PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 4
3
b) Let us find the magnitude and direction of the electric field at the origin.
By the azimuthal symmetry of the problem, the electric field must point in the direction
ẑ defined by θ = 0. Computing rather directly, we see
¯
¯
∞
¯
∂ϕ(r, θ) ¯¯
Q X ` r`−1
¯
|E | = −
=−
[P
(cos
α)
−
P
(cos
α)]
P
(cos(0))
,
¯
`+1
`−1
`
¯
`+1
¯
∂r
8π²
2`
+
1
R
0
r=0
`=1
r=0
Q 1
=−
P1 (1) [P2 (cos α) − P0 (cos α)] ,
8π²0 3R2
µ
¶
¢
Q 1
1¡
2
3
cos
=−
α
−
1
−
1
,
8π²0 3R2 2
¡ 2
¢
Q
cos α − 1/3 − 2/3 ,
=−
16π²0 R2
∴E =−
Q sin2 α
ẑ.
16π²0 R2
‘
’
óπ²ρ
²́δ²ι
δ²ιξαι
c) Let us briefly discuss the limiting cases of the above results when α → 0, π.
Because sin2 (α) = α2 −
approaches
a4
3
+ O(α6 ), and is π-periodic, it is clear that the electric field
E =−
Qα2
ẑ + O(α4 ),
16π²20
for α → 0 and
E =−
Q(π − α)2
ẑ + O((π − α)4 ),
16π²20
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 on
the 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, the
potential 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 a
point charge of magnitude Q/8πR2 (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
Z
∞ X̀
³ r ´`
X
a(a2 − r2 )
ϕ(θ0 , φ0 )dΩ0
ϕ(r, θ, φ) =
=
A
Y`m (θ, φ),
`m
4π
a
(r2 + a2 − 2ar cos γ)3/2
`=0 m=−`
where cos γ = cos θ cos θ0 + sin θ sin θ0 cos(φ − φ0 ) and A`m =
R
∗
(θ0 , φ0 )ϕ(θ0 , φ0 ).
dΩ0 Y`m
From our work in Jackson’s second chapter, we know that the first expression for the potential is
that obtained from the Green’s function
G(r, r0 ) =
1
a
−
.
|~r − ~r0 | r0 |~r − ra022 ~r0 |
Therefore, it is sufficient for us to show that the second expression for the potential is obtainable from
the 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, x0 ) in terms of spherical harmonics. Following the now ‘standard’
procedure used above, we see that
Ã
!
∞ X̀
X
1
r`
a
r`
0
∗
0
0
G(r, r ) = 4π
Y (θ , φ )Y`m (θ, φ)
− 0 ¡ 2 ¢`+1 ,
2` + 1 `m
r0`+1
r a r0
`=0 m=−`
r 02
µ
¶
∞
`
0`
X X̀
r
r r`
1
∗
0
0
Y (θ , φ )Y`m (θ, φ)
− 2`+1 .
= 4π
2` + 1 `m
r0`+1
a
`=0 m=−`
To find the potential, we must compute the normal derivative of the Green’s function in the direction
of r0 , evaluated at r0 = a. Let us spend a moment and compute this.
¯
µ
¶¯
∞ X̀
X
∂G(r, r0 ) ¯¯
1
r`
r0`−1 r` ¯¯
∗
0
0
= 4π
Y (θ , φ )Y`m (θ, φ) −(` + 1) 0`+2 − ` 2`+1 ¯
,
¯
∂r0 ¯r0 =a
2` + 1 `m
r
a
`=0 m=−`
= 4π
∞
X
µ
¶
X̀
1
r`
r`
∗
Y`m
(θ0 , φ0 )Y`m (θ, φ) −(` + 1) `+2 − ` 2`+2 ,
2` + 1
a
a
r 0 =a
`=0 m=−`
= −4π
∞ X̀
X
`=0 m=−`
= −4π
∞ X̀
X
³ r ´` 1
1
∗
Y`m
(θ0 , φ0 )Y`m (θ, φ) (2` + 1)
,
2` + 1
a a2
∗
Y`m
(θ0 , φ0 )Y`m (θ, φ)
`=0 m=−`
³ r ´` 1
.
a a2
Now, we can directly compute the potential ϕ(r, θ, φ) using
¯
Z
1
∂G(r, r0 ) ¯¯
ϕ(r, θ, φ) = −
ϕ(θ0 , φ0 )
a2 dΩ0 ,
4π
∂r0 ¯r0 =a
Z
∞ X̀
³ r ´` 1
X
1
0
0
∗
ϕ(θ , φ )4π
Y`m
(θ0 , φ0 )Y`m (θ, φ)
a2 dΩ0 ,
=
4π
a a2
`=0 m=−`
Z
=
ϕ(θ0 , φ0 )
∞ X̀
X
∗
Y`m
(θ0 , φ0 )Y`m (θ, φ)
`=0 m=−`
=
∞ X̀
X
Y`m (θ, φ)
`=0 m=−`
∴ ϕ(r, θ, φ) =
³ r ´` Z
a
∞ X̀
X
`=0 m=−`
³ r ´`
a
dΩ0 ,
∗
ϕ(θ0 , φ0 )Y`m
(θ0 , φ0 )dΩ0 ,
A`m
³ r ´`
a
Y`m (θ, φ).
‘
’
óπ²ρ
²́δ²ι
δ²ιξαι
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
µ
¶
1
1
q
−
,
ϕ(x) =
4π²0 |x − ~a| |x + ~a|
where ~a ≡ (a, 0, 0) in spherical coordinates. As before, we can expand the function
1/|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