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Homework Help: Potential question

  1. Apr 6, 2010 #1


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    1. The problem statement, all variables and given/known data
    Electric charges are distributed on a spherical surface of radius a so as to produce the potential

    [tex] \Phi(\vec r)=A(x^2-y^2)+Bx[/tex]

    in the region r<a. Find the potential in the region r>a (hint: use the table of spherical harmonics).

    2. Relevant equations
    I am unsure, but I think I should start with the following general potential expression (solution to laplace's eqn in terms of the spherical harmonics Y).

    [tex] \Phi(r, \theta, \phi)= \sum_{l=0}^{\infty} \sum_{m=-l}^l \left[ A_{lm}r^l+B_{lm}r^{-(l+1)} \right] Y_l^m(\theta, \phi)[/tex]

    3. The attempt at a solution
    If I am to use the above potential eqn, I need to utilize boundary conditions to find the coefficients A and B, but the surface potential is not specified so I'm not sure where to start...can someone point me in the right direction?

    Thanks for your comments.
  2. jcsd
  3. Apr 6, 2010 #2
    First write out the coefficients in that expansion for r<a. This can be done easily by expanding out the x and y in their angular components and finding combinations of spherical harmonics that match them.

    Once you have that, you will need BC's. You know at the surface of the sphere the potential is continuous. You also know how the potential behaves at infinity. This should be enough information to solve for that expansion in the region r>a.
    Last edited: Apr 6, 2010
  4. Apr 12, 2010 #3


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    i'm having trouble finding the combination of spherical harmonics which describes:

    [tex]\Phi(\vec r)=A(x^2-y^2)+Bx=Ar^2(cos^2 \theta -sin^2 \theta )+Brcos \theta=Ar^2cos 2 \theta + Brcos \theta[/tex]

    any hints/advice?
  5. Apr 13, 2010 #4
    This isn't 2 dimensions. So there should be a [tex]\phi[/tex] term in there. Remember:

    [tex]x = r cos(\phi)sin(\theta)[/tex]
    [tex]y = r sin(\phi)sin(\theta)[/tex]
    [tex]z = r cos(\theta)[/tex]

    Also this might be useful:

    [tex] sin(\phi) = \frac{e^{i\phi}-e^{-i\phi}}{2i}[/tex]
    [tex]cos(\phi) = \frac{e^{i\phi}+e^{-i\phi}}{2}[/tex]
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