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Potential of spherical dipole

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data
    (Griffith' electrodynamics, problem 3.22): a spherical shell (radius R) carries uniform charge sigma0 on the "northern" hemisphere, –sigma0 on the "southern" one. Find the potential inside and outside the sphere, calculating coefficients explicitly up to A6 and B6.

    2. Relevant equations
    Attached document: i don't know how to paste equations on this page, it's my first time.. for reference, it's the same basic solution developed for example 3.9 of the book, p.142

    3. The attempt at a solution
    the above makes reference the method of separation of variables (in spherical coordinates), with solution in the form of a Fourier series with coefficients involving legendre polynomials (variable = "cosine theta"): since the charge density is constant over each hemisphere, i end-up with no polynomial at all.. While an infinite series seems expected (this would be more clear if i could paste my equations); i guess i don't know where to plug-in the right boundary condition.. Anybody could enlight me?

    Attached Files:

  2. jcsd
  3. Jul 7, 2008 #2
    For a tutorial in LaTeX coding:


    Well, in getting to the point where you are at, you have already used all the boundary conditions that you can (maybe you should review what Griffiths does to get to the A_l and B_l correlation).

    The good news is that you're doing fine. Since it is crappy to integrate a legendre polynomial between 0 and pi, you will want to use a trick to change it so it is between -1 and 1. If you want to see it then look down, but if you want to work on it yourself then don't look.

    [tex]A_l = \frac{\sigma_0}{2 \epsilon_0 R^{l-1}}(\int_0^{\pi/2}P_l(cos \theta)sin \theta d \theta - \int_{pi/2}^\pi P_l(cos \theta)sin \theta d \theta)[/tex]
    and let x=cosØ
  4. Jul 7, 2008 #3
    Hi, thank you for the answer! i was thinking about this kind of solution but the reasoning of the text (that's probably what i got wrong) seems to suggest that since:
    Code (Text):

    [tex] \sigma\_0 (cos\theta\) = constant = \sigma\_0 P_0 (cos\theta\) [tex]
    (of course that's for theta between 0 and pi/2, otherwise it's the same, but negative constant)
    Then all coefficients are zero except for l = 0, but then the two integrals just reduce to:

    Code (Text):

    [tex] A_l =\sigma\_0 /\epsilon\_0 [tex]
    Now if i'm not supposed to kill off the polynomials, i guess i just end up with the whole infinite series, for which the formula is straightforward...
    (take you for the LaTex link, too, i hope i used it right..)
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