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Given a grounded conducting sphere, $V=0$ and $\text{radius} = R$, centered at the origin with a pure electric dipole (dipole moment $\vec p$) situated at the origin and pointing along the positive $z$ axis, I should be able to solve Laplace's equation in spherical coordinates to find the potential everywhere in the sphere. I can separate the differential equations and use Legendre polynomials, but I'm having trouble defining and using my boundary conditions. What I (think I) know so far:

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

$$r\to R \Rightarrow V\to 0$$ $$\theta \to \frac{\pi}{2} \Rightarrow V\to 0$$ $$\theta \to 0 \space or \space \pi \Rightarrow V\to \frac{\hat{r}\cdot \vec{p}}{4\pi\varepsilon_0r^2}$$

At least I think those boundary conditions work. I need to define a condition as well for the case of $$r \to 0$$ but that would seem to make the potential explode. How can I use these boundary conditions to solve for the potential everywhere inside the sphere?

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1 Answer 1

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As $r \to 0$, we know that the potential approaches $\frac{1}{4 \pi \epsilon _0} \frac{|\vec p|\cos{\theta}}{r^2}$.

Therefore(just by comparing terms, which usually the easiest of solving these equations):

$$n \neq 1 \Rightarrow B_n = 0\\ B_1\ P_1 \left(\cos(\theta)\right)=B_1\ \cos\left(\theta \right)=\frac{p\ \cos\theta}{4\pi \epsilon_0}\\ B_1=\frac{p}{4\pi \epsilon_0}$$.

Now for $r=R$ we have:

$$\sum_{n=0}^\infty (A_n R^n ) P_n(\cos{\theta})+ \frac{B_1}{R^{2}} P_1(\cos{\theta})=0 \\ A_1 = -\frac{B_1}{R^3}, n \neq 1\Rightarrow A_n = 0$$

So the potential will be:

$$V\left(r,\theta \right)=\frac{p\cos(\theta)}{4\pi \epsilon_0}\left(\frac{1}{r^2}-\frac{r}{R^3} \right)$$

I hope I didn't make any mistakes.

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Thanks for responding! Can you elaborate why for $n \neq 1 \Rightarrow B_n = 0$? –  Art M May 27 '13 at 19:39
@ArtM If they existed, as $r \to 0$ they would become more important than the first term we found. –  Ali May 28 '13 at 5:26
Am I correct in asserting that adding a thin hollow spherical dielectric of radius $0<R'<R$would not change this solution, because the boundary conditions and total charge enclosed would not change? –  Art M May 28 '13 at 19:50
@ArtM I think that will change the boundary condition I used, and replace it by something more complicated. –  Ali May 28 '13 at 19:59
But $r\to R \Rightarrow V=0$ and $r\to 0 \Rightarrow V\to \frac{1}{4 \pi \epsilon _0} \frac{|\vec p|\cos{\theta}}{r^2}$ still apply. What other boundary condition was there? –  Art M May 28 '13 at 20:04

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