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Homework Help: 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


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    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]
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