Azimuthally Symmetric Potential for a Spherical Conductor

[Loonuh]

Homework Statement

Homework Equations

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The Attempt at a Solution

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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

<br /> \varphi(r,\theta) = \sum_{n=0}^\infty (A_n r^n + \frac{B_n}{r^{n+1}})P_n(\cos\theta),<br />

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

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

and

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

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

<br /> \varphi = 0 \;\;\;\; \frac{3}{2} &lt; \cos\theta &lt; 1\\<br /> \varphi = \varphi_0 \;\;\;\; \frac{-\sqrt{3}}{2} &lt; \cos\theta &lt; \frac{\sqrt{3}}{2}\\<br /> \varphi = 0 \;\;\;\; -1 &lt; \cos\theta &lt; -\frac{3}{2}<br />

then the above two integrals reduce to

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

and

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

respectively.

Now, since P_n are odd functions for odd n, A_n^{in} and B_n^{out} 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?

 

[TSny]

Note that cos(60^o) ≠ √3/2.

You say the integrals appear to be intractable. Did you write out the integrands explicitly for n = 0 and n = 2?

 

[Loonuh]

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.

 

[TSny]

OK. Good!