PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 4 3
b)
Let us ﬁnd the magnitude and direction of the electric ﬁeld at the srcin.By the azimuthal symmetry of the problem, the electric ﬁeld must point in the directionˆ
z
deﬁned by
θ
= 0. Computing rather directly, we see

E

=
−
∂ϕ
(
r,θ
)
∂r
r
=0
=
−
Q
8
π
0
∞
=1
2
+ 1
r
−
1
R
+1
[
P
+1
(cos
α
)
−
P
−
1
(cos
α
)]
P
(cos(0))
r
=0
,
=
−
Q
8
π
0
13
R
2
P
1
(1)[
P
2
(cos
α
)
−
P
0
(cos
α
)]
,
=
−
Q
8
π
0
13
R
2
12
3cos
2
α
−
1
−
1
,
=
−
Q
16
π
0
R
2
cos
2
α
−
1
/
3
−
2
/
3
,
∴
E
=
−
Q
sin
2
α
16
π
0
R
2
ˆ
z.
‘
´
oπρ
’
´
δι δ
ιξαι
c)
Let us brieﬂy discuss the limiting cases of the above results when
α
→
0
,π
.Because sin
2
(
α
) =
α
2
−
a
4
3
+
O
(
α
6
), and is
π
periodic, it is clear that the electric ﬁeldapproaches
E
=
−
Qα
2
16
π
20
ˆ
z
+
O
(
α
4
)
,
for
α
→
0 and
E
=
−
Q
(
π
−
α
)
2
16
π
20
ˆ
z
+
O
((
π
−
α
)
4
)
,
for
α
→
π.
This is expected. Notice that as the spherical charge distribution closes,
α
→
0, we approach the ﬁeld inside a closed sphere, which vanishes by Gauß’ law.The symmetric situation as
α
→
π
also begins to vanish because the total charge onthe sphere decreases like (
π
−
α
)
2
near
α
∼
π
. Therefore, the ﬁeld will decrease like(
π
−
α
)
2
.Similarly, for
α
→
0, cos
α
∼
1 and so
P
+1
(1)
−
P
−
1
(1) vanishes. Therefore, thepotential will vanish for
α
→
0 as expected for the interior of a charged sphere. For
α
→
π
, the potential will decrease like (
π
−
α
)
2
and will approach the potential of apoint charge of magnitude
Q/
8
πR
2
(
π
−
α
)
2
.
Problem 3.5
Let us consider the potential inside a sphere of radius
a
where the potential at the surface is speciﬁed.We are to demonstrate that
ϕ
(
r,θ,φ
) =
a
(
a
2
−
r
2
)4
π
ϕ
(
θ
,φ
)
d
Ω
(
r
2
+
a
2
−
2
ar
cos
γ
)
3
/
2
=
∞
=0
m
=
−
A
m
ra
Y
m
(
θ,φ
)
,
where cos
γ
= cos
θ
cos
θ
+ sin
θ
sin
θ
cos(
φ
−
φ
) and
A
m
=
d
Ω
Y
∗
m
(
θ
,φ
)
ϕ
(
θ
,φ
).From our work in Jackson’s second chapter, we know that the ﬁrst expression for the potential isthat obtained from the Green’s function
G
(
r,r
) = 1

r
−
r
 −
ar

r
−
a
2
r
2
r

.
Therefore, it is suﬃcient for us to show that the second expression for the potential is obtainable fromthe above Green’s function to show that the two expressions for the potential are equivalent.
4 JACOB LEWIS BOURJAILY
Let us begin by expressing
G
(
x,x
) in terms of spherical harmonics. Following the now ‘standard’procedure used above, we see that
G
(
r,r
) = 4
π
∞
=0
m
=
−
12
+ 1
Y
∗
m
(
θ
,φ
)
Y
m
(
θ,φ
)
r
r
+1
−
ar
r
a
2
r
2
r
+1
,
= 4
π
∞
=0
m
=
−
12
+ 1
Y
∗
m
(
θ
,φ
)
Y
m
(
θ,φ
)
r
r
+1
−
r
r
a
2
+1
.
To ﬁnd the potential, we must compute the normal derivative of the Green’s function in the directionof
r
, evaluated at
r
=
a
. Let us spend a moment and compute this.
∂G
(
r,r
)
∂r
r
=
a
= 4
π
∞
=0
m
=
−
12
+ 1
Y
∗
m
(
θ
,φ
)
Y
m
(
θ,φ
)
−
(
+ 1)
r
r
+2
−
r
−
1
r
a
2
+1
r
=
a
,
= 4
π
∞
=0
m
=
−
12
+ 1
Y
∗
m
(
θ
,φ
)
Y
m
(
θ,φ
)
−
(
+ 1)
r
a
+2
−
r
a
2
+2
,
=
−
4
π
∞
=0
m
=
−
12
+ 1
Y
∗
m
(
θ
,φ
)
Y
m
(
θ,φ
)(2
+ 1)
ra
1
a
2
,
=
−
4
π
∞
=0
m
=
−
Y
∗
m
(
θ
,φ
)
Y
m
(
θ,φ
)
ra
1
a
2
.
Now, we can directly compute the potential
ϕ
(
r,θ,φ
) using
ϕ
(
r,θ,φ
) =
−
14
π
ϕ
(
θ
,φ
)
∂G
(
r,r
)
∂r
r
=
a
a
2
d
Ω
,
= 14
π
ϕ
(
θ
,φ
)4
π
∞
=0
m
=
−
Y
∗
m
(
θ
,φ
)
Y
m
(
θ,φ
)
ra
1
a
2
a
2
d
Ω
,
=
ϕ
(
θ
,φ
)
∞
=0
m
=
−
Y
∗
m
(
θ
,φ
)
Y
m
(
θ,φ
)
ra
d
Ω
,
=
∞
=0
m
=
−
Y
m
(
θ,φ
)
ra
ϕ
(
θ
,φ
)
Y
∗
m
(
θ
,φ
)
d
Ω
,
∴
ϕ
(
r,θ,φ
) =
∞
=0
m
=
−
A
m
ra
Y
m
(
θ,φ
)
.
‘
´
oπρ
’
´
δι δ
ιξαι
Problem 3.6
Let us consider a system of two point charges of charge
±
q
located at
z
=
±
a
, respectively.
a)
Let us ﬁnd the electrostatic potential as an expansion in spherical harmonics and powers of
r
.First notice that the electrostatic potential is trivially given by
ϕ
(
x
) =
q
4
π
0
1

x
−
a
 −
1

x
+
a

,
where
a
≡
(
a,
0
,
0) in spherical coordinates. As before, we can expand the function1
/

x
−
a

following Jackson’s equation (3.70). Furthermore, by azimuthal symmetry,it is clear that only
m
= 0 spherical harmonics contribute and so we can substitute Legendre polynomials in their place. Azimuthal symmetry also implies that