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For a conductor in electrostatic equilibrium, the interior field is zero and the surface is equipotential. The tangential exterior field is zero. A thin Gaussian pillbox gives the normal jump $E_{\mathrm{out}}\cdot \mathbf{n}-E_{\mathrm{in}}\cdot \mathbf{n}=\sigma /{ϵ}_0$. Thus, in vacuum and with $\mathbf{n}$ pointing out of the conductor,These are the electrostatic boundary conditions at a conductor.

To obtain electrostatic energy in field form, integrate over the space outside the conductors, cutting it off first at a large sphere. From the Maxwell equations, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ and $\mathbf{E}=-\nabla \varphi$, soOn a conductor boundary, the outward normal of the integration region points into the conductor, so its flux contribution is $-{\varphi }_i{Q}_i/{ϵ}_0$. At the remote outer boundary the flux integral tends to zero: for a localized charge distribution, $\varphi =O(1/r)$ and $\mathbf{E}=O(1/r^2)$. The divergence theorem therefore givesSince the field vanishes inside conductors,The field integral is over all space, although the charges occupy a finite volume. For a finite outer integration boundary there is additionally the term $({ϵ}_0/2)\oint \varphi \mathbf{E}\cdot d\mathbf{S}$.

For the electrostatic energy of alternating-charge concentric shells, the total enclosed charge after shell $n$ isBy spherical symmetry and Gauss's law, the field is zero for $r<r_1$, and in every subsequent gap, including outside the last shell,Although its direction alternates, its squared magnitude is the same in every gap. ConsequentlyThe coefficient of $1/r_1$ is $q^2/(8\pi {ϵ}_0)$, independent of the other radii and of the number of shells.

The method of images uses the real line charge at the origin and the opposite image charge at $(2d,0)$. LetSuperposing their logarithmic electrostatic potentials gives, in the physical region $x<d$,At $x=d$, $r_1=r_2$, so $\varphi =0$. Its tangential derivatives also vanish on that plane, giving $E_y=E_z=0$, exactly the electrostatic boundary conditions at a conductor. The image lies outside the physical domain, so it creates no extra physical source there. Together with the appropriate behaviour at infinity, uniqueness of Poisson equation identifies this field with the grounded-conductor solution.

The flat interface is an equilibrium reservoir with $\mu =0$, while the droplet has the constant surface value ${\mu }_s=\gamma /({\varphi }_{B}R)$. The exterior chemical potential satisfies the Laplace equation, tends to zero far away, and obeys these two surface values. These are exactly the electrostatic boundary conditions at a conductor for a sphere held at potential ${\mu }_s$ above a grounded infinite plane.

Introduce an arbitrary permittivity $ϵ$. In the electrostatic problem, $Q=-ϵ{\int }_{S_R}{\partial }_n\mu dS=C(R,h){\mu }_s$. Therefore the diffusion-capacitance analogy gives the outward currentFor an isolated sphere, its capacitance is $C(R,\infty )=4\pi ϵR$, so $I(R,\infty )=4\pi MR{\mu }_s=4\pi M\gamma /{\varphi }_B$. Dividing givesThus sphere-plane capacitance determines the evaporation enhancement. The assumed spherical shape gives the corresponding law $\dot{R}=-M\gamma c(R/h)/(2{\varphi }_B^2{R}^2)$ for droplet evaporation near a planar reservoir. The local surface current is generally nonuniform, even though its total is described by a single capacitance.

In electrostatic equilibrium, a nonzero electric field inside a conductor would move its free charges. Hence $E=0$ in the conducting material and the potential is constant throughout each connected conductor. Immediately outside its surface the tangential electric field is zero. A Gaussian pillbox across the surface givesSince $E_{\mathrm{in}}=0$, the electrostatic boundary conditions at a conductor give

For the widely separated shells, let their charges be $Q_1,Q_2$. Their common potential requiresThus the charge sharing between distant connected spheres is

For the charge on connected concentric spherical shells, the potentials at the two radii areEquality forces $Q_1=0$. Henceso all charge lies on the exterior of the outer shell.

For the neutral sphere in the uniform field, the far-field condition gives $\alpha =-E$. Constancy of the potential on $r=R$ gives $\beta =E{R}^3$. ThereforeThe outward normal field at the surface isThe induced charge on a conducting sphere in a uniform electric field is consequentlyFinally,confirming neutrality.