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

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

where $$A_n$$ and $$B_n$$ are determined by the boundary condition of the conductor. For $$r < R$$ we should have that $$B_n = 0$$ to make sure the potential is finite and for $$r > R$$ 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

$$\varphi = 0 \;\;\;\; \frac{3}{2} < \cos\theta < 1\\ \varphi = \varphi_0 \;\;\;\; \frac{-\sqrt{3}}{2} < \cos\theta < \frac{\sqrt{3}}{2}\\ \varphi = 0 \;\;\;\; -1 < \cos\theta < -\frac{3}{2}$$

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!