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Potential for Electric Charge over Spherical Shell using Legendre Functions

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Electric Charge is distributed over a thin spherical shell with a density which varies in proportion to the value of a single function [tex] P_l(cos \theta) [/tex] 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 [tex] r^l P_l(cos \theta) [/tex] at a point [tex] (r, \theta) [/tex] inside the sphere and [tex] r^{-(l+1)} P_l(cos \theta) [/tex] at a point [tex] (R, \theta) [/tex] outside.

    2. Relevant equations

    2.26:

    [tex] \frac {1} {|R-r|} = \frac {1} {R} + \frac {r} {R^2} P_1 + \frac {r^2} {R^3} P_2 + ... [/tex]


    2.27:

    [tex] 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) [/tex]


    Orthogonality relations for Legendre Functions:

    [tex] \int_{-1}^{+1} P_{n}(x) P_{n'}(x) dx = 0 for n \not= n' [/tex]
    [tex] \int_{-1}^{+1} P_{n}(x) P_{n'}(x) dx = \frac {2} {2n + 1} for n = n' [/tex]


    3. The attempt at a solution


    [tex] dV = \frac {1} {4 \pi \epsilon_o} \frac {1} {|R-r|} dq [/tex]

    Now consider ring of charge on sphere with area [tex] 2 \pi R^2 sin(\theta) d\theta [/tex]

    Therefore charge on the ring is [tex] dq = P_l(cos \theta) (2 \pi R^2 sin(\theta) d\theta) [/tex]

    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:

    [tex] 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 [/tex]

    Since question not reliant on [tex] \phi [/tex] due to spherical symmetry, m = 0 which means all the [tex] C_{l,m} [/tex] become just [tex] P_l(cos \theta) [/tex]:

    [tex] 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 [/tex]

    Now we integrate theta from 0 to pi as mentioned before:

    [tex] 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 [/tex]

    If we use substitution [tex] u = cos \theta [/tex] :

    [tex] 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 [/tex]

    Using the orthogonality relation:

    [tex] 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} [/tex]

    Cleaning up:

    [tex] V = \frac {1} {\epsilon_o} \sum_{l=0}^{\infty} r^l R^{1-l} P_l(cos \theta') \frac {1} {2l +1} [/tex]

    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 [tex] R^{1-l} [/tex] but I know this must be wrong as the final answer still has [tex]l[/tex]s in it.

    Any help would be much appreciated. Am I close or have I totally missed the mark?

    Thanks in advance.
     
  2. jcsd
  3. Sep 29, 2008 #2
    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:

    [tex] 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 [/tex]

    If we use substitution [tex] u = cos \theta [/tex] :

    [tex] 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 [/tex]
     
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