Expressing a potential inside a spherical shell as

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
dikmikkel
Messages
160
Reaction score
0

Homework Statement


The potential inside a spherical shell is given by:
[itex]V_{-}(x,y,z)= \frac{V_0}{R^2}(6z^2-3x^2-3y^2)[/itex]
[itex]P_n(\cos(\theta ))[/itex] where [itex]\theta[/itex] is the polar angle.

The potential on the surface carries a surface charge density [itex]\sigma[/itex]. 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

Homework Equations


In spherical coordinates i have:
[itex]V(r,\theta ) = \sum\limits_{l=0}^{\infty}(A_lr^lP_l(\cos(\theta)) = V_0(\theta)[/itex]

The Attempt at a Solution


This is how far i made it. Now i suppose i could multiply it with [itex]P_{l'}(\cos( \theta ))\sin(\theta)[/itex] and integrate, but i can't 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:
Physics news on Phys.org
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.
 
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
[itex]\frac{3V_0r^2}{R^2}(3\cos(\theta)^2-1)[/itex]