0 . The displacement field, however,
surface), so the field in the cavity is just E
~ = 0 E
~ 0.
must be different. It becomes D
~ 0 are insignifc In this case, the ends dominate and the walls parallel to E
~
~
~ =D
~0
icant. Therefore, E is most certainly not continuous, but D is. Thus, D
1 ~
~
and E = 0 D0 .
ALTERNATE METHOD
Following the hint in the problem, treat the cavities as superpositions of
oppositely polarized dielectrics. Then, the needle like cavity resembles a single
~ 0 . The disc-like cavity resemdipole, cancelling the displacement but leaving E
~
bles a capacitor with charge density given by the polarization. In that case D
~
is the same, but E is not. You get the same results either way you look at it.
Problem 8 (Griffiths 4.18 (a),(d) & (e))
a The (free) charge density on the plates is fixed, and there are no free
~ is uniform throughout both dielectrics and
charges in the dielectrics. Thus, D
is given by
I
~ · dA
~ = Qf ree−enclosed
D
DA = σf ree A
~ = σâ
D
6
where â is the vector pointing from the positively charged plate to the negatively
charged plate (−ẑ).
~ and we know how E
~ is related to D.
~
d We know how V is related to E,
So, we can just use the result of part (a) to show that
Z 2a
1 ~
1 ~
D · d~z −
D · ~z
2
a
0 1
Z a
Z 2a
1
1
=
σdz +
σdz
2
1.5
0
0
0
a
7a
=
σ
60
Z
a
∆V = −
e
In slab 1 we have
~ = 0 E
~ + P~
D
1
⇒ P~ = σẑ.
2
Therefore, inside slab 1 we have
~ · P~ = 0
ρb = −∇
and at the lower and upper interfaces respectively we get
1
σb = P~ · (∓)ẑ = ± σ.
2
In slab 2 we have
~ = 0 E
~ + P~
D
1
⇒ P~ = σẑ.
3
So,
~ · P~ = 0
ρb = −∇
and at the lower and upper interfaces respectively we get
1
σb = P~ · (∓)ẑ = ± σ.
3
Problem 9 (Griffiths 4.21)
Place an arbitrary linear charge density λ on the inner wire and calculate
the potential difference between the wire and the outer shell. This free charge
~ everywhere between the two conductors as seen in
gives rise to a uniform D
numerous examples:
~ r) = λ r̂.
D(~
2πr
7
The potential difference between the conductors is then
b
Z c
λ
λ
dr +
dr
2π
r
2π
0
0 r r
a
b
λ
b
1 c
.
=
ln
+ ln
2π0
a
r
b
Z
∆V =
So, the capacitance per unit length is
Cl =
=
λ
Vλ
ln
b
a
8
2π0
+ 1r ln
c
b