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 directionz defined by θ = 0. Computing rather directly, we see

|E | = − ∂ϕ(r, θ)∂r

∣∣∣∣r=0

= − Q

8πε0

∞∑

`=1

`

2` + 1r`−1

R`+1[P`+1(cos α)− P`−1(cosα)] P`(cos(0))

∣∣∣∣∣r=0

,

= − Q

8πε0

13R2

P1(1) [P2(cos α)− P0(cos α)] ,

= − Q

8πε0

13R2

(12

(3 cos2 α− 1

)− 1)

,

= − Q

16πε0R2

(cos2 α− 1/3− 2/3

),

∴ E = −Q sin2 α

16πε0R2z.

‘oπερ ’εδει δειξαι

c) Let us briefly discuss the limiting cases of the above results when α → 0, π.

Because sin2(α) = α2 − a4

3 + O(α6), and is π-periodic, it is clear that the electric fieldapproaches

E = − Qα2

16πε20z +O(α4),

for α → 0 and

E = −Q(π − α)2

16πε20z +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. Forα → π, the potential will decrease like (π − α)2 and will approach the potential of apoint charge of magnitude Q/8πR2(π − α)2.

Problem 3.5Let 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(a2 − r2)

4π

∫ϕ(θ′, φ′)dΩ′

(r2 + a2 − 2ar cos γ)3/2=

∞∑

`=0

∑

m=−`

A`m

( r

a

)`

Y`m(θ, φ),

where cos γ = cos θ cos θ′ + sin θ sin θ′ cos(φ− φ′) and A`m =∫

dΩ′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′| −a

r′|~r − a2

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′) = 4π

∞∑

`=0

∑

m=−`

12` + 1

Y ∗`m(θ′, φ′)Y`m(θ, φ)

(r`

r′`+1− a

r′r`

(a2

r′2 r′)`+1

),

= 4π

∞∑

`=0

∑

m=−`

12` + 1

Y ∗`m(θ′, φ′)Y`m(θ, φ)

(r`

r′`+1− r′`r`

a2`+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

= 4π

∞∑

`=0

∑

m=−`

12` + 1

Y ∗`m(θ′, φ′)Y`m(θ, φ)

(−(` + 1)

r`

r′`+2− `

r′`−1r`

a2`+1

)∣∣∣∣∣r′=a

,

= 4π

∞∑

`=0

∑

m=−`

12` + 1

Y ∗`m(θ′, φ′)Y`m(θ, φ)

(−(` + 1)

r`

a`+2− `

r`

a2`+2

),

= −4π

∞∑

`=0

∑

m=−`

12` + 1

Y ∗`m(θ′, φ′)Y`m(θ, φ) (2` + 1)

( r

a

)` 1a2

,

= −4π

∞∑

`=0

∑

m=−`

Y ∗`m(θ′, φ′)Y`m(θ, φ)

( r

a

)` 1a2

.

Now, we can directly compute the potential ϕ(r, θ, φ) using

ϕ(r, θ, φ) = − 14π

∫ϕ(θ′, φ′)

∂G(r, r′)∂r′

∣∣∣∣r′=a

a2dΩ′,

=14π

∫ϕ(θ′, φ′)4π

∞∑

`=0

∑

m=−`

Y ∗`m(θ′, φ′)Y`m(θ, φ)

( r

a

)` 1a2

a2dΩ′,

=∫

ϕ(θ′, φ′)∞∑

`=0

∑

m=−`

Y ∗`m(θ′, φ′)Y`m(θ, φ)

( r

a

)`

dΩ′,

=∞∑

`=0

∑

m=−`

Y`m(θ, φ)( r

a

)`∫

ϕ(θ′, φ′)Y ∗`m(θ′, φ′)dΩ′,

∴ ϕ(r, θ, φ) =∞∑

`=0

∑

m=−`

A`m

( r

a

)`

Y`m(θ, φ).

‘oπερ ’εδει δειξαι

Problem 3.6Let 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

4πε0

(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 substi-tute Legendre polynomials in their place. Azimuthal symmetry also implies that