Electrostatic boundary value problem with radial dielectrics

1. Apr 14, 2014

Fernbauer

1. The problem statement, all variables and given/known data
A unit sphere at the origin contains no free charge or conductors in its interior or on its boundary. It is, however, embedded in a dielectric medium. The dielectric is linear, but the permitivity varies by angle about the origin. It is constant along any radial direction. So we are given the dielectric distribution $\epsilon(\theta, \phi)$.

We are also given the potential at every point on the surface of the unit sphere, $V(\theta, \phi).$

What is the potential at the origin?

2. Relevant equations
This is a clearly solvable problem, since the potential everywhere inside any enclosed volume can be determined if we know the potential everywhere on the boundary. This is just a Dirichlet boundary-value problem. Our geometry of a unit sphere at the origin means the complexity is simply from the angular dielectric effect.

By linearity, the potential at the origin will be some weighted function of the potential on the spherical surface. So $V(0,0,0)=\int\int w(\theta,\phi) V(r=1, \theta,\phi) \sin \theta\ d\theta\ d\phi$. So the problem is really to find $w(\theta, \phi)$ given $\epsilon(\theta, \phi)$

Gauss's law with dielectrics gives us the important relation:

$0=\int\int {\partial V\over \partial r}(r=1,\theta,\phi) \sin \theta\ d\theta\ d\phi$

As a special case where $\epsilon(\theta, \phi)$=C, the mean value property of harmonic functions tells us the potential at the origin is just the mean value of the potential on the sphere surface.

As a weight function, $w$ is normalized, so $\int\int_S w=1$.

Finally, this can be viewed from the viewpoint as a Green's function. If a unit charge is added to the origin, what charge density is induced on the sphere's surface if the init sphere was grounded? By Green's Reciprocity Theorem, that density is the same $w(\theta,\phi)$ as our weighted potential question.

3. The attempt at a solution

If I could find a Green's weight function for the case where the potential was given by point charge outside the sphere that did not depend on the charge's location, that weight would also be the solution for the arbitrary case since any potential could be thought of as an infinite superposition of point charges in some distribution.
Unfortunately, solving for the potential of a point in the radial dielectric medium seems intractable.

Thinking of qualitative behavior, imagine that one area of the sphere has an especially high value of $\epsilon$. Very high permittivity regions, at the limit, act almost as conductors. Thus the surface of the sphere in such a high permittivity region should have a high weighting since that potential is well "conducted" by the dielectric.

We can also think about the angular differential behavior of $w$ and $\epsilon$.
If ${\partial^2\over {\partial\theta \partial \phi}}w$ is proportional to ${\partial^2\over {\partial\theta \partial \phi}}\epsilon$ then by double integration we find that $w$ is just a scaled version of $\epsilon$ plus some integration constant. The normalization requirement that $\oint w=1$ and the $\epsilon=C$ example tells us the ratio and constant (0) directly, so the answer would be super-simple:

$w(\theta, \phi)={1\over{\int\int_S \epsilon(\theta, \phi)}}\epsilon(\theta,\phi)$

I do believe this is the final answer, but this leaves the question of how to prove the differential relation.. it doesn't seem to fall out of the Gauss's law equation.

Thanks much for the help and teaching me what I'm missing! I suspect there's an elegant and intuitive way to prove that differential relationship.

2. Apr 19, 2014

MaxMelvin

Your analysis fails since it does not take into account any polarization of the dielectrics.
That polarization would be radial, by symmetry, but it's still unknown.

So therefore there is no Greens function that will tell you the potential at the center of the sphere, even when you know the potential everywhere on the sphere's surface.