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Potential of a sphere

  1. Dec 8, 2007 #1
    [SOLVED] Potential of a sphere

    1. The problem statement, all variables and given/known data
    The potential at the surface of a sphere of a radius R is given by [itex] V_{0} = k\cos 3\theta [/itex]. Find the potential inside and outside the sphere.

    2. Relevant equations
    Solution to Laplace's equation in spherical coords is given by
    [tex] V(r,\theta)=\sum_{l=1}^{\infty}\left(A_{l}r^l + \frac{B_{l}}{r^{l+1}}\right)P_{l}(\cos\theta) [/tex]
    [tex] \cos 3\theta = 4\cos^3\theta-3\cos\theta [/tex]

    3. The attempt at a solution

    The only problem really is finding the coefficient A. B is related to A like
    [tex] B_{l}=-A_{l}R^{2l+1}[/tex]
    I used the expansion of the cosine and found a linear combination of the Legendre Polynomials such that
    [tex] V_{0} = \frac{8}{5}kP_{3}(\cos\theta)-\frac{3}{5}kP_{1}(\cos\theta) [/tex]

    From here can i just say that
    [tex] A_{3} = \frac{8k}{5R^3} [/tex] and
    [tex] A_{1} = \frac{-3k}{5R} [/tex]

    or do i have to solve for A using
    [tex] A_{l} = \frac{2l+1}{2R^l} \int_{0}^{\pi} V_{0}(\theta) P_{l}(\cos\theta)\sin\theta d\theta [/tex]

    Thanks for your help!
  2. jcsd
  3. Dec 8, 2007 #2


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    Homework Helper
    Gold Member

    The way you've written the solution, the first boundary condition should be
    [tex] B_{l}=+A_{l}R^{2l+1}[/tex]

    The other BC you need to use is that
    [tex] V(R,\theta) = V_{\theta}[/tex]

    So, both the integral and equating the coefficients once you find the linear combination will give the same thing.
    Last edited: Dec 8, 2007
  4. Dec 9, 2007 #3
    awesome thanks
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