1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expressing a potential inside a spherical shell as

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant 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]

    3. 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 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. jcsd
  3. Sep 16, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  4. Sep 17, 2011 #3
    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook