Baluncore said:
The good or bad conductivity of the soil is not really an issue as it is an equipotential.
I'm not sure i can buy that.
It's only equipotential if it conducts, else there's an electric field across it which isn't possible if it's a conductor.
Let me change the shape from a flat plate to a sphere.
I need to do that because the capacitance of a single sphere is straightforward, see
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html
but the capacitance of a single plate is fearsome ,
http://arxiv.org/pdf/1304.6624v1.pdf
Abstract
The capacitance of arbitrarily shaped objects is reformulated in terms of the Neumann-Poincar ́e operator. Capacitance is simply the dielectric permittivity of the surrounding medium multiplied by the area of the object and divided by the squared norm of the Neumann-Poincar ́e eigenfunction that corresponds to its largest eigenvalue. The norm of this eigenfunction varies slowly with shape changes and allows perturbative calculations.
and i know my limits way too well to tangle with Poincaire operators and eignenfunctions .
So here goes:
If the soil is nonconductive and has permittivity not much different from that of free space
then an electric field exists between the surface of the sphere and the 'ground' wire that's also buried in the soil
and that volume holds whatever energy is stored in the capacitor.
If the distance to that ground wire is infinite then it's an isolated sphere with capacitance 4πε
0R , per that hyperphysics link
If the soil is conductive, then
free charges in it migrate to the surface of the insulator and the electric field is constrained to the volume of the insulator
and capacitance will be εA/d = ε4πR
2/d
so I hypothesize that conductive soil probably increases capacitance by ratio (ε4πR
2/d ) / ( 4πεR ) = R/d
For OP's flat plate offering the difference between conducting and nonconducting soil might simplify to some reasonable function resembling √area / d
and that's why electrolytic capacitors work so well with one plate of metal, one plate of conductive semi-liguid paste, and insulation just a microscopically thin layer of oxide on the metal plate's surface. Gives a very small denominator.
That's what i think
and i'll welcome corrections. (See recent General Discussion thread "Have you ever failed"..)
old jim