1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.