# non-symmetric spherical capacitor

The figures below illustrate (in cross-sectional view) three separate spherical capacitors: an inner solid conducting sphere is surrounded by a hollow thin conducting shell, concentric with the sphere. The space between the conductors is filled with dielectrics, which do not contain any net charge. For simplicity I would like to think of the "top'' dielectric as air, and the one at the "bottom'' as some sort of liquid, so that the situations would represent a basic spherical capacitor with some amount of liquid sitting at the bottom.

The system remains symmetric under rotations about the vertical: there is no azimuthal dependence (i.e. $\varphi$ dependence) in the problem.

The question is: how does one go about finding the capacitance of the two arrangements on the right?

There is no "closed form'' solution to this problem (there is a similar problem in Griffith's E\&M book, which the comment that this cannot be solved in closed form), i.e. I'm not looking so much for the answer as to a method of analysis.

The configuration on the left is equivalent to two spherical capacitors in parallel. To solve for this configuration (or any configuration where the dielectric occupies a wedge-shaped volume defined by a solid angle), one can solve Laplace's equation in spherical coordinates $$\nabla^2 V=0$$ inside the dielectrics since there is no free charge anywhere. As the surfaces of the inner sphere and the outer shell are spherical equipotential, the potential $V$ is a function of $r$ only, and so $$\nabla^2 V = \frac{1}{r^2}\frac{d}{dr}\left(r^2 \frac{dV}{dr}\right)=0 \qquad \Rightarrow \qquad V(r)=\frac{C_1}{r}+C_2$$ with $C_1$ and $C_2$ integrating constants. From this one can extract the electric field $\vec E=-\vec\nabla V(r)$ and then the surface charge densities $\sigma_s=\vert \vec D\vert$ on each portions of the spherical surfaces through the electric displacement $\vec D=\epsilon\vec E$.

Although the "boundary conditions'' of the other two situations are apparently similar to the first, there is something quantitatively different: if $V$ is radial only, then $\vec E$ will be radial only, but this is not possible because - for instance in the middle situation - $\vec E$ will not in general cross the interface between the two dielectrics at normal incidence, so by the dielectric-dielectric boundary conditions (Snell's law for dielectrics) it will be "deflected'' at the surface and thus no longer be radial in the bottom dielectric.

I concluded that even given this, the change of direction of $\vec E$ at the interface implies that cannot be radial only.

Is there a difference between the physics of the two situations on the right? i.e. does it matter that, in one case, one of the dielectric reaches all the way up to the conducting spherical core? Can I say $V$ in air for the middle configuration will be radial only since air completely surrounds the central conducting sphere?

So: how does not approach the solution?

I was going to assume in both sections (air and liquid) that $$V=V(r,\theta)= f(r)g(\theta)= \sum_{\ell}\left(A_\ell r^\ell + B_{\ell} r^{-\ell-1}\right)P_\ell(\cos\theta)$$ with $P_\ell(z)$ a Legendre polynomial, and then "slug it out'' by solving numerically for the coefficients $A_\ell$ and $B_\ell$ so the potential is continuous everywhere and the boundary conditions on the conducting surfaces are met.

I'm not sure how to mathematically describe the dielectric-dielectric boundary of the last two situations.) The horizontal boundary can be defined through a condition on the polar angle (with the vertical defining $\theta=0$). Presumably I need the equation of a straight line segment in polar coordinates.

Does $V$ in air even remain radial only until some polar angle $\theta_c$ defined by the surface of the second dielectric? Is there a simplification in one of the two cases on the right?