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The integral and differential forms of Gauss's law in vacuum areFor an isolated spherical conductor, symmetry gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$ outside and zero inside. Taking electric potential zero at infinity, $V(R)={\int }_R^{\infty }E(r)dr=Q/(4\pi {ϵ}_{0}R)$. Thus the capacitance of an isolated conducting sphere isCharging from zero gives the electrostatic potential energyThis is also the integral of electric field energy density ${ϵ}_0{E}^2/2$ outside the sphere. The later insertion into a coaxial cylindrical capacitor conserves its electric charge, not its original isolated electric potential.

Put electric charge $+Q$ on the inner electrical conductor and $-Q$ on the outer one. In the gap, spherical symmetry makes the electric field radial and constant on a spherical Gaussian surface. Gauss's law givesThe inner-to-outer electric potential difference is thereforeThus the capacitance of the spherical capacitor isAs $b\to \infty$ this gives the capacitance of an isolated conducting sphere, $4\pi {ϵ}_{0}a$; for $b-a\ll a$ it approaches the parallel-plate capacitor result with area $4\pi a^2$.

For a smooth isolated conducting sphere, Gauss's law gives the surface electric field $E_s=Q/(4\pi {ϵ}_0{a}^2)$ and electric potential $V=Q/(4\pi {ϵ}_{0}a)=a{E}_s$. At the specified dielectric strength, $V_{\max }=3.0\times 10^3\mathrm{V}$. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}a$, so its capacitor energy isThis is the ideal breakdown-limited energy of a conducting sphere, taking air's permittivity approximately equal to vacuum permittivity and neglecting nearby objects and surface irregularities. The vacuum permittivity used numerically is $8.8542\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$.^[ref]

For charge sharing between distant connected spheres, use the standard approximation that the spheres are sufficiently far apart for mutual electrostatic effects to be negligible, and that the wire adds negligible capacitance. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}r$. ThusThe initial charges are $Q_i=C_i{V}_i$. Connecting the spheres allows charge to flow until their potentials are equal. Charge conservation givesThe initial and final electrostatic energies are $U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$ and $U_f=\frac{1}{2}(C_1+C_2)V_f^2$. Their difference becomes Joule heating in the resistive wire:Numerically,The resistance determines how fast this happens, not the final energy difference. Sufficient damping suppresses sustained charge-current oscillations; the calculation also neglects energy radiated away. At finite sphere separation the capacitance matrix would be needed, and the radii and initial potentials alone would not specify an exact correction.

For a conducting sphere of radius $R$ at center height $h>R$ above a grounded infinite plane,The distance $h$ is measured to the center, so the surface gap is $h-R$. Relative to the capacitance of an isolated conducting sphere, the enhancement is $c(R/h)=C(R,h)/(4\pi ϵR)$.

The method of images gives a constructive proof. At fixed sphere potential $V_s$, start with $q_0=4\pi ϵR{V}_s$ at height $z_0=h$ and place its opposite image below the plane. Cancel that image's nonconstant sphere potential by the successive chargeswith opposite mirror charges below the plane at every step. The resulting sphere potential is $V_s$ and the plane potential is zero. Since $q_n=4\pi ϵR{V}_s\sinh \alpha /\sinh ((n+1)\alpha )$, summing the sphere charges gives the displayed capacitance. In particular $c(y)=1+y/2+O(y^2)$ for small $y$.

The exact sphere-plane boundary-value problem is also treated in Behunin and colleagues' analysis of sphere-plane electrostatics.