Homework #5 Solutions
Q1.
The general solution has the following form
Φ(r ,
θ)=
∑
l=0
∞
(
A
lr
l+
B
lr
ll+1)
P
l(
cos
θ)
for
θ>π/
2 .
The potential beneath the disk has the same potential form due to symmetry.
Let's separate the space into two regions:
for
r
≥a
,
Φ(
r ,
θ)=
∑
l=0
∞
B
lr
ll+1P
l(
cos
θ)
and
for
r
<
a
Φ(
r ,θ)=
∑
l=0
∞
A
lP
l(
cos
θ)
On the other hand, we are able to derive the potential on the z-axis, which has the following form:
Φ(
r ,
0
)=
Q
2
πϵ
0a
(
√
1
+
(
r
a
)
2−
r
a
)
Now, we know that the Maclaurin series of
√
1
+
x
2=
1
+
∑
n=1
∞
(−
1
)
n(
2n
−
3
)! !
(
2n
)! !
x
2nhere
(
2n
−
3
)
! !=
1
⋅
3
⋅
5...
⋅(
2n
−
3
)
and
(
2n
)
!!
=
2
⋅
4
⋅
6 ...
⋅(
2n
)
√
1
+
(
r
a
)
=
1
+
∑
n=1∞
(
(−
1
)
n(
2n
−
3
)
! !
(
2n
)! !
(
r
a
)
2n
)
For
r
≥
a
,
Φ(
r ,
0
)=
Q
2
πϵ
0a
(
r
a
√
1
+
(
a
r
)
2
−
r
a
)
, there
Φ(
r ,
0
)=
Q
2
πϵ
0a
[
−
r
a
+
r
a
+
r
a
∑
n=0∞
(
(−
1
)
n(
2n
−
3
)! !
(
2n
)! !
(
r
a
)
−2n
)
]
=
Q
2
π ϵ
0a
∑
n=1∞
(
(−
1
)
n(
2n
−
3
)
! !
(
2n
)
!!
(
r
a
)
−2n+1
)
In the region of
r
≥
a
and
θ>π/
2
,
Φ(r ,
θ)=
Q
2
πϵ
0a
∑
n=1∞
(
(−
1
)
n(
2n
−
3
)
! !
(
2n
)
!!
(
a
r
)
2n−1Likewise, in the region
r
≤
a
and
θ>π/
2 ,
Φ(
r ,
θ)=
Q
2
πϵ
0a
[
1
−
r
a
cos
θ+
∑
n=1∞
(
(−
1
)
n(
2n
−
3
)
!!
(
2n
)
! !
(
r
a
)
2n
P
2n(
cos
θ)
)
]
Q2.
We use the Green function method to solve this problem. Recall that for a point charge within a conducting sphere, the Green function is (see Jackson, page 119)
G
D=
4
π
∑
l=0
∞
∑
m=−ll
1
2
l
+
l
[
r
<lr
l>+1−
(
r r '
)
lb
2l+1]
Y
lm*
(θ
' ,
ϕ
'
)
Y
lm(θ
,
ϕ)
Use the addition theorem of spherical harmonics
P
l(
cos
γ)=
4
π
2
l
+
1
m=−l∑
l
Y
lm*(θ
' ,
ϕ
'
)
Y
lm(θ
,
ϕ)
and consider the symmetry of the problem, the Green functioncan be written in terms of Legendre polynomials.
G
D=
∑
l=0
∞
[
r
<lr
>l+1−
(r r '
)
l
b
2l+1]
P
l(
cos
θ
'
)
P
l(
cos
θ)
The integral solution to the Direchlet boundary value problem is:
Φ(⃗
x
)=
1
4
π ϵ
0∫
ρ(⃗
x '
)
G
D(⃗
x ,
⃗
x '
)
d
τ−
1
4
π
∮
Φ
∂G
D∂n '
d a
in this problem, the potential on the boundary (
r
=
b
) is 0. Therefore, the potential everywhere isΦ(⃗
x
)=
1
4
π ϵ
0∫
ρ( ⃗
x '
)G
Dd
τ
The line charge distribution within the sphere can be written as:
ρ(
r ,
θ
,
ϕ)=
3
Q
8
π
d
3d
2−
r
2r
2[δ(
cos
θ−
1
)+δ (
cos
θ+
1
)]
if r < d andρ(
r ,
θ
,
ϕ)=
0
if r > d.The solution becomes:
Φ(⃗
x
)=
1
4
π ϵ
∫
0 2π∫
0
π
∫
0
∞
ρ(⃗
x '
)
G
Dr '
2sin
θ
' d r ' d
θ
' d
ϕ
'
Φ(⃗
x
)=
3
Q
16
πϵ
0d
3[
∫
0d
(
d
2−
r
2)
G
D(θ
'
= 0
)
d r '
+
∫
0
d
(
d
2−
r
2)
G
D(θ
'
=
π)
d r '
]
G
D(θ
'
=
0
)=
∑
l=0
∞
[
r
<l
r
>l+1−
(
r r '
)
l
G
D(θ
'
=
0
)=
∑
l=0
∞
[
r
<lr
>l+1−
(
r r '
)
l
b
2l+1]
(−
1
)
l
P
l(
cos
θ)
The final solution is evaluated as:
Φ(⃗
x
)=
3
Q
8
πϵ
0d
3∑
l=0,2,4..
∞
P
l(
cos
θ)
∫
0d
dr '
(
d
2−
r '
2)
[
r
l<r
>l+1
−
(
r r '
)
lb
2l+1]
if r > d
Φ(⃗
x
)=
3
Q
8
πϵ
0d
3∑
l=0,2,4..
∞
P
l(
cos
θ)
∫
0
d
dr '
(
d
2−r '
2)
[
r '
lr
l+1−
(
r r '
)
lb
2l+1]
Φ(⃗
x
)=
3
Q
8
πϵ
0d
3∑
l=0,2,4..
∞
P
l(
cos
θ)
d
l
(
l+
1
)(
l+
3
)
b
l+1[
(
b
r
)
l+1
−(
r
b
)
l
]
if r < d
Φ(⃗
x
)=
3
Q
8
πϵ
0d
3∑
l=0,2,4..
∞
P
l(
cos
θ)
[
d
2(
2
l
+
1
l
(l
+
1
)
)
+
r
2(
2
l
+
1
(
2
−l
)(
3
+l
)
)
+
r
l
(
1
d
l−2(
2
l
(l+
2
)
)
−
(
2
(l
+
3
)(
l+
1
)
d
l+3b
2l+1)
)
]
(b) The surface-charge density induced on the shell :
σ=
[
ϵ
0∂ Φ
∂
r
]
r=b
The potential is needed at the surface
r
=
b
which is greater than d, so we use ther
>
d
solutionσ=−
3
Q
4
π
b
2∑
l=0,2,4.. ∞
P
l(
cos
θ)
2
l
+
1
(l
+
1
)(
l+
3
)
(
d
b
)
l
(c)
in the limit thatd
≪
b
,d
/
b
≪
1
, so(d
/b)
l≈
0
except l = 0. The potential becomes:Φ(⃗
x
)=
Q
4
π ϵ
0[
1
r
−
1
b
]
This is just the potential due to a point charge Q at the center of the sphere. The induced charge is
σ=−
Q
4
π
b
2This is equivalent to a total charge -Q uniformly distributed on a sphere with radius b.
Q3.
V
(ϕ
, z
)=
∑
m , n
I
m(
n
π
L
b
)
(
a
nmsin
m
ϕ+b
mncos
m
ϕ
)
sin
(
n
L
π
z
)
The Fourier coefficients are:
a
mn=
2
π
L I
m(
n
π
b
/
L
)
∫
0 2πd
ϕ
∫
o L
d z V
(ϕ
, z)
sin
m
ϕ
sin
(
n
π
L
z
)
b
mn=
2
π
L I
m(
n
π
b
/
L
)
∫
0 2πd
ϕ
∫
o L
d z V
(ϕ
, z)
cos
m
ϕ
sin
(
n
π
L
z
)
(
b
0,n=
1
π
L I
0(
n
π
b
/
L
)
∫
0 2πd
ϕ
∫
o L
d z V
(ϕ
, z
)
sin
(
n
π
L
z
)
)
now, given the potential:
V
(ϕ
, z
)=
V for
−π/
2
<ϕ<π /
2
and
V
(ϕ
, z
)=−
V for
π /
2
<ϕ<
3
π /
2
all the coefficients for
a
mn vanish.b
mn=
2
V
π
L I
m(
n
π
b
/
L
)
[
−π /∫
2π /2
−
∫
π/2 3π /2
]
d
ϕ
cos
m
ϕ
∫
0L
d z
sin
n
π
z
L
=
2
V
π
2I
m(
n
π
b
/
L
)
4 sin
(
m
π/
2
)
m
1
−(−
1
)
nn
(
m≠
0
)
for m, n=even numbers
b
mn vanishes.b
mn can be written asb
2k+1,2l+1=
16
V
π
2I
2k+1
((
2l
+
1
)π
b
/
L)
(−
1
)
k(
2k
+
1
)(
2l
+
1
)
Therefore
Φ=
16
V
π
2∑
k=0
∞
∑
l=0
∞
(−
1
)
k(
2
k
+
1
)(
2
l
+
1
)
I
2k+1(
(
2
l+
1
)πρ
L
)
I
2k+1(
(
2
l
+
1
)π
b
L
)
cos
(
2
k
+
1
)ϕ
sin
(
2
l
+
1
)π
z
L
(b) For
L
≫
b
bothρ/
L
andb
/
L
are much less than 1.0. We utilize a small argument expansion of the modified Bessel function:L
ν(
x
)=
1
Γ(ν+
1
)
(
x
2
)
ν
and
sin
(
2
l
+
1
)π
z
L
=
sin
(
l
+
1
2
)π=(−
1
)
l
Hence in this limit, the potential form becomes
Φ=
16
V
π
2∑
k ,l
(−
1
)
k2
k
+
1
(−
1
)
l2
l
+
1
(
ρ
b
)
2k+1
cos
(
2
k
+
1
)ϕ=
16
V
π
2∑
l=0
∞
[
(−
1
)
l2
l
+
1
]
ℝ
[
(−
1
)
k2
k
+
1
(
ρ
b
e
iϕ
)
2k+1]
Taylor expansion for arctan is:
tan
−1(
z
)=
∑
n
(−
1
)
n2
n
+
1
z
2n+1ϕ=
16
V
π
2tan
−1
(
1
)ℝ
tan
−1(
ρ
b
e
iϕ
)=
4
V
π ℝ
tan
−1(
ρ
b
e
iϕ)
Use the following formula to calculate
ℝ
tan
−1(
z
)
tan
−1a
+
tan
−1b
=
tan
−1a
+
b
1
−
a b
ℝ
tan
−1(
z)=
1
2
(
tan
−1
z+
tan
−1z
*
)=
1
2
tan
−1
z
+z
*
1
−
z
∗
z
*
,henceΦ=
2
π
V
tan
−12
(ρ/
b
)
cos
ϕ
1
−(ρ/b)
2=
2
V
π
tan
−12
b
b
2ρ
cos
ϕ
−ρ
2Q4.
From lecture notes (Boundary-Value Problems in Electrostatics II) page 58, we know that the general solution to the Laplace equation in cylindrical coordinates has the form:
Φ(
r ,
ϕ
, z
)=
∑
k1,m
(
A
k1, msin
m
ϕ+
B
k1,mcos
m
ϕ)(
C
k1,me
k1z+
D
k1,m
e
−k1z
)[
F
k1,m
J
m(
k
1r
)+
G
k1,mN
m(
k
1r
)]
+
∑
k2,m
(
A
k2,msin
m
ϕ+
B
k2,mcos
m
ϕ)(
C
k2,msin
k
2z
+
D
k2, mcos
K
2z
)[
F
k2, mI
m(
k
2r
)+
G
k2, mK
m(
k
2r
)]
As z approaches infinity and
r
=
0
,Φ
must be finite. Together with the azimuthal symmetry of the solution, the potential becomes:Φ(r ,
ϕ
, z
)=
∑
k
A
kJ
0(k r
)e
−k z ,now in this problem, with no constraint on the radius, k becomes a continuous spectrum and not a discrete set, so
Φ(r ,
ϕ
, z
)=
∫
0∞
A
(k
)
J
0(k r
)e
−kzd k
Apply the last boundary condition
Φ(
z
=
0
)=
V
(
r
)
V
(
r
)=
∫
0∞
A(k
)
J
0(
kr)d k
Use the orthogonality of Bessel function:
∫
0
∞
x J
0(
k x
)
J
0(k ' x
)d x=
1
k
δ(
k '
−k
)
A
(
k
)=
k
∫
0∞
V
(
r
)
r J
0(
k r
)
d r
A(
k
)=V a J
1(k a)
The final solution is:
Φ(r ,
ϕ
, z
)=V a
∫
0∞
J
1(
k a)
J
0(k r
)e
−k zd k
(b) The potential a perpendicular distance z above the center of the disc is:
Φ(r=
0
)=V a
∫
0∞
J
1(k a
)
J
0(
0
)
d k
=V a
∫
0inivity
J
k(
ka)e
−k zdk
Use
J
1(
x
)=
2
1
π
i
∫
02π
Φ(
r
=
0
)=
1
2
π
i
v
∫
0 2πe
iθ∫
0
∞
e
(icosθ−z/a)xdx d
θ=
1
2
π
V
∫
0 2πe
iθ1
(
cos
θ+
i z
/
a
)
d
θ
Φ(
r
=
0
)=
i
2
π
V
[
∫
0∞
cos
θ+
sin
θ
z
/
a
(
cos
2θ+
z
2/
a
2)
d
θ+
i
∫
0∞
−
cos
θ
z
/
a
+
sin
θ
cos
θ
(
cos
2θ+
z
2/
a
2)
d
θ
]
FinThe potential can be simplified to:
Φ(r=
0
)=
1
2
π
V
∫
0 2πcos
2θ
cos
2θ+
z
2/
a
2d
θ
Substitute
u=
1
√
1
+a
2/
z
2tan
θ
Φ(
r
=
0
)=
V
[
1
−
2
π
2
√
z
2+
a
2∫
0
∞
1
1
+
u
2d u
]
Φ(r=
0
)=V
[
1
−
z
