# Potential of Sphere, given potential at suface

The sphere of radius $R$ has the potential at the surface equal to $$V_0 = \alpha \sin^2(\theta) + \beta$$ where $\alpha, \beta$ are some constants. Find the potential inside, and outside the sphere. $$V(r,\theta) = \sum_{l=0}^{\infty}(A_l r^l + \frac {B_l} {r^{(l+1)}})P_l(\cos(\theta)).$$ $P_l$ is the legendre polynomials given by $P_0(x)=1$, $P_1(x)=x$, $P_2(x) = \frac {3x^2-1} {2}$.

I'm not really too sure how to solve these problems having looked through the examples, so I tried to solve using equal coefficients, this is just my work for the inside of the sphere. Getting rid of $B_l$ so it doesn't blow up inside at 0.

$$\alpha \sin^2(\theta) + \beta = \sum_{l=0}^{\infty}(A_l r^l)P_l(\cos(\theta)$$

I write out enough legndre polynomials to match the highest term of the given potential, and rewrite $\sin^2(\theta)$ in terms of $\cos^2(\theta)$.

$$\alpha + \beta - \alpha \cos^2(\theta) = A_0 + A_1r\cos(\theta) + A_2r^2(\frac {3\cos^2(\theta)-1} {2})$$

The A_1 term doesn't effect the potential, given the fact that it is not of the power of any of the given potential, so I get rid of it, then expand the A_2 term.

$$\alpha + \beta - \alpha \cos^2(\theta) = A_0 + \frac 3 2 A_2r^2\cos(\theta)- \frac 1 2 A_2 r^2$$

Now setting $\alpha = \frac 3 2 A_2r^2\cos(\theta)$, and $\alpha + \beta = A_0 - \frac 1 2 A_2 r^2$, and get $$A_2 = \frac {2\alpha} {3r^2\cos(\theta)}$$ $$A_0 = \alpha + \beta + \frac {\alpha} {3\cos^2(\theta)}$$

Adding these into our original Legendre Polynomial expression we get: $$V_{inside} = \alpha(\frac {3cos^2(\theta)+1} {3cos^2(\theta)}) + \beta + \frac {2\alpha} {3r^2\cos^2(\theta)}r^2(\frac {3\cos^2(\theta) -1} {2})$$

Simplifying: $$V_{inside}= \alpha(\frac {3\cos^2(\theta)+1 +3\cos^2(\theta) - 1} {3\cos^2(\theta)}) + \beta$$ $$V_{inside} = 2\alpha + \beta$$

I just want to know if I am doing this correctly, as I can't find any similar problems in the book, and this answer seemed very simplified, so I'm not sure if correct process, or not.

• I edited the MathJax to make this more readable. Check that I didn't change the meaning. – garyp Oct 1 '16 at 20:51