How Do You Calculate the Potential Inside and Outside a Spherical Surface?

[stunner5000pt]

[SOLVED] Potential of a sphere

Homework Statement

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.

Homework Equations

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

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!

 

[siddharth]

The only problem really is finding the coefficient A. B is related to A like
$$B_{l}=-A_{l}R^{2l+1}$$

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

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!

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.

 

[stunner5000pt]

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.

awesome thanks