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For a parallel-plate capacitor of area $A$ and separation $d$, filled by a spatially varying positive scalar permittivity, neglect edge fields. At fixed terminal voltage, the physical electric potential minimizes $\int \epsilon |\nabla V{|}^{2}d\mathcal{V}/2$. A linear trial potential gives the arithmetic upper bound. At fixed plate charge, minimizing $\int |\mathbf{D}{|}^2/(2\epsilon )d\mathcal{V}$ over divergence-free electric displacement fields gives the harmonic lower bound by using a uniform trial displacement. The averages are volume averages. Layers parallel to the field attain the arithmetic bound; layers perpendicular to the field attain the harmonic bound. Equal volumes of vacuum and relative permittivity $ϵ$ therefore give $2ϵC_0/(1+ϵ)\le C\le (1+ϵ)C_0/2$, where $C_0={\epsilon }_{0}A/d$.

Define side $P$ as the higher proton electrochemical potential reservoir and side $N$ as the synthesis-side reservoir. For transfer $P\to N$ at common temperature $T$, define the positive downhill proton-motive force magnitude $\Delta p_{\mathrm{drive}}$ byHere $a$ is thermodynamic activity, $\psi$ is electric potential, $R$ is the molar gas constant and $F$ is the Faraday constant. Thus both a higher proton thermodynamic activity and a more positive electric potential on side $P$ contribute to the driving magnitude. This definition fixes the sign; conventions for a signed proton-motive force differ. The calculation under chemiosmosis separates these two contributions numerically.

Assume fixed reservoirs, tight mechanical energy coupling, one transported proton per $c$ subunit per revolution, and three ATP produced per revolution. Neglect leaks, slipping and membrane transport of the chemical reagents. Let $g_{\mathrm{ATP}}>0$ be the local molar Gibbs free energy cost of ATP formation at the stipulated composition, pH and magnesium-binding convention. It is not necessarily the standard state value. Adding changes in the state function gives, per mole of complete turns,For this model net synthesis is thermodynamically favoured when $\Delta G_{\mathrm{turn}}<0$. Equality is a reversible balance, not a promise of a useful reaction rate. A negative value establishes an available driving margin; it does not specify how fast the enzyme runs.

Choose $n=10$, $g_{\mathrm{ATP}}=50.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, $F=96485\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$ and $\Delta p_{\mathrm{drive}}=0.180\mathrm{V}$. These are illustrative inputs, not measurements of the preparation in the experiment below. Thenand $\Delta G_{\mathrm{turn}}=-23.673\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Per ATP, the ideal input is $57.891\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, leaving a driving margin of $7.891\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The balance point isAt $0.140\mathrm{V}$ the same synthesis cycle instead has $\Delta G_{\mathrm{turn}}=+14.921\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$: the chosen reservoirs favour its reverse.

The machine is reversible: sufficient ATP hydrolysis can drive opposite rotation and pump protons uphill. Actual direction also depends on inhibition and available substrates; a weak proton-motive force need not make every intact enzyme immediately run backward. Respiratory uncoupling supplies a different failure mode: a parallel proton leak dissipates the gradient without productive shaft rotation.^[ref]

The motor budget $10/3$ is not the whole cost of ATP delivered outside a mitochondrion. Import of ADP and inorganic phosphate, and export of ATP, also draw on the electrochemical gradient.^[ref] For a deliberately simplified system budget, assign these transports together one additional downhill-proton equivalent per delivered ATP, valued at the same $F\Delta p_{\mathrm{drive}}$. This is a chosen accounting convention, not a claim of one extra literal proton through $F_0$, nor a universal transport stoichiometry. The total becomes $10/3+1=13/3$ equivalents and $75.258\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ of gradient expenditure per delivered ATP at $0.180\mathrm{V}$. The ratio of stored chemical free energy to that expenditure is $50/75.258\approx 66.4\%$, compared with $50/57.891\approx 86.4\%$ for the motor alone. This system budget changes the resource expenditure, not the motor's balance point $\Delta p_{\ast }$. Real transport terms require their own electrochemical potentials, and leaks add further cost.

For an annulus of inner radius $a>0$, outer radius $b>a$ and uniform surface charge density $\sigma$, integrate circular rings of electric charge to get $V(x)=\sigma [\sqrt{x^2+b^2}-\sqrt{x^2+a^2}]/(2{ϵ}_0)$ with zero electric potential at infinity. Differentiation gives the displayed axial electric field. Its magnitude is greatest at $|x|=(ab{)}^{2/3}/\sqrt{a^{2/3}+b^{2/3}}$. Near the centre, $E_x=Qx/[2\pi {ϵ}_{0}ab(a+b)]+O(x^3)$, where $Q=\pi \sigma (b^2-a^2)$; an oppositely charged particle constrained to the axis consequently undergoes small-amplitude simple harmonic motion.

Spatial variation of the valence band and conduction band edges associated with a varying electric potential. Near a semiconductor boundary it can deplete mobile charge carriers or create a semiconductor inversion layer.

The diagrams solve an explicitly bounded Dirichlet problem in a uniform charge-free medium of constant permittivity. Take a width scale $W=1$ and terminal electric potential difference $V=1$. Two zero-thickness ideal plate segments occupy $|x|\le W/2$ at $y=\pm d/2$, with electric potentials $\pm V/2$. Fix $\varphi =0$ on the entire rectangle $x=\pm 2W$, $y=\pm 1.5W$. This boundary is an additional enclosing equipotential, rather than a condition at infinity. The model is invariant along the omitted third direction. The connection between the electric potential, its gradient, and the charge-free Laplace equation follows from electrostatics.^[ref]

On a square mesh of spacing $h=W/160$, the finite difference method imposesat free nodes, while plate and outer nodes retain their prescribed electric potentials. There are $641\times 481$ nodes. Each plotted gap aligns exactly with the mesh; the smallest gap has 16 mesh intervals. The generators solve the resulting linear system by the conjugate gradient method, stopping when the maximum dimensionless five-point residual is below $2\times 10^{-10}$, and use central finite differences to approximate $\mathbf{E}=-\nabla \varphi$ away from plate nodes.

The discrete maximum principle keeps the computed electric potential between $-V/2$ and $+V/2$. Reflection checks require evenness in $x$ and oddness in $y$; halving $h$ tests discretization effects, while enlarging the outer rectangle tests sensitivity to its artificial boundary conditions. These are different checks: a small residual only establishes that the discrete equations were solved. Differentiating near zero-thickness endpoints is especially sensitive to the mesh, so the plots do not report a converged edge peak. No total electric charge, exact open-domain three-dimensional capacitance, or measured material property is extracted from this model.

For a bounded numerical check, compare $h=W/80$ with the plotted $h=W/160$ at midplane positions $x/W=0,0.25,0.40,0.50,0.60,0.80$. At $d/W=0.10$, the largest change in $-E_{y}d/V$ is approximately $0.0265$, at $x/W=0.50$; the other five sampled changes are below $0.00472$. At $d/W=0.25$, moving the outer rectangle from $\pm 2W\times \pm 1.5W$ to $\pm 3W\times \pm 2W$ while retaining $h=W/80$ changes these samples by at most approximately $0.00622$. These calculated sensitivities support the qualitative comparison and expose its limits; they are not experimental uncertainties or evidence that the artificial boundary has converged to infinity.

The built-in potential of a p-n junction is the electrostatic potential difference established when carrier diffusion exposes ionized dopants until the opposing drift and diffusion currents balance. The equilibrium electrochemical potential is constant even though the electric potential and band edges vary. In a nondegenerate homojunction with neutral-edge majority densities $p_p,n_n$, $e{V}_b=k_{B}T\log (p_p{n}_n/n_i^2)$. This is band bending within the same material, not necessarily a discontinuous offset of different materials' band edges.

A capacitor stores electrostatic energy by separating electric charge between two electrical conductors. In the ideal two-terminal model their electric charges are $+Q$ and $-Q$, with a vacuum or insulating gap preventing direct transfer between them. Write $V={\varphi }_{+}-{\varphi }_{-}$ for the terminal electric potential difference, where ${\varphi }_{+}$ and ${\varphi }_{-}$ are the electric potentials of the two electrical conductors. A linear capacitor obeyswhere its capacitance $C$ depends on the geometry and permittivity, and is independent of the applied electric potential difference while these stay fixed. ^[ref].

For example, the parallel-plate capacitor below has capacitance $C=88.54\mathrm{pF}$. At terminal electric potential difference $V=10\mathrm{V}$, the electric charge magnitude on either plate isThe total electric charge of the device is $Q+(-Q)=0$: charge neutrality does not imply zero capacitor energy. This capacitor stores $4.427\mathrm{nJ}$ in its electric field. The schematic in parallel-plate capacitor shows the separation of electric charge; the plots in capacitor energy show how storage grows with terminal electric potential difference.

The useful distinction is between storing net electric charge and storing separated electric charge. A capacitor can release electrostatic energy into an electrical circuit; in an ideal RLC circuit, energy moves between the capacitor's electric field and the magnetic field associated with inductance. An ideal disconnected capacitor retains its separated electric charges and capacitor energy because there is no path for discharge.

The capacitor gap aspect ratio compares the gap $d$ with a lateral plate width $W$. For rectangular plates of widths $W_x,W_z$, require both $d/W_x\ll 1$ and $d/W_z\ll 1$ for the leading parallel-plate capacitor approximation. Large area alone is insufficient: a long narrow strip can have large area and still have $d$ comparable with its smaller width. In addition, the observation point must be many gap widths from an edge. Reducing $d/W$ improves the interior approximation without making the edge itself disappear.

In a two-dimensional straight gap, write the electric potential as the infinite-plate solution plus a perturbation:The perturbation satisfies the Laplace equation and vanishes on the two ideal plates. Separation of variables in a semi-infinite strip, with $s$ the distance inward from its entrance, gives transverse modesThus edge disturbances have a natural penetration scale set by $d$. Modes forbidden by a particular symmetry have zero coefficients; this expression is an explanation of localization in an ideal strip, not a universal numerical correction to finite-plate capacitance. The outside boundary conditions determine the coefficients.

There is no single aspect-ratio cutoff guaranteeing an assigned accuracy: the requested observable, plate shape and thickness, surroundings, dielectric distribution, and guard geometry all matter. The small-gap approximation is local and is not uniform across an edge. It also does not justify extrapolating $C\simeq ϵA/d$ to arbitrarily large $d$, where finite electrical conductors and their external electric field dominate. At fixed $V$, formally shrinking an ideal gap makes the interior electric field grow like $1/d$; real electric currents across an imperfectly insulating gap, surface roughness and contact eventually invalidate that ideal limit.

A cavity in a conductor is a void surrounded by an electrical conductor; completely enclosed is the essential qualification in exact electrostatic shielding arguments. At electrostatic equilibrium, the surrounding connected metal is an equipotential. An empty closed cavity in a conductor has constant electric potential and zero electric field, even if the metal is isolated and charged. An occupied closed cavity in a conductor generally has a nonzero electric field; its inner-wall total electric charge cancels the total electric charge of the internal sources. These are distinct statements, illustrated by Faraday's ice pail experiment.^[ref]

For an electron packet, a sufficiently positive gate electric potential attracts electrons beneath the insulating layer of a MOS capacitor. The corresponding potential energy is $U=-e\varphi$, where $e>0$ is the magnitude of elementary charge: a larger electric potential gives a deeper potential well for electrons. A drawing of potential energy must therefore reverse the vertical sense of a drawing of electric potential.

A simplified three-phase transfer raises the adjacent gate's electric potential before lowering the old holding gate. This overlaps the accessible regions, then leaves the packet under the new gate. Repeat with the next gate to move it forward. Isolated wells with an intervening barrier would store the packet without transferring it.^[ref]

Place an infinite sheet of surface charge density $\sigma$ at $z=0$ between grounded conducting planes $z=\pm d/2$, in vacuum. Reflection symmetry and Gauss's law give $E_z=\sigma \mathrm{sgn}(z)/(2{\epsilon }_0)$ in the two gaps. Integrating the electric field with zero electric potential at the conductors gives $V(z)=\sigma (d/2-|z|)/(2{\epsilon }_0)$. Each facing conducting surface carries induced surface charge density $-\sigma /2$. For $\sigma >0$, an electron released from rest at a plate reaches the sheet with kinetic energy $e\sigma d/(4{\epsilon }_0)$; its nonrelativistic speed follows from $m_e{v}^2/2$.

An isolated conducting sphere of radius $R$ initially at potential $V_0$ has charge $Q=4\pi {ϵ}_{0}R{V}_0$. Inserting it without contact into the hollow inner conductor of a long coaxial cylindrical capacitor conserves that charge. The neutral inner conductor acquires $-Q$ on its cavity wall and $+Q$ outside. With the outer conductor grounded, the annular electric field is $Q/(2\pi {ϵ}_{0}lr)$ and the inner potential is the displayed result. The original electric potential is not conserved. End effects are neglected; exact cavity fields are geometry-dependent.

A surface can change its surface charge density as its local electric potential and solution composition change, for example through reversible ionization or ion adsorption. This couples the surface reaction equilibrium to the surrounding electric double layer. Fixed-charge and fixed-potential models are limiting boundary conditions rather than interchangeable descriptions. Salt or pH changes can therefore alter both electrostatic screening and the repulsion amplitude.

For externally tangent conducting spheres of radii $a,b$, centre separation $D=a+b$, and common electric potential $V$, the method of images givesAlternating spherical image reflections cancel the nonconstant boundary electric potential. Pairing terms makes each series convergent. The ratio is generally not $a:b$: that rule uses isolated capacitances and applies to widely separated conducting spheres connected by a wire. For $b=2a$, the smaller sphere carries the fraction $1/2-\pi /(6\sqrt{3}\ln 3)\simeq 0.22483$ of the total electric charge.

Chemiosmosis couples transmembrane ion movement down an electrochemical gradient to a chemical transformation. For mitochondrial ATP synthesis, protons return from the intermembrane space to the mitochondrial matrix through ATP synthase.^[ref]

An original thermodynamic example separates the electrical and chemical contributions. At fixed temperature $T$, use ideal dilute thermodynamic activities $a$ and the electric potential $\psi$ on each side. The molar electrochemical potential difference for inward transfer of a singly charged proton isHere $R$ is the molar gas constant, $F$ the Faraday constant, and $\ln$ the natural logarithm. The sign convention is destination minus origin: a negative difference makes inward transfer favourable. Both lower proton activity and a more negative electric potential inside favour inward movement.

Stipulate $T=300\mathrm{K}$, $a_{\mathrm{matrix}}/a_{\mathrm{outside}}=0.1$, ${\psi }_{\mathrm{matrix}}-{\psi }_{\mathrm{outside}}=-0.12\mathrm{V}$, $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $F=96485\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$. These illustrative inputs give approximately $-5.743$ and $-11.578\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for the chemical and electrical terms, hence $-17.321\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ per transferred proton. For a hypothetical device coupling three inward protons per ATP to an ATP-formation cost of $45\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$,The sign follows by adding changes in the state function Gibbs free energy. It establishes thermodynamic feasibility for this chosen model, not its reaction rate. Three is a stipulated coupling count, not a universal ATP synthase stoichiometry; transport costs and leaks can also consume the available free energy.

A grounded hemispherical boss of radius $a$ on the grounded plane $z=0$ in an otherwise uniform downward electric field has the displayed exterior electric potential, with $r^2=x^2+y^2+z^2$. It vanishes both on the hemisphere $r=a$ and on the surrounding plane, and tends to $E_{0}z$ at large distance. The outward normal field gives surface charge $-3{\epsilon }_0{E}_0\cos \theta$ on the hemisphere and $-{\epsilon }_0{E}_0(1-a^3/s^3)$ on the plane at cylindrical radius $s>a$. The apex field is $3E_0$, so a field-limited breakdown voltage is reduced by a factor three. For a capacitor with finite spacing $d\gg a$, this is a leading local approximation; it is not an exact solution on the distant second plate.

A spherical electrical conductor has constant electric potential throughout its connected interior in electrostatic equilibrium. For an isolated sphere in a homogeneous medium, spherical symmetry and Gauss's law give the exterior field of a point electric charge at its center.

A conductor in electrostatic equilibrium is an electrical conductor at rest whose mobile charge carriers have finished redistributing, with no maintained bulk electric current. In the ideal macroscopic model,A residual bulk electric field would drive further electric charge redistribution. The electric potential difference between any two points joined through the metal is $-\int \mathbf{E}\cdot d\mathbf{l}=0$, so each connected piece is an equipotential. Gauss's law then gives zero net bulk charge density. This is a statement about net macroscopic electric charge, not the absence of positive atomic nuclei and negative electrons. Excess electric charge is represented by surface charge density; atomic surface structure is outside this continuum model. ^[ref].

The word inside must distinguish metal from a cavity in a conductor. A completely enclosed, charge-free cavity in a conductor has constant boundary electric potential $V_c$. The constant solution of Laplace's equation and Uniqueness of the Dirichlet problem give $\Phi =V_c$ and $\mathbf{E}=0$ throughout the cavity in a conductor, regardless of its shape or external electric charges. Grounding is unnecessary for this inward electrostatic shielding; an isolated enclosure can change its overall electric potential without producing an internal electric field. Zero enclosed net electric charge alone would only give zero electric flux. A positive and negative pair inside a cavity in a conductor can have zero total electric charge and a nonzero electric field. ^[ref].

If fixed electric charges inside a cavity in a conductor sum to $q$, a Gaussian surface lying entirely in the metal and enclosing that cavity in a conductor has zero electric flux. Thus the total induced inner-wall electric charge is $-q$. For one such cavity in a conductor and an isolated enclosure with total metal electric charge $Q_c$, the outer-wall electric charge is $Q_c+q$. In particular, an initially neutral isolated enclosure generally has an exterior electric field when $q\ne 0$. For fixed internal sources, external changes only shift the constant electric potential inside an isolated enclosure, leaving the internal electric field unchanged. ^[ref].

A grounded enclosure fixes $V_c=0$ and can acquire metal electric charge $-q$; with no exterior sources and $\Phi \to 0$ at infinity, Uniqueness of the Dirichlet problem gives zero exterior electric field. With external sources present, the exterior electric field need not vanish, but it is independent of the internal sources at fixed grounded boundary electric potential. ^[ref].

Faraday's ice pail experiment makes the distinction between redistribution and transfer visible. In his letter of 4 February 1843, Faraday described an insulated pewter pail connected by wire to a gold-leaf electrometer, and a brass ball suspended on a long dry silk thread to reduce the hand's influence. He discharged the pail and charged the ball away from it. Lowering the ball without contact produced leaf divergence, which grew during entry and became approximately independent of further depth; withdrawing it collapsed the leaves. Contact with the bottom transferred electric charge, and the withdrawn ball was discharged. The readout was leaf divergence, not a direct map of the internal electric field. These observations distinguish electrostatic induction from contact transfer of electric charge. ^[ref].

For a modern version of Faraday's ice pail experiment, connect an insulated inner metal pail to the signal input of a high-input-impedance electrometer, with its reference lead and an outer shield grounded. Zero and momentarily ground the pail, then isolate the signal side. Lower a charged paddle deeply without touching and record the electric potential difference; withdraw it and check return to zero. Repeat with contact, then withdrawal: a residual reading indicates transferred electric charge. Reverse the paddle's sign and repeat. The electrometer reads volts; inferring electric charge requires the effective capacitance of pail, cable and input, so a stable reading at fixed geometry is a relative measure, not automatically a value in coulombs. Keep supports clean, nearby charged objects distant, and the operator grounded; humidity can cause leakage. ^[ref].

A complementary electrostatic shielding demonstration uses a charged hollow metal enclosure, a small metal sampling ball on an insulating handle, and an electrometer. Discharge the sampling ball between trials; touch it to the exterior, withdraw it, and transfer its collected electric charge to the electrometer. Repeat by touching the inner wall without touching the exterior during withdrawal. The exterior sample produces a response, whereas the inner sample gives a null within instrument sensitivity. This compares accessible surface responses rather than assuming a perfectly spherical enclosure. ^[ref]. Useful additional controls are an uncharged enclosure baseline, a repeat exterior sample to verify sensitivity, and repeated trials at different depths. An access hole, charged handle, leakage or disturbed surface charge density limits the inference; a null bounds detectable transfer, rather than proving an exactly zero electric field. An open pail's depth plateau tests good enclosure empirically. Neither demonstration establishes shielding of arbitrary time-dependent electromagnetic fields.

A contact potential difference is an electrostatic electric potential difference associated with different conducting surfaces brought into electrical equilibrium. Surface condition and different work functions can create a vacuum gap field even when an external voltmeter reads zero. Connecting the metals establishes equilibrium of their electronic chemical potentials including electrostatic energy; it need not equalize the electrostatic electric potential in the gap.

The Coulomb potential energy is the interaction potential energy of two point charges in vacuum. With separation $r>0$, electric charges $q_1,q_2$, $k=1/(4\pi {\epsilon }_0)$ and zero potential energy at infinite separation,This is mutual interaction energy; it does not include the divergent self-energies of ideal point charges. Unlike electric potential, which has units of volts, $U$ has SI units of joules. With charge 1 fixed and positive radial direction away from it, the conservative force in Coulomb's law givesThus the $1/r$ potential energy and the inverse-square law for force are two descriptions of the same interaction. ^[ref].

For $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$ and $r=0.20\mathrm{m}$, using the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$,At $r=0.40\mathrm{m}$, $U=-1.3485\times 10^{-7}\mathrm{J}$. Separating the pair from $0.20$ to $0.40\mathrm{m}$ therefore requires external work $\Delta U=+1.3485\times 10^{-7}\mathrm{J}$ if the initial and final kinetic energies agree; the electric force does $-1.3485\times 10^{-7}\mathrm{J}$ of work. Separating them all the way to infinity requires $+2.697\times 10^{-7}\mathrm{J}$. The negative initial potential energy means that the pair lies below the chosen separated reference; it does not mean a negative kinetic energy.

Like-signed electric charges have positive $U$, which decreases as they separate, so their electric force is repulsive. Opposite-signed electric charges have negative $U$, which decreases as they approach, so their electric force is attractive. This attractive interaction supplies the Coulomb potential used in the hydrogen atom; determining its stationary quantum states also requires quantum mechanics. ^[ref]. Dividing $U$ by a nonzero test electric charge $q_2$ gives the source's electric potential ${\Phi }_1=k{q}_1/r$. More generally, the electrostatic potential energy of point charges sums each unordered charge pair once. This energy description is useful because work between two separations follows from endpoint values, without integrating a complicated trajectory.