# Expressing a potential inside a spherical shell as

1. Sep 16, 2011

### dikmikkel

1. The problem statement, all variables and given/known data
The potential inside a spherical shell is given by:
$V_{-}(x,y,z)= \frac{V_0}{R^2}(6z^2-3x^2-3y^2)$
$P_n(\cos(\theta ))$ where $\theta$ is the polar angle.

The potential on the surface carries a surface charge density $\sigma$. Besides this, ther's no other charges and no outher field. The potential is rotational symmetric around the z-axis inside and outside, and goes to 0 far away from the sphere.

b) express the potential inside the spherical shell using a LegendrePolynomial

2. Relevant equations
In spherical coordinates i have:
$V(r,\theta ) = \sum\limits_{l=0}^{\infty}(A_lr^lP_l(\cos(\theta)) = V_0(\theta)$

3. The attempt at a solution
This is how far i made it. Now i suppose i could multiply it with $P_{l'}(\cos( \theta ))\sin(\theta)$ and integrate, but i cant figure out how to simplify it and extract the solution.
I'm aware that the functions' are orthogonal, but the integral of a sum is something i've never done before.

Last edited: Sep 16, 2011
2. Sep 16, 2011

### vela

Staff Emeritus
You're making the problem more complicated than you need to. First, express the given V in terms of spherical coordinates and write all the trig functions in terms of cos θ. Then compare what you get to the Legendre polynomials.

3. Sep 17, 2011

### dikmikkel

Ha ha, yeah now i get it :) It's because normally we want to find the potential. But now i get it as the Legendre from 0 to 2, which makes sense by the order of the cartesian polynomial. So i am getting
$\frac{3V_0r^2}{R^2}(3\cos(\theta)^2-1)$