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VortexLattice
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Homework Statement
Hey all, I want to find the potential due to a point-like (mathematical, not physical) dipole inside and outside of a sphere made of dielectric material with dielectric ε. It's of radius R. Outside the sphere is vacuum.
Homework Equations
The Attempt at a Solution
Inside there sphere is easy (I think). For the potential of a dipole, we just use the standard
[itex]V(r,\theta) = \frac{1}{4\pi \epsilon} \frac{\vec{p}\bullet\vec{x}}{\left|x\right|^3}[/itex]
with the dielectric's [itex]\epsilon[/itex], of course. Outside the sphere, I try to build a solution using Legendre polynomials. I want them to match the boundary condition of the potential on the surface of the sphere. To find this, I note that for a point on the surface of the sphere, x is [itex]\vec{x} = sin(\theta)cos(\phi)\hat{x} + sin(\theta)sin(\phi)\hat{y} + cos(\theta)\hat{z}[/itex] and I say the dipole p is pointing in the z direction, with magnitude p. So on the surface of the sphere:
[itex]V(R,\theta) = \frac{1}{4\pi \epsilon}\frac{p cos(\theta)}{R^3}[/itex]
So this is what I need my solution outside to match at r = R.
So outside I want something of the form [itex]V(r,\theta) = \sum_{m = 0} ^\infty [(A_m r^m + B_m R^{-(m+1)})P_m(cos(\theta))][/itex]
But I'm looking at the outside, as r goes to infinity, so I know A = 0 for all m. So I get rid of these, and do the standard little trick to figure out the coefficients B:
[itex]B_m = \frac{2m + 1}{2}R^{m + 1} \int_0^{\pi} P_m(cos(\theta)) \frac{1}{4\pi\epsilon}\frac{p cos(\theta)}{R^3} sin(\theta) d\theta [/itex]
Is this all I can do? Do I just find the coefficients and then that's as far as I can go? I don't think I can integrate the Legendre polynomials generally.
Also, is there a nicer way of doing it? I was thinking of something involving that trick of viewing the dipole as two point charges very close together (at a distance d), finding the field of each by using method of images, and then taking the limit as d->0. But doing method of images with dielectrics is tricky. Any advice?
Thanks!