Potential for Electric Charge over Spherical Shell using Legendre Functions

[CaptainMarvel]

Homework Statement

Electric Charge is distributed over a thin spherical shell with a density which varies in proportion to the value of a single function P_l(cos \theta) at any point on the shell. Show, by using the expansions (2.26) and (2.27) and the orthongonality relations for the Legendre functions, that the potential varies as r^l P_l(cos \theta) at a point (r, \theta) inside the sphere and r^{-(l+1)} P_l(cos \theta) at a point (R, \theta) outside.

Homework Equations

2.26:

\frac {1} {|R-r|} = \frac {1} {R} + \frac {r} {R^2} P_1 + \frac {r^2} {R^3} P_2 + ...


2.27:

P_l(cos \theta_{AB}) = \sum_{m=-l}^{+l} (-1)^{|m|} C_{l,m}(\theta_A , \phi_A) C_{l,-m}(\theta_B , \phi_B)


Orthogonality relations for Legendre Functions:

\int_{-1}^{+1} P_{n}(x) P_{n'}(x) dx = 0 for n \not= n'
\int_{-1}^{+1} P_{n}(x) P_{n'}(x) dx = \frac {2} {2n + 1} for n = n'

The Attempt at a Solution

dV = \frac {1} {4 \pi \epsilon_o} \frac {1} {|R-r|} dq

Now consider ring of charge on sphere with area 2 \pi R^2 sin(\theta) d\theta

Therefore charge on the ring is dq = P_l(cos \theta) (2 \pi R^2 sin(\theta) d\theta)

To get the full charge we will have to integrate this ring dq from 0 to pi.

Subbing in using the dq and the relations 2.26 and 2.27:

dV = \frac {1} {4 \pi \epsilon_o} \sum_{l=0}^{\infty} \sum_{m=-l}^{+l} (-1)^{|m|} \frac {r^l} {R^{l+1}} C_{l,m}(\theta_A , \phi_A) C_{l,-m}(\theta_B , \phi_B) P_l(cos \theta) 2 \pi R^2 sin(\theta) d\theta

Since question not reliant on \phi due to spherical symmetry, m = 0 which means all the C_{l,m} become just P_l(cos \theta):

dV = \frac {2 \pi} {4 \pi \epsilon_o} \sum_{l=0}^{\infty} r^l R^{1-l} P_l(cos \theta') P_l(cos \theta) P_l(cos \theta) sin(\theta) d\theta

Now we integrate theta from 0 to pi as mentioned before:

V = \frac {2 \pi} {4 \pi \epsilon_o} \sum_{l=0}^{\infty} r^l R^{1-l} P_l(cos \theta') \int_0^{\pi} P_l(cos \theta) P_l(cos \theta) sin(\theta) d\theta

If we use substitution u = cos \theta :

V = \frac {2 \pi} {4 \pi \epsilon_o} \sum_{l=0}^{\infty} r^l R^{1-l} P_l(cos \theta') \int_1^{-1} P_l(u) P_l(u) du

Using the orthogonality relation:

V = \frac {2 \pi} {4 \pi \epsilon_o} \sum_{l=0}^{\infty} r^l R^{1-l} P_l(cos \theta') \frac {2} {2l +1}

Cleaning up:

V = \frac {1} {\epsilon_o} \sum_{l=0}^{\infty} r^l R^{1-l} P_l(cos \theta') \frac {1} {2l +1}

Now I'm stuck though. I have no idea how to get rid of the summation. I'd like to do something let set l = 1 to get rid of the R^{1-l} but I know this must be wrong as the final answer still has ls in it.

Any help would be much appreciated. Am I close or have I totally missed the mark?

Thanks in advance.

 

[CaptainMarvel]

I don't know if it's just my browser but there's a bit of my previous post that is doesn't seem to be able to process correctly. Here are those two lines again:

Now we integrate theta from 0 to pi as mentioned before:

V = \frac {2 \pi} {4 \pi \epsilon_o} \sum_{l=0}^{\infty} r^l R^{1-l} P_l(cos \theta') \int_0^{\pi} P_l(cos \theta) P_l(cos \theta) sin(\theta) d\theta

If we use substitution u = cos \theta :

V = \frac {2 \pi} {4 \pi \epsilon_o} \sum_{l=0}^{\infty} r^l R^{1-l} P_l(cos \theta') \int_1^{-1} P_l(u) P_l(u) du