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The potential at the surface of an insulating sphere (radius R) is given by
$$V(R,\theta) = k \cos(3\theta)$$ where $k$ is a constant. Use separation of variables to find the potential inside the sphere (r $\leq$ R) and outside the sphere (r $\geq$ R), and then use your answer to determine the surface charge density $\sigma(\theta)$ on the sphere. Assume there is no charge inside or outside the sphere.

My approach: I will set my reference point at infinity, therefore the potential at infinity is 0, to set that as a boundary condition. Essentially, the approach in finding the surface charge density would be to find the potential function outside the sphere, then integrate that with the surface area. I am following Griffiths for E&M, but he doesn't include a specific example I can refer to for the separation of variables, and how to apply it. I can follow his examples for following potentials pretty easily, but cannot understand the first steps to set up this problem with SOV.

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"I don't understand how to do the problem" is not sufficiently clear and does not focus on the physics that is confusing you. The answer is that you know what conditions the electric potential obeys in regions of no charge not no time varying magnetic fields, and you know the boundary conditions, so away you go. Note that you do have to solve the exterior and interior separately. –  dmckee Mar 14 '13 at 19:54
    
Actually, Griffiths does give an example: 3.6 for the inside region, and 3.7 for the outside region. Pages 139-141. –  Colin McFaul May 15 '13 at 4:33
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1 Answer 1

Started with Griffith's potential form, for inside and outside the sphere.

$$\Phi (r,\theta) = E_{l=0}^\infty A_lr^lP_lcos(\theta), r \leq R$$ $$\Phi (r,\theta) = E_{l=0}^\infty B_lr^{-(l+1)}P_lcos(\theta), r \geq R$$

The potential is continuous for the tangential component, therefore:

$$B_l = A_l R^{2l+1}$$

But the normal component is discontinuous (Griffiths explains this to converge to $\sigma\over\epsilon_0$) , therefore:

$$\sum((2l+1)A_lR^{-1-2}-A_llR^{l-1})P_lcos(\theta)=-\sigma(\theta)/\epsilon_0$$

and find $A_l$ as: $$A_l = 1/(2\epsilon_0R^{l-1})*\int\sigma(\theta)P_lcos(\theta)sin(\theta)d\theta $$

I used this identity along with the Legendre Polynomial I solved for: $$ V(\theta) = kcos(3\theta) = (k/5)(8P_3cos(\theta)-3P_1cos(\theta))$$

and this Legendre can be analyzed in the two different conditions.

The charge density: $$\sigma(\theta) = \epsilon_o[d\Phi/dr(R_+)-d\Phi/dr(R_-)$$

$$\epsilon_0(k/5R)(56P_3cos(\theta)-9P_1cos(\theta))$$

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