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An isolated conducting sphere of radius $R$ in a medium of permittivity $ϵ$ has surface electrostatic potential $V=Q/(4\pi ϵR)$ relative to infinity. Thus its capacitance is $C=Q/V=4\pi ϵR$.

A quasi-static chemical potential outside a fixed-shape droplet solves the same Laplace equation as an electrostatic potential outside an equipotential conductor. If $\mu ={\mu }_s$ on the droplet and zero on the surrounding reservoirs, its outward diffusive current iswhere $C$ is the corresponding electrostatic capacitance in a medium of permittivity $ϵ$. The arbitrary $ϵ$ cancels against the capacitance's proportionality to $ϵ$. Thus electrostatic geometry directly determines evaporation rates.

Across a charged dielectric interface the tangential electric field is continuous and the normal displacement jump is the fixed surface charge density. Without a dipole sheet the electrostatic potential is continuous. For a charge sheet with equal permittivity on both sides, its normal potential derivative has jump $-{\sigma }_{\mathrm{fixed}}/\epsilon$. Setting the field to zero on one side is an additional physical boundary assumption, as in a conductor or a negligible-exterior-admittance idealization.

Let the electrolyte permittivity be $\epsilon$ and let ${\kappa }_D=1/{\ell }_D$ be the inverse Debye–Hückel screening length. We reserve $k$ for the imposed lateral wavevector. In the Debye–Hückel approximation, linearization about an electrically neutral bulk electrolyte givesaway from fixed charges. For ionic species of valence $z_i$ and bulk number density $n_i$, ${\kappa }_D^2={\sum }_i{n}_i{z}_i^2{e}^2/(\epsilon k_{B}T)$; the approximation requires $|z_{i}e\varphi |\ll k_{B}T$. Substitution of a lateral cosine into this screened Poisson equation leaves a vertical decay constantThe screened sinusoidal surface-charge mode thus decays faster than a laterally uniform charge mode. A nonzero $k$ retains a finite decay length even in the zero-salt limit.

The exterior media and their boundary conditions are not specified in the paper. Specifying the two surface charge densities alone does not fix the normal derivative on the inside unless the outside response is also fixed. We first take the confined-gap idealization, with the prescribed charge supplying all the displacement flux into the electrolyte between the sheets. The electrostatic interface boundary conditions are thenThis is, for example, the limit of negligible exterior displacement admittance. The electrostatic potential solving these conditions and the Debye–Hückel approximation equation isDifferentiation checks the two surface signs directly. This solution allows a sine as well as a cosine lateral component when the phase is nonzero.

In linear screened electrostatics, the quadratic charging free energy is $\frac{1}{2}\int \sigma \varphi dA$. It is equivalently $\frac{\epsilon }{2}\int [(\nabla \varphi {)}^2+{\kappa }_D^2{\varphi }^2]dV$ with the stated boundary conditions. The second term represents the linearized ionic response; integrating only the bare electric field energy would omit it. If $\mathcal{A}$ denotes membrane area, the wavelength-averaged energy density isUsing $⟨{\cos }^2(kx){⟩}_x=1/2$ and $⟨\cos (kx)\cos (kx+\theta ){⟩}_x=\cos \theta /2$ gives the phase registration of screened charged sheets energyThe positive cosine coefficient shows that, for nonzero charge amplitude and finite separation,At the minimum the electrostatic potential simplifies to $\varphi =\alpha \cos (kx)\sinh (pz)/[\epsilon p\cosh (pd/2)]$. Positive and negative charge patches face opposite signs across the gap. The two charge patterns therefore favor a half-wavelength lateral offset rather than like-charge alignment.

For comparison, if identical electrolyte fills all of space on both sides of infinitesimally thin charge sheets, the electrostatic interface boundary conditions are continuity of $\varphi$, decay at infinity, and the charge-induced derivative jumpsSuperposing the two screened sinusoidal surface-charge modes gives, in particular between the sheets,Its full-space continuation replaces each vertical distance by the corresponding absolute distance. Evaluating the electrostatic potential on both sheets now yieldsThe constant term is the two isolated-sheet self energies, and the cosine term is their screened interaction. The energy depends on the exterior convention, but both stated physical idealizations give the same phase registration of screened charged sheets: unlike charge patches oppose each other. In the full-space convention the optimum interaction energy is $-{\alpha }^2{e}^{-pd}/(4\epsilon p)$, giving an attractive normal force density $-{\alpha }^2{e}^{-pd}/(4\epsilon )$. Phase sensitivity becomes exponentially weak for $pd\gg 1$; if $\alpha =0$ or the separation tends to infinity there is no selected phase.

One can also display how an exterior dielectric response interpolates between these cases. Let $Y_o\ge 0$ be its normal displacement admittance for this lateral Fourier mode: $Y_o=0$ for the confined-gap idealization, $Y_o=\epsilon p$ for identical exterior electrolyte, and $Y_o={\epsilon }_o|k|$ for an ion-free exterior dielectric. DefineThe symmetric and antisymmetric surface-charge combinations giveSince $D_s<D_a$, its cosine coefficient is positive. This makes the phase-minimizing conclusion robust while exposing the boundary information needed for an absolute electrostatic energy.

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.