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

$\endgroup$

closed as off-topic by ACuriousMind, Jim, Prahar, Kyle Kanos, Brandon Enright Nov 25 '14 at 19:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Jim, Prahar, Kyle Kanos, Brandon Enright
If this question can be reworded to fit the rules in the help center, please edit the question.

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.

$\endgroup$
  • $\begingroup$ yes please I need more help ! $\endgroup$ – A.khalaf Nov 25 '14 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 '14 at 18:51
  • $\begingroup$ I just wrote the general solution, then I did not do anythings ! $\endgroup$ – A.khalaf Nov 25 '14 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 '14 at 19:01
  • $\begingroup$ from the table, for $l=0$ $\endgroup$ – A.khalaf Nov 25 '14 at 19:30

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