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!

Homework Help: 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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook