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Azimuthally Symmetric Potential for a Spherical Conductor

  1. Dec 5, 2015 #1
    1. The problem statement, all variables and given/known data
    BSWCL.jpg

    2. Relevant equations/3. The attempt at a solution
    I am trying to solve problem 2-13 from my textbook "Principles of Electrodynamics" (see image below).

    I believe that I should be solving the potential as

    [tex]
    \varphi(r,\theta) = \sum_{n=0}^\infty (A_n r^n + \frac{B_n}{r^{n+1}})P_n(\cos\theta),
    [/tex]

    where [tex]A_n[/tex] and [tex]B_n[/tex] are determined by the boundary condition of the conductor. For [tex]r < R[/tex] we should have that [tex]B_n = 0[/tex] to make sure the potential is finite and for [tex]r > R[/tex] we should have that [tex]A_n = 0[/tex] to ensure that the potential goes to 0 at infinity. That being said, at [tex]r=R[/tex], we can apply the boundary conditions such that we

    [tex]A_n^{in} = \frac{2n+1}{2a^n}\int_{-1}^{1}\varphi(R,\theta)P_n(\cos\theta)d\cos\theta[/tex]

    and

    [tex]B_n^{out} = \frac{2n+1}{2}a^{n+1}\int_{-1}^{1}\varphi(R,\theta)P_n(\cos\theta)d\cos\theta[/tex].

    Now, my problem comes when I want to solve these two integrals because it appears to me that if I apply the boundary condition, namely that

    [tex]
    \varphi = 0 \;\;\;\; \frac{3}{2} < \cos\theta < 1\\
    \varphi = \varphi_0 \;\;\;\; \frac{-\sqrt{3}}{2} < \cos\theta < \frac{\sqrt{3}}{2}\\
    \varphi = 0 \;\;\;\; -1 < \cos\theta < -\frac{3}{2}
    [/tex]

    then the above two integrals reduce to

    [tex]A_n^{in} = \frac{2n+1}{2a^n}\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\varphi_0P_n(\cos\theta)d\cos\theta[/tex]

    and

    [tex]B_n^{out} = \frac{2n+1}{2}a^{n+1}\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\varphi_0P_n(\cos\theta)d\cos\theta[/tex]

    respectively.

    Now, since P_n are odd functions for odd n, [tex]A_n^{in}[/tex] and [tex]B_n^{out}[/tex] appear to be 0 for all of the odd values of n. However, the problem arises now when I am trying to evaluate these integrals in general for even values of n, in this case the results seem intractable because I am trying to integrate the Legendre Polynomials over a symmetric interval where the limits are not -1 and 1.

    From my understanding, this problem should be as straightforward as applying the boundary conditions as I have done to solve for the potential, is there something that I am overlooking?
     
  2. jcsd
  3. Dec 5, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Note that cos(60o) ≠ √3/2.

    You say the integrals appear to be intractable. Did you write out the integrands explicitly for n = 0 and n = 2?
     
  4. Dec 6, 2015 #3
    Whoops, limits should be +-1/2. I meant that
    the general solutions to the integrals seem intractable, but I turns out that you only need
    the 0 and 2 terms to answer all of the parts of the problem, so then this is actually very easy.
     
  5. Dec 6, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK. Good!
     
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