# Potential of a sphere

1. Dec 8, 2007

### stunner5000pt

[SOLVED] Potential of a sphere

1. The problem statement, all variables and given/known data
The potential at the surface of a sphere of a radius R is given by $V_{0} = k\cos 3\theta$. Find the potential inside and outside the sphere.

2. Relevant equations
Solution to Laplace's equation in spherical coords is given by
$$V(r,\theta)=\sum_{l=1}^{\infty}\left(A_{l}r^l + \frac{B_{l}}{r^{l+1}}\right)P_{l}(\cos\theta)$$
$$\cos 3\theta = 4\cos^3\theta-3\cos\theta$$

3. The attempt at a solution

The only problem really is finding the coefficient A. B is related to A like
$$B_{l}=-A_{l}R^{2l+1}$$
I used the expansion of the cosine and found a linear combination of the Legendre Polynomials such that
$$V_{0} = \frac{8}{5}kP_{3}(\cos\theta)-\frac{3}{5}kP_{1}(\cos\theta)$$

From here can i just say that
$$A_{3} = \frac{8k}{5R^3}$$ and
$$A_{1} = \frac{-3k}{5R}$$

or do i have to solve for A using
$$A_{l} = \frac{2l+1}{2R^l} \int_{0}^{\pi} V_{0}(\theta) P_{l}(\cos\theta)\sin\theta d\theta$$

Thanks for your help!

2. Dec 8, 2007

### siddharth

The way you've written the solution, the first boundary condition should be
$$B_{l}=+A_{l}R^{2l+1}$$

The other BC you need to use is that
$$V(R,\theta) = V_{\theta}$$

So, both the integral and equating the coefficients once you find the linear combination will give the same thing.

Last edited: Dec 8, 2007
3. Dec 9, 2007

### stunner5000pt

awesome thanks