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 $$l$$s 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 $$u = cos \theta$$ :

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