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Potential between 2 concentric spheres

  1. Sep 5, 2008 #1
    The problem: Find the potential between 2 concentric spheres, of radii a & b, where the potential of the spheres are held at Ca*P3(cos(t)) & Cb*P5(cos(t)), where P3 & P5 are the 3rd & 5th Legendre functions, t is theta & Ca & Cb are constants.

    The general solution to Laplace with azimuthal symetry:

    V = sum over l of (Al*(r^l)+Bl*r^(-(l+1)))*Pl(cos(t))

    I can use the orthoganality of the Legendre functions to elimimate all but one of the series terms when I plug in the boundry conditions, giving:

    V(a,t) = (A*a^3 + B*a^(-4))P3(cos(t)) = Ca*P3(cos(t))

    V(b,t) = (C*b^5 + D*b^(-6))P5(cos(t)) = Ca*P5(cos(t))

    The solution should be a combination of these, but I'm a bit stumped about how to find the constants when I can't use r=zero or r=infinity to zero one of them (as is done in single shell problems). I have a feeling this may become obvious after a good sleep, but I'd be grateful for a nudge in the right direction.
  2. jcsd
  3. Sep 5, 2008 #2
    When using the orthogonality above, you see that you can just get rid of the polynomials and you are left with 4 constants; but if you go back and use orthogonality again, this time multiplying by say, P0(cost), you can get rid of the RIGHT sides, which reduce to zero. With that, you can solve A in terms of B and C in terms of D: when you plug the values into your first equations, this leaves you with 2 constants to determine...
  4. Sep 5, 2008 #3
    Sorry, disregard the above!!! i went a little too fast.. will get back to you
  5. Sep 5, 2008 #4
    we know that potentials obey the law of superposition and that there should be only one solution satisfying the boundary conditions. Thus, if we have a potential V1 at radius a and a potential V2 at radius b, we can think that V1 vanishes at b and V2 vanishes at a, so:
    V1(a) = Ca*P3(cos(t)), V1(b) = 0
    V2(b) = Cb*P5(cos(t)), V2(a) = 0
    right? I think this gives you enough boundary conditions to solve for the constants; your final solution is then in terms of: V(r) = V1(r) + V2(r), for a<r<b
  6. Sep 5, 2008 #5
    Thanks Goddar, as soon as I left it alone for a while the same answer came to me. Interestingly, your first answer sounds similar to what I was initially thinking as well, great minds & all that ;)
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