Electrostatics

Electrostatics studies time-independent electric charge distributions and their electric fields. In vacuum, its local equations in SI units areHere $\rho$ is charge density, ${ϵ}_0$ is vacuum permittivity, and $\Phi$ is electric potential. Gauss's law supplies the divergence equation; the static form of Faraday's law of induction supplies the zero curl. The Poisson equation becomes Laplace's equation where there is no electric charge. Prescribed electric charges give electric fields through Coulomb's law and the superposition principle; mobile electric charges on an electrical conductor instead require solving for their distribution together with the electric potential. ^[ref].

The useful distinction is between prescribed sources and prescribed boundary conditions. On a grounded electrical conductor, an external connection fixes its electric potential relative to an electrical reference potential and allows electric charge to enter or leave. On an isolated electrical conductor, total electric charge is fixed and its constant electric potential is an unknown. A neutral isolated electrical conductor can therefore have a nonzero local surface charge density through electrostatic induction. Equal electric potentials also do not imply equal electric charges: charge sharing between distant connected spheres gives a simple geometric example. ^[ref].

Electrostatics is useful because a spatial boundary value problem replaces the need to follow every moving charge carrier during charge relaxation. It describes the settled state of a capacitor and explains electrostatic shielding of sensitive equipment. Its assumptions must match the apparatus: stationary boundaries and sources, negligible time variation of the magnetic field, and enough time for charge relaxation. A metal carrying a maintained electric current through finite electrical resistance generally needs a nonzero internal electric field by Ohm's law; it is not a conductor in electrostatic equilibrium. Static macroscopic electric charge also does not mean that every electron is motionless. ^[ref].

Electrical charging associated historically with rubbing materials. It supplies ordinary electric charge, not a separate electrical substance. Compared with a chemical source, a friction machine can have a very different voltage, charge delivery and internal impedance.

In 1833 Michael Faraday compared electrical effects from different sources and then their quantitative delivery. His galvanometer test discharged a machine-charged jar battery through a wet thread and the detector, connecting frictionally produced charge to magnetic action.^[ref]

An interaction between electric charges mediated by an electric field. In a uniform dielectric the pair potential energy is $q_1{q}_2/(4\pi ϵr)$; mobile ions can modify it through electrostatic screening.

For $a<r<a(1+\delta )$, a charge density ${\rho }_0{a}^2{h}^{\prime }(r/a)/(\delta r^2)$ has total electric charge $4\pi {\rho }_0{a}^3[h(1+\delta )-h(1)]/\delta$. As $\delta \to 0$, it tends to uniform surface charge on $r=a$ with surface charge density $\sigma ={\rho }_{0}a{h}^{\prime }(1)$. The limiting normal electric field has jump $\sigma /{ϵ}_0$, although every finite-thickness layer has continuous electric field at its inner edge. Taking the thin-layer limit and taking a one-sided spatial limit need not commute.

A charge distribution invariant under rotations about one center. For its spherically symmetric potential, Gauss's law gives the radial electric field $\mathbf{E}=Q(r){\mathbf{e}}_r/(4\pi {ϵ}_0{r}^2)$, where $Q(r)$ is the enclosed electric charge.

For positive scalar permittivities and specified free charges, fixed conductor potentials, dielectric matching conditions and decay at infinity, the electrostatic field is unique. Subtract two proposed potentials and integrate $\nabla \cdot (ϵ\delta \varphi \nabla \delta \varphi )$. The boundary terms vanish or cancel across dielectric interfaces, leaving $\int ϵ|\nabla \delta \varphi {|}^{2}dV=0$. Hence the difference field is zero, with any potential constant fixed by the boundary convention.

In a vacuum domain with specified charge density, fixed boundary electric potential and the appropriate condition at infinity, two solutions of the Poisson equation have difference $w$ satisfying ${\nabla }^{2}w=0$ with zero boundary data. Integration by parts gives $\int |\nabla w{|}^{2}dV=0$; hence their electric fields coincide, and their potentials coincide after fixing the constant. This is the Dirichlet case of the more general dielectric uniqueness result. It justifies the exterior field constructed by the method of images.

Mobile electric charges redistribute in response to an imposed electric potential, partly compensating its field. The induced density response and the Poisson equation must be solved self-consistently. Classical dilute mobile ions lead to the Debye–Hückel approximation; a degenerate electron system instead has a response set by its electronic compressibility and interactions. A high charge response can pin a partly occupied Landau level near the Fermi energy, whereas an incompressible gap suppresses that response. electrostatic screening is not a claim that every material perfectly excludes every static field.

In a charge-free wedge $0<\theta <\alpha$ bounded by equal-potential conductor faces, separation of Laplace's equation gives the displayed local harmonic expansion. Bounded potential at a finite corner removes negative powers. Outer boundary conditions determine the remaining coefficients; the conductor potential alone does not determine a nonconstant solution in an unbounded domain.

If the lowest allowed angular harmonic has a nonzero coefficient, differentiating its potential gives the displayed field scale. The field vanishes at a right-angle recess, stays finite at a smooth face, and diverges at a protruding conducting corner whose exterior vacuum angle exceeds $\pi$. A rounded tip cuts off the ideal power law at its radius scale. If the first coefficient vanishes, the first nonzero higher harmonic determines the exponent instead.

The outward normal electric traction on an equipotential conductor adjoining a linear isotropic dielectric, for the ideal normal-field configuration, is $p_e=ϵE_n^2/2=\mathbf{E}\cdot \mathbf{D}/2$. Integrating this normal traction over a surface gives the electric force. It is a field-stress contribution; mechanical pressure and dielectric boundary conditions must be included when determining an actual fluid-interface equilibrium.

An equipotential sphere centred in a flat interface between vacuum and a linear dielectric admits the same potential $C/r$ in both media. The radial field is tangential at the interface, so tangential E and normal D boundary conditions are satisfied. Displacement flux through the hemispheres gives $Q=2\pi {ϵ}_0(1+{ϵ}_r)C$, yielding the displayed sphere potential. E has equal magnitude at equal radius; D and free surface charge are larger by ${ϵ}_r$ on the dielectric side.

Integrating electrostatic pressure gives a force toward the dielectric of magnitude $Q^2({ϵ}_r-1)/[8\pi {ϵ}_0(1+{ϵ}_r{)}^2{a}^2]$. A deforming liquid interface or surface tension needs additional mechanical conditions beyond this ideal flat-interface solution.

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.

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.

A charge cloud $\rho =A/r$ for $0<r<a$ encloses $Q(r)=2\pi A{r}^2$. With no central point charge and potential zero at infinity, Gauss's law gives $\varphi =A(2a-r)/(2{ϵ}_0)$ inside and $A{a}^2/(2{ϵ}_{0}r)$ outside. The density singularity is integrable, and the potential is finite at the origin.

An electric double layer consists of an interfacial electric charge and the compensating distribution of mobile ions in the adjacent chemical solution. Counterions are enriched and coions depleted near a simple charged interface. A diffuse layer is described approximately by the Poisson-Boltzmann equation; a compact specifically adsorbed layer requires additional surface chemistry. Overlap of diffuse layers can produce a screened repulsion between like-charged colloids.

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.

A dimensionless ratio comparing surface conduction with bulk conduction over a characteristic length. With pore radius r and surface conductance $K_s$, one convention is $\mathrm{Du}=K_s/(\sigma r)$; the cylindrical surface-to-bulk conductance ratio is twice this value.

A model separating a compact interfacial layer from a diffuse ion distribution. Specified adsorption and boundary conditions determine the potential at the outer compact-layer boundary; a diffuse-layer equation alone does not determine specific adsorption.

Overcompensation of an initial interfacial charge by adsorbed species or correlated ions, so the charge or potential inferred outside the compact interfacial region has the opposite sign. Screening of a fixed boundary charge alone is not the same mechanism.

The electrokinetic potential at a hydrodynamic slipping plane relative to bulk solution. It need not equal the potential at the physical solid surface.

The spatially extended ionic atmosphere screening a charged surface beyond the compact Stern layer.

An ion with the same charge sign as the surface or fixed charge being screened. In a simple diffuse layer it is depleted near a similarly charged surface.

The compact interfacial region associated with finite ion closest approach and specifically adsorbed charge. A diffuse electrolyte layer lies beyond it.

Electro-osmosis is fluid motion driven by an applied electric field acting on mobile electric charge near a surface. A charged interface attracts a diffuse population of counterions in an electric double layer. A field component tangent to the interface exerts a force density ${\rho }_e\mathbf{E}$ on that mobile charge; momentum transfer to the surrounding solvent drives flow opposed by viscous stress. In a capillary electrophoresis measurement, this flow advects the sample as well as its electrical drift relative to the fluid. Locally, the measured axial velocity is $v_{\mathrm{observed}}=u_{\mathrm{eo}}+{\mu }_{\mathrm{ep}}E$, where $u_{\mathrm{eo}}$ is the electroosmotic fluid velocity and ${\mu }_{\mathrm{ep}}$ is the electrophoretic mobility. ^[ref]

A homogeneous charged channel can add a constant electro-osmotic drift to the ideal electrophoretic velocity at fixed field. It changes the coefficient of a linear length-versus-entry-time law without generating a new length exponent. Charge-dependent adsorption, heterogeneous trapping or cancellation of drift followed by diffusion lies outside that simple claim; wall charge cannot be excluded as an indirect cause in every real experiment.

With $\psi (0)=\zeta$, $\psi \prime (d)=0$ and no slip at either wall, linear screening gives $\psi /\zeta =\cosh [\kappa (d-x)]/\cosh (\kappa d)$. The velocity is $v/U=1-(1-\mathrm{sech}\kappa d)x/d-\cosh [\kappa (d-x)]/\cosh (\kappa d)$. This electrically neutral-wall boundary assumes no external normal field; it differs from a grounded wall.

Zero wall charge constrains the jump in normal electric displacement, while fixing a wall potential is a separate electrical condition. Taking the external normal field to vanish gives a zero normal derivative at an ideal uncharged wall. A grounded wall may carry induced charge even though its imposed potential is zero.

Differentiating the screened-channel velocity gives $x_{\max }=d-{\kappa }^{-1}\mathrm{arsinh}[(\cosh \kappa d-1)/(\kappa d)]$. For $\kappa d\gg 1$, $x_{\max }\simeq \log (\kappa d)/\kappa$. The outer flow declines almost linearly to the passive no-slip wall rather than forming a symmetric plug.

In a half-space with no pressure gradient, Stokes balance and Poisson electrostatics give $\eta v\prime \prime =\epsilon E\psi \prime \prime$. No slip at the wall and zero bulk shear imply $v=\epsilon E(\psi -\zeta )/\eta$. The bulk velocity is $U=-\epsilon \zeta E/\eta$; it is not the fluid speed at the no-slip wall.

A coion has electric charge of the same sign as the reference interface or particle. Its number density is depleted near that interface in the simplest Poisson-Boltzmann equation description. A counterion has the opposite sign. These roles depend on the reference charge: near a positively charged sphere in magnesium chloride, magnesium ions are coions and chloride ions are counterions.

For a point electric charge in a charge-free neighbourhood, a static electric potential obeys Laplace's equation. The maximum principle for harmonic functions excludes a strict interior minimum or maximum of that electric potential, hence a strict minimum of the charge's potential energy. A static electric field alone therefore cannot provide stable three-dimensional confinement under these assumptions.

Faraday's ice pail experiment distinguishes redistribution of electric charge by electrostatic induction from transfer of electric charge by contact. Lower a charged body into an initially neutral isolated metal pail: an externally connected electrometer responds even though the body never touches the metal. Withdraw the body without contact and the response disappears. Repeat with conductive contact before withdrawal and a residual response can remain. The striking point is that an exterior response does not establish that the pail has gained net electric charge.

Apparatus and readout. A modern implementation uses two concentric wire-mesh cylinders: a 10 cm diameter, 15 cm high inner pail on insulating rods and a grounded outer shield. Connect the electrometer's signal lead to the inner pail and its reference lead to the shield. Momentarily ground and zero the signal assembly, then isolate it; the shield remains grounded. A charged disk on an insulating handle goes into the lower half without touching. The electrometer measures electric potential difference in volts, with sign determined by the lead connections. It does not directly measure electric charge in coulombs.^[ref]

A charge ledger before contact. To separate exact reasoning from the open apparatus, first idealize the pail as a closed cavity in a conductor with one internal body of total electric charge $q$. Let the isolated metal have total electric charge $Q_0$. The metal's electric field is zero at electrostatic equilibrium.^[ref] A Gaussian surface enclosing the cavity in a conductor but lying wholly in the metal therefore has zero electric flux. Gauss's law and charge conservation giveFor $Q_0=0$ and $q>0$, the inner-wall total is $-q$ and the outer-wall total is $+q$: the metal remains neutral. The exterior electric field can therefore be nonzero without any transfer from the body. The connected electrometer and signal cable belong to the same isolated conducting assembly and share its exterior electric charge. The zero-electric field assertion concerns metal, not the occupied cavity in a conductor.

This ledger fixes totals, not the local surface charge density. Moving an off-center body changes the inner-wall distribution while its integral remains $-q$. With fixed outer geometry, fixed $Q_0+q$ and unchanged exterior sources, Uniqueness of the Dirichlet problem together with the fixed total outer electric charge determines the same exterior solution: the outside cannot reveal the body's position or detailed internal distribution. For example, internal bodies with stipulated electric charges $+2q$ and $-q$ give the same total exterior response as a single body with $+q$. Internal bodies with $+q$ and $-q$ can give a null exterior response while retaining a nonzero internal electric field. This is why a null reading cannot establish an empty cavity in a conductor.

Withdrawal is the control. Record a baseline, insert without contact, withdraw, and check return to baseline. After a fresh zero, repeat with the charged disk touching the inner wall, then withdraw and record the residual response. Finally zero the pail and reinsert the disk without contact to test remaining electric charge. Keep the operator grounded, discharge unwanted handle electric charge, and keep the unused charged disk distant. Dirty supports and humidity promote leakage. These controls separate transfer from electrostatic induction and stray coupling.^[ref]

In the ideal contact test, the internal body is itself an electrical conductor and makes a conducting connection to the wall. After charge relaxation, body and enclosure become a single equipotential; the remaining charge-free void has zero electric field. The body then carries no excess electric charge on surfaces facing that void. Withdrawal without subsequent charging leaves total metal electric charge $Q_0+q$. A coated paddle can retain electric charge on insulating patches, so merely touching it does not guarantee complete transfer. A residual reading establishes some transfer under the controls; the disk retest checks the stronger claim of complete transfer. The ideal argument assumes no leakage and no relevant contact-generated charging.

Grounding changes the ledger. With no exterior sources and zero electric potential at infinity, a closed grounded enclosure has zero exterior electric field, inner-wall electric charge $-q$ and zero outer-wall electric charge.^[ref] Consequently, disconnecting ground before withdrawal retains total metal electric charge $-q$, which moves to the exterior after withdrawal. Keeping ground connected during withdrawal instead restores zero electric charge. A lasting opposite-sign response can therefore arise through grounding without body–wall contact.

For the real open pail, no Gaussian surface lying wholly in metal closes over the mouth. Increasing insertion depth and seeking a stable response tests whether enclosure is a good approximation; mesh openings, rim proximity and nearby objects limit it. In a fixed-geometry approximation, the positive insertion response obeys $\Delta V\simeq q/C_{\mathrm{eff}}$, where $C_{\mathrm{eff}}$ is the effective capacitance of the pail, shield, cable and electrometer input. Changed geometry changes that conversion. Wait for charge relaxation and instrument settling at each step, while keeping the observation short compared with leakage. The experiment probes settled electrostatics; it does not establish shielding against arbitrary changing electromagnetic fields.

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.

The Debye–Hückel approximation linearizes mobile-ion Boltzmann densities about a neutral bulk electrolyte. It is valid for ionic potential energies small compared with $k_{B}T$. The inverse Debye–Hückel screening length satisfies ${\kappa }_D^2={\sum }_i{n}_i{z}_i^2{e}^2/(\epsilon k_{B}T)$. The quadratic screened charging free energy includes both electric field energy and the linearized ionic-response term $\epsilon {\kappa }_D^2{\varphi }^2/2$.

For a plane bounding a symmetric electrolyte in $x>0$, assume no displacement field on the other side, bulk potential zero, and $|ze\psi |\ll k_{B}T$. The Debye–Hückel approximation gives ${\psi }^{\prime \prime }={\kappa }^2\psi$, whose decaying solution is ${\psi }_0{e}^{-\kappa x}$. The electrostatic interface boundary conditions give $\sigma =-\epsilon {\psi }^{\prime }(0^{+})=\epsilon \kappa {\psi }_0$. Equivalently, its diffuse charge density is ${\rho }_e=-\epsilon {\kappa }^2\psi$ and its integral is $-\sigma$. Here $\epsilon ={\epsilon }_0{\epsilon }_r$ in SI units and ${\kappa }^2=2z^2{e}^2{n}_{\infty }/(\epsilon k_{B}T)$ when $n_{\infty }$ counts each ionic species. An isolated sheet with identical electrolyte on both sides has two diffuse layers, so the total sheet charge is $2\epsilon \kappa {\psi }_0$ instead. A surface geometry and the field on its other side must therefore be specified.

For a uniformly charged sphere of radius $a$, total electric charge $Q$, and an exterior dilute electrolyte of permittivity $\epsilon$, the Debye–Hückel approximation gives $(r^2{\varphi }^{\prime }{)}^{\prime }={\kappa }^2{r}^2\varphi$. Decay at infinity and Gauss's law, $-\epsilon {\varphi }^{\prime }(a)=Q/(4\pi a^2)$, select the displayed electric potential. Substituting $u=r\varphi$ reduces the proof to $u^{\prime \prime }={\kappa }^{2}u$. The screening parameter is ${\kappa }^2={\sum }_i{z}_i^2{e}^2{n}_{i,\infty }/(\epsilon k_{B}T)$, and the ion profiles are $n_i\simeq n_{i,\infty }(1-z_{i}e\varphi /(k_{B}T))$. The validity condition is $|z_{i}e\varphi (a)|\ll k_{B}T$ for every relevant species. Integrating the diffuse charge density using the radial equation gives total screening charge $-Q$.

A lateral sinusoidal surface charge density with wave number $k$ in the Debye–Hückel approximation generates vertical exponential or hyperbolic functions with inverse length $p=\sqrt{k^2+{\kappa }_D^2}$. In a homogeneous electrolyte filling all space, an isolated sheet at $z_s$ produces $\varphi ={\sigma }_k(x)e^{-p|z-z_s|}/(2\epsilon p)$. Boundary conditions determine the normalization when the electrolyte occupies only a gap.

Two equally modulated charge sheets in identical full-space electrolyte have charging energy per area ${\alpha }^2[1+e^{-pd}\cos \theta ]/(4\epsilon p)$, where $p=\sqrt{k^2+{\kappa }_D^2}$. Its phase-dependent interaction favors a half-wavelength offset, with opposite charge patches facing each other. For an electrolyte gap with negligible exterior displacement response, the energy is instead ${\alpha }^2[\coth (pd)+\cos \theta /\sinh (pd)]/(2\epsilon p)$, with the same minimizing phase. Absolute energies require specified exterior electrostatic interface boundary conditions.

For a chosen reference charge $q$, the separation at which its Coulomb interaction energy equals $k_{B}T$. The charge-$q$ convention is ${\ell }_B=q^2/(4\pi {ϵ}_{\mathrm{abs}}{k}_{B}T)$ in SI, or ${\ell }_B=\beta q^2/ϵ$ in Gaussian units. The commonly tabulated value uses the elementary charge; changing counterion valence requires tracking that reference convention. It sets the strength of counterion condensation.

A mean-field coupling of the Poisson equation to an ideal Boltzmann distribution of mobile ions. In Gaussian units with uniform dielectric constant $ϵ$, ${\nabla }^2\varphi =-(4\pi /ϵ)[{\rho }_f+{\sum }_i{q}_i{n}_{i,0}{e}^{-\beta q_i\varphi }]$. The potential reference and chemical potentials determine the normalization factors. It neglects discrete-ion correlations and assumes local thermal equilibrium. A counterion-only model is not the same equation as a symmetric salt reservoir.

For an ideal planar electric double layer at fixed surface charge density, let $v=e{\varphi }_0/(k_{B}T)$. Adding bulk divalent chloride concentration $\delta c$ adds two monovalent coions per counterion. Differentiating the squared Grahame equation gives the displayed expression, with all denominator concentrations evaluated at the surface. The numerator concentrations are bulk increments. Dropping depleted coions at strongly negative potential gives the leading counterion approximation. A large finite increment instead requires solving the fixed-charge equation at the new composition; exponentiating a first-order voltage estimate does not make that estimate exact.

A negatively charged plane with monovalent positive counterions in $z>0$ and no salt reservoir has the mean-field profile displayed above, where $b=e/(2\pi {\ell }_B|\sigma |)$ and Bjerrum length ${\ell }_B=e^2/(4\pi \epsilon k_{B}T)$. Its potential relative to the wall is $2k_{B}T\log (1+z/b)/e$. The positive-ion charge density integrates to $|\sigma |$, neutralizing the plane; the field decays algebraically. The potential has no finite zero-at-infinity reference, unlike the screened symmetric-reservoir problem.

For a symmetric monovalent point-ion electrolyte next to a plane, define $u=e\psi /(k_{B}T)$ and ${\kappa }_D^2=2e^2{n}_{\infty }/(\epsilon k_{B}T)$, with $n_{\infty }$ the density of each sign. The Poisson-Boltzmann equation becomes $u^{\prime \prime }={\kappa }_D^2\sinh u$. Multiplying by $u^{\prime }$ and matching zero field and potential in the bulk gives $(u^{\prime }{)}^2=2{\kappa }_D^2(\cosh u-1)$. The decaying branch integrates to the displayed profile, reducing to $u_0{e}^{-{\kappa }_{D}z}$ at weak potential. This solution assumes a salt reservoir and is distinct from a counterion-only diffuse layer.

A strong negative surface potential creates exponentially enhanced positive counterion density and exponentially suppressed negative co-ion density. The co-ion population remains positive, unlike an extrapolated linear approximation. Finite ion size and correlations eventually limit the applicability of point-ion mean-field theory.

For $|e\psi |\ll k_{B}T$, $\psi \prime \prime ={\kappa }^2\psi$ with ${\kappa }^2=2e^2{n}_0/(\epsilon k_{B}T)$ for a symmetric monovalent salt with per-species reservoir density $n_0$. The half-space solution with $\psi (0)=\zeta$ and vanishing bulk potential is $\zeta e^{-\kappa x}$. If density denotes the sum of species, the written factor of two changes.

For point ions of charge $\pm e$ in potential $\psi$, equilibrium electrochemical potentials give $n_{\pm }=n_0\exp [\mp e\psi /(k_{B}T)]$, with $n_0$ the reservoir density of each species. The charge density is $-2e{n}_0\sinh [e\psi /(k_{B}T)]$. A negative wall enriches positive counterions and depletes negative co-ions.

For a symmetric electrolyte with mobile charges $\pm q$, bulk density $n_0$ of each sign, and constant permittivity $\epsilon$, the Poisson-Boltzmann equation gives $n_{\pm }=n_0{e}^{\mp q\psi /(k_{B}T)}$. In planar geometry let $E=-{\psi }^{\prime }$ and $\rho =q(n_{+}-n_{-})$. Then $k_{B}T(n_{+}+n_{-}{)}^{\prime }=\rho E$ and $\epsilon E{E}^{\prime }=\rho E$ by Gauss's law, proving that the displayed bulk-subtracted normal stress is constant. At the midpoint of identical equally charged plates, symmetry gives $E=0$ and $\Pi =2n_0{k}_{B}T[\cosh (q{\psi }_{\mathrm{mid}}/(k_{B}T))-1]$. Its ionic osmotic pressure and electrostatic contributions together transmit the force; bare surface-charge repulsion alone omits essential terms.

For a planar surface contacting a symmetric monovalent ideal electrolyte on one side, the Poisson-Boltzmann equation is ${\psi }^{\prime \prime }=(2e{n}_{\infty }/\epsilon )\sinh (e\psi /(k_{B}T))$. Multiplication by ${\psi }^{\prime }$ and integration using the zero field and zero potential in the bulk givesThe solution decays towards the bulk, and the boundary condition $\sigma =-\epsilon {\psi }^{\prime }(0^{+})$ therefore yields the displayed charge-potential relation, with ${\kappa }^2=2e^2{n}_{\infty }/(\epsilon k_{B}T)$. Its small-potential limit is the planar Debye-Hückel double layer. It is nonlinear but still a mean-field model with point-like ions and constant permittivity; it does not include a compact interfacial layer or finite-ion-size saturation.

Differentiate the Grahame equation for a planar symmetric monovalent electrolyte with respect to surface potential at fixed reservoir concentration and temperature. The result is the displayed capacitance per unit area. It approaches $\epsilon {\kappa }_D$ in the Debye–Hückel approximation. A compact surface layer can contribute a separate series capacitance; point-ion packing assumptions also fail at sufficiently high near-surface concentration.^[ref]

Confine the counterions of a uniformly charged infinite cylinder within a coaxial cylinder of radius $R$, with total mobile line charge $\lambda$. The electric field vanishes at $R$, while its surface value at $a$ is fixed by the bare line charge. This gives $a{\varphi }^{\prime }(a)=2\lambda /ϵ$ and $R{\varphi }^{\prime }(R)=0$ in Gaussian units. Sending $R/a\to \infty$ defines the infinite-dilution limit of cylindrical counterions.

In a neutral counterion-only cylindrical cell, send the outer radius to infinity at fixed cylinder charge. Below the Manning parameter threshold, number density tends to zero at each fixed radius. Above threshold, the cylindrical counterion-only Poisson-Boltzmann profile retains line charge $\lambda -q/{\ell }_B$ locally, while the rest recedes to infinity. Therefore the limit of the full finite-cell charge integral is not the charge integral of the pointwise limiting density.

For $u=\log (r/a)$ and $\Phi =\beta q\varphi -2u$, a counterion-only Poisson-Boltzmann equation reduces to ${\Phi }^{\prime \prime }=-C{e}^{-\Phi }$. Conditions $\Phi (0)=0$, $\Phi \to \infty$ and ${\Phi }^{\prime }\to 0$ give $\Phi =2\log (1+bu)$, $b=\sqrt{C/2}>0$. Matching surface Gauss's law gives $b=\eta -1$, so this branch requires the Manning parameter $\eta >1$. Its number density is $n(r)=(\eta -1{)}^2(a/r{)}^2/[2\pi {\ell }_B{a}^2(1+(\eta -1)\log (r/a){)}^2]$. It leaves residual line-charge magnitude $q/{\ell }_B$.

Localization of a finite fraction of oppositely charged ions near a charged object even as the confining volume grows without bound. In the salt-free infinite-cylinder mean-field model, the Manning parameter must exceed one for a nonzero limiting local density. The condensed charge does not include ions that escape to infinity; finite-cell charge neutrality and the local infinite-volume limit must not be interchanged.

In the mean-field infinite-dilution cylinder model, the fraction of neutralizing charge that remains localized is $f_c=\max (1-1/\eta ,0)$, where $\eta$ is the Manning parameter. For $\eta >1$, integrating the cylindrical counterion-only Poisson-Boltzmann profile gives $Q_c=\lambda -q/{\ell }_B$. Below threshold, the local limiting density vanishes even though every finite neutral cell contains the full counterion charge.

The positive mobile charge per unit axial length accumulated from cylinder radius $a$ to $r$ is $Q(r)=2\pi {\int }_a^{r}s\rho (s)ds$. For fixed negative line charge $-\lambda$, Gauss's law gives signed radial electric field $E_r=2[-\lambda +Q(r)]/(ϵr)$ in Gaussian units. Its absolute value is the field magnitude. An integral over a local infinite-dilution profile measures condensed charge, rather than every ion in a finite neutral cell.

The dimensionless line-charge coupling $\eta =\lambda {\ell }_B/q$, with negative cylinder line charge $-\lambda$ and counterions of charge $q>0$. Here the Bjerrum length uses the same reference charge $q$. The mean-field infinite-dilution threshold is $\eta =1$. If ${\ell }_B$ is instead tabulated for elementary charge $e$ and counterions have charge $ze$, the corresponding coupling is $z\lambda {\ell }_B/e$.

Coulomb's law gives the mutual electric force between stationary point charges in vacuum. Put electric charges $q_1$ and $q_2$ at ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, and define the displacement from charge 1 to charge 2 by $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$, with separation $r=|\mathbf{r}|>0$ and unit vector $\hat{\mathbf{r}}=\mathbf{r}/r$. In SI units, the force on charge 2 due to charge 1 isHere ${\epsilon }_0$ is the vacuum permittivity. The magnitude is $k|q_1{q}_2|/r^2$. If $q_1{q}_2>0$, the force on charge 2 points away from charge 1; if $q_1{q}_2<0$, it points toward charge 1. Interchanging the charges gives ${\mathbf{F}}_{1\leftarrow 2}=-{\mathbf{F}}_{2\leftarrow 1}$, consistent with Newton's third law in this electrostatic setting. ^[ref].

For a fully computed example, stipulate $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$, ${\mathbf{r}}_1=0$, ${\mathbf{r}}_2=0.20\mathrm{m}\hat{\mathbf{x}}$ and the approximate value $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. ThenThe force is attractive, of magnitude $1.3485\times 10^{-6}\mathrm{N}$. The source electric field at charge 2 is ${\mathbf{E}}_1({\mathbf{r}}_2)=449.5N/C\hat{\mathbf{x}}$, pointing away from the positive source. Multiplication by the negative $q_2$ reverses the direction to give the attractive force. This field excludes charge 2's own self-field.

Holding the electric charges fixed while changing the separation from $R$ to $2R$ or $3R$ gives the normalized inverse-square law valuesFor the stipulated example, separations $0.20$, $0.40$ and $0.60\mathrm{m}$ give magnitudes $1.3485\times 10^{-6}$, $3.37125\times 10^{-7}$ and $1.498333\dots \times 10^{-7}\mathrm{N}$. The diagram computes these ratios and uses one common arrow-length scale for both sign choices.

The inverse-square law has a geometric explanation through Gauss's law: a source point charge gives fixed total electric flux through a concentric sphere, whose surface area is $4\pi r^2$. Dividing that flux by the area gives a radial electric field proportional to $1/r^2$. This connects a pairwise force calculation to a global conservation-like accounting of field sources. For several stationary point charges, the superposition principle adds the individual electric fields as vectors, and ${\mathbf{F}}_i=q_i{\mathbf{E}}_{\mathrm{other}}({\mathbf{r}}_i)$ uses all sources except charge $i$ itself. The Coulomb potential energy gives an alternative route when the desired quantity is work rather than instantaneous force.

Place $+q$ at $-L$ and $-q$ at $+L$. For a nonzero point charge $q_3$ between them, the signed force at $x=fL$ is $2k_{e}q{q}_3(1+f^2)/[L^2(1-f^2{)}^2]$. It is nonzero at $f=0$ and even in $f$, so its first nonconstant term is quadratic. The midpoint is not an equilibrium and cannot be treated as a harmonic restoring centre.

Three identical like point charges, initially at rest at an equilateral triangle's vertices, remain equilateral as they move away from the fixed centroid. Each resultant force is radial by symmetry. If the centroid radius is $r$, its magnitude is $k_e{q}^2/(\sqrt{3}{r}^2)$. In a nonrelativistic model neglecting radiation, energy conservation gives the final speed $v=\sqrt{2k_e{q}^2/(ma)}$ for initial side $a$ and charge mass $m$.

Interaction described by the Coulomb potential and its associated Coulomb force.

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.

A thin disk of radius $a>0$ in $z=0$ carries uniform surface charge density $\sigma =Q/(\pi a^2)$. Summing the Coulomb potential of concentric rings gives the displayed axis potential, with zero reference at infinity. Put $A=\sigma /(2{ϵ}_0)$, use spherical coordinates $z=r\cos \theta$, and let $P_2(t)=(3t^2-1)/2$. The near-center expansion, separately harmonic above and below the charged sheet, isThe absolute value represents the jump in normal electric field across the sheet; a single regular solid harmonic series cannot describe both half-spaces simultaneously. The constant and quadratic coefficients follow from the positive-axis Taylor series, and reflection symmetry supplies the other half-space. The far-field multipole expansion isThe monopole is the total charge, the dipole vanishes by reflection symmetry, and the next term is the axisymmetric electric quadrupole. These expansions converge for $r<a$ within each source-free half-space and for $r>a$ in the exterior, respectively; at a point of the sheet their potentials have continuous limiting values but their normal derivatives differ.

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.

A state with a stationary electric charge distribution and no net drift of mobile charge driven by the electric field. A conductor in electrostatic equilibrium has zero interior electric field, while bound charges in an insulating material need not move to eliminate its interior field.

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].