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Potential inside a hollow sphere

  1. Sep 30, 2006 #1
    I have a problem finding the potential inside a hollow sphere. The problem is formulated as:
    "Determine the electric potential inside a hollow sphere of radius a given that the potential on its surface is f(theta)"
    I have found a function V(r,theta) through separation of variables and all I have left is to apply the boundary conditions. One is given, but I guess I need another one when my function has two unknown constants. How do I find that second boundary condition?
     
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  3. Sep 30, 2006 #2

    siddharth

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    Your radial solution will be of the form

    [tex] f(r) = Ar^n + B/r^{n+1} [/tex]

    So, what happens when r goes to 0? What does that tell you about B?

    You'll need only one boundary condition for [itex] V(r,\theta) [/itex] as a result. (Although you'll have infinite constants. Use the orthogonality property of Legendre polynomials to calculate them)
     
    Last edited: Sep 30, 2006
  4. Sep 30, 2006 #3
    So this forces B to equal zero, right? Then my potential function looks like V(r,theta) = A_n r^n P-n(cos(theta)) and I should somehow use that the integral over P_m* P_n equals zero if m is different from n, ok? I just don't know how..?
    (Damn, it would be easier if I knew how to use LaTeX) ;)
     
  5. Sep 30, 2006 #4

    siddharth

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    Yeah

    You've almost got it. Your potential function will be
    [tex] V(r,\theta)= \sum_{n=0}^\infty A_n r^n P_n(\cos \theta) [/tex]

    and so at r=R

    [tex] f(\theta)= \sum_{n=0}^\infty A_n R^n P_n(\cos \theta) [/tex]

    (Check out this LaTeX tutorial, or click the image to see the code)

    Since the legendre polynomials are orthogonal, multiply both sides of the above equation by [tex] P_m (\cos \theta) \sin \theta [/tex] (this is important. Do you see why?) and integrate (can you tell me the limits?). So, can you take it from here?
     
    Last edited: Sep 30, 2006
  6. Sep 30, 2006 #5
    I understand why I should multiply both sides with P_m(cos theta) but not why to include sin theta. The limits should be from 0 to pi, right?
     
  7. Sep 30, 2006 #6
    I got it now. sin(theta) is for the substitution cos(theta) = x to work out. Then I get an expression for A_m with an integral over P_m(cos theta)f(theta)sin(theta)d(theta). But then what?
     
  8. Sep 30, 2006 #7

    siddharth

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    Yes. Now, what about the other side of the equation? Remember that your aim is to get the value of [itex]A_n[/itex].

    If you found that as an integral, then you're done! You know the solution to the potential at any point in the sphere.
     
    Last edited: Oct 1, 2006
  9. Oct 1, 2006 #8
    But the correct answer should be
    [tex]V(r,\theta) = -\frac{u_0}{3} + \frac{4u_0}{3}(\frac{r}{R})^2P_2(\cos \theta)[/tex].

    The given potential at R is
    [tex]V(R,\theta) = u_o \cos(2\theta)[/tex].

    (I'm using LaTeX!)
     
  10. Oct 1, 2006 #9

    siddharth

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    Since you know [itex]f(\theta)[/itex], evaluate the integral using the orthogonal property!

    Where exactly are you stuck at? Can you post that here?
     
  11. Oct 1, 2006 #10
    I substituted [tex]\cos \theta = x[/tex] and used the orthogonal property. Then I re-substituted to get the [tex]\theta[/tex]-dependence again and now my expression looks like

    [tex]A_m = -\frac{1+2m}{2} a^\textit{-m} \int_{0}^{\pi} P_m(\cos \theta) u_0 cos(2\theta) sin \theta d\theta[/tex]

    I dunno where to go from here.
     
    Last edited: Oct 1, 2006
  12. Oct 1, 2006 #11

    siddharth

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    To simplify the integral on the RHS, as I've said before, you're going to need the orthogonal property . Have you tried writing [itex]\cos(2\theta) [/itex] in terms of [itex]P_2(\cos \theta) [/itex]? (Also, [itex]1=P_0(\cos \theta)[/itex])
     
  13. Oct 1, 2006 #12
    Hey, I solved it! Thanks alot! This has really helped me in my understanding of Legendre polynomials.
     
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