1
$\begingroup$

How to find the potential in region $a<r<b$

I know that the general solution for Laplace's equation is $$V(r,\theta)=\sum_{l=0} \left[A_l r^l +\frac{B_l}{r^{l+1}} \right]P_l(\cos{\theta}).$$

But I don't know how to use the boundary conditions to solve this problem. So how to? enter image description here

$\endgroup$

1 Answer 1

1
$\begingroup$

Your boundaries are at $r=a$ and $r=b$. Notice that the potentials at these two surfaces are independent of $\theta$ (they are spherically symmetric). Look at a list of the first few Legendre Polynomials $P_{l}(\cos{\theta})$. For what value of $l$ does $P_{l}(\cos{\theta})$ not depend on $\theta$? Further, notice that $V(a) = V(r=a,\theta) = V_{0}$, and $V(b) = V(r=b,\theta) = -V_{0}$. I will expand on this answer if you need more help, but these are the vital clues.

$\endgroup$
5
  • $\begingroup$ yes please I need more help ! $\endgroup$
    – A.khalaf
    Nov 25, 2014 at 18:49
  • $\begingroup$ Can you first add some more to your question to show the work you have already done? $\endgroup$
    – G. Paily
    Nov 25, 2014 at 18:51
  • $\begingroup$ I just wrote the general solution, then I did not do anythings ! $\endgroup$
    – A.khalaf
    Nov 25, 2014 at 18:57
  • $\begingroup$ Try answering this first: Look at a table showing the first few Legendre Polynomials. For what value of $l$ does $P_{l}(\cos\theta)$ not depend on $\theta$? $\endgroup$
    – G. Paily
    Nov 25, 2014 at 19:01
  • $\begingroup$ from the table, for $l=0$ $\endgroup$
    – A.khalaf
    Nov 25, 2014 at 19:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.