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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.

An electrical circuit connects components that carry electric current and sustain electric potential difference differences. Its dynamics may couple electrical resistance, inductance and capacitance.

For a quasistatic change of capacitance at fixed voltage, the capacitor energy is $U=C{V}^2/2$ and charge is $Q=CV$. Electrical work supplied to the field is $VdQ$, so mechanical work is $dU-VdQ=-V^{2}dC/2$. Increasing the separation of parallel plates makes $dC<0$: mechanical work is positive even though field energy decreases, because electrical energy returns to the source.

In electrostatics, Gauss's law is ${\oint }_{\partial \Omega }\mathbf{E}\cdot \mathbf{n}dS=Q_{\mathrm{enclosed}}/{ϵ}_0$, or locally $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, where ${\rho }_e$ is charge density and ${ϵ}_0$ is vacuum permittivity. Spherical symmetry makes the electric field radial, and Gauss's law givesChoose the additive constant of the electrostatic potential by $\varphi (\infty )=0$. Using $\mathbf{E}=-\nabla \varphi$ and continuity of potential across each charged shell givesThe field jumps at the shell surfaces, so the displayed formulas are for the requested open regions. The potential difference of the spherical capacitor is $\Delta \varphi =Q(1/a-1/b)/(4\pi {ϵ}_0)$. Thus its capacitance and electrostatic energy areThe energy follows either by charging work ${\int }_0^{Q}q/Cdq$, or directly from $W=({ϵ}_0/2)\int |\mathbf{E}{|}^2{d}^{3}x=Q^2(1/a-1/b)/(8\pi {ϵ}_0)$.

An H2RG detector is a hybrid infrared detector: a photosensitive array is bonded, pixel by pixel through indium contacts, to a separate silicon readout circuit. In infrared use the absorber is normally mercury cadmium telluride, whose composition sets the absorption cutoff. The H2RG readout architecture can also be used with other absorbing materials, so the readout name does not fix the detector’s wavelength response. The manufacturer’s H2RG specification gives a $2048\times 2048$ grid and $18\mu \mathrm{m}$ pixel pitch, together with reference-pixel and guide-window capabilities.

Each illuminated pixel contains a photodiode connected to its own readout node. Absorbed photons generate carriers; their accumulated charge changes the node voltage, with $|\Delta V|=|\Delta Q|/C$ for node capacitance $C$. A reset establishes the baseline; row and column addressing and multiplexed output amplifiers sample the pixel voltage. This is not the serial transfer of charge packets through adjacent pixels used by a charge-coupled device.

Nondestructive detector readout allows repeated measurements during an integration. Correlated double sampling removes the reset baseline, Fowler sampling differences groups of reads at the beginning and end, and up-the-ramp sampling estimates the signal from its slope and can flag cosmic-ray steps or saturation. The “R” denotes detector reference pixels, which monitor electronic offsets; “G” denotes a programmable guide window that can be read rapidly between full-array operations. Reference pixels do not measure sky background. Cooling suppresses dark current and thermal background. Read noise, nonlinearity, persistence and interpixel coupling still require calibration.

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.

A capacitor consists of two conductors carrying equal and opposite charges. Its capacitance iswhere $Q$ is the magnitude of the charge on either conductor and $V$ is their potential difference.

For $a<r<b$, a coaxial Gaussian cylinder of length $\ell$ encloses charge $\lambda \ell$. Gauss's law givesTaking $V$ to mean the inner potential minus the outer potential,Since $Q=\lambda L$,

The field energy isThese are the standard coaxial cylindrical capacitor formulas.

An RLC circuit combines electrical resistance, inductance and capacitance. Stored energy can oscillate between magnetic and electric forms, while Joule heating dissipates it.

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.

Two concentric conducting spherical shells of radii $a<b$ with charges $Q,-Q$ have an electric field only in the intervening vacuum: $E_r=Q/(4\pi {ϵ}_0{r}^2)$ by Gauss's law. Integrating gives a potential difference $Q(1/a-1/b)/(4\pi {ϵ}_0)$, and hence the displayed capacitance. The electrostatic energy is $Q^2/(2C)$, either by charging work or by integrating ${ϵ}_0|E{|}^2/2$ over the gap.