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For a localized charge distribution, the far electric potential is an inverse-radius expansion whose first terms are the total charge monopole, the electric dipole moment, and the electric quadrupole moment.

An electric potential difference compares the electric potentials at two points. In an electrical circuit it is the voltage across the terminals of a component. A vertex potential in an electrical network is specified relative to a reference vertex; terminal differences are independent of an additive change of reference.

Chemical systems whose reactions, transport or equilibrium couple to electric charges and electric potential. An electrolyte carries mobile ions; a Poisson-Boltzmann equation can approximate their equilibrium distributions.

The four-potential combines the electric potential and magnetic vector potential as $A^{\mu }=(\varphi /c,\mathbf{A})$. The electromagnetic four-potential $A_{\mu }$ transforms underwhile the antisymmetric field tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is unchanged.

For distinct point charges $q_i$ at positions $r_i$, the electrostatic potential energy isWhen some charges are fixed, the terms depending on a movable charge form its potential energy in their electric potential.

The grounded conducting plane supplies the induced charge needed for a uniform field between the plates and zero field above the upper plate. Gauss's law gives $E_z=-\sigma /{ϵ}_0$ in the gap. With electric potential zero at the grounded plate,Integrating the electrostatic energy density ${ϵ}_0|\mathbf{E}{|}^2/2$ across the gap gives the capacitor energy per unit areaFor an isolated upper plate, charge is fixed, so increasing $d$ to $(1+\alpha )d$ gives exactly $\Delta u=\frac{1}{2}\sigma V\alpha$, with $V$ its initial value. The positive increase equals the mechanical work done against attraction.

At fixed voltage, the new charge density is $\sigma /(1+\alpha )$, and the energy becomes $u/(1+\alpha )$. Consequently $\Delta u=-\frac{1}{2}\sigma V\alpha +O({\alpha }^2)$. Charge $\sigma \alpha +O({\alpha }^2)$ per unit area flows back to the voltage source, which receives energy $\sigma V\alpha +O({\alpha }^2)$. The externally supplied mechanical work is still $\frac{1}{2}\sigma V\alpha +O({\alpha }^2)$: the field loses half this returned electrical energy and the mechanical work supplies the other half. Thus the fixed-voltage capacitor energy balance includes both field and source energy.

Let $A_s$ denote the surface area of an n-sphere for the unit sphere $S^s$. In this notation $s$ is the intrinsic dimension of the sphere. Cartesian factorization of the Gaussian integral gives $I_N={\pi }^{N/2}$. In hyperspherical coordinates,The substitution $t={\rho }^2$ supplies the defining Gamma function integral. Scaling the sphere radius multiplies its surface volume by $r^{N-1}$, henceThis is the volume of the sphere itself, not of its enclosed ball.

Take $D$ to denote spacetime dimension and $d=D-1$ spatial dimensions. In uncompactified flat space, Gauss's law for a charge $q$ and permittivity ${\epsilon }_d$ gives $E_r{A}_{d-1}{r}^{d-1}=q/{\epsilon }_d$. Since $E_r=-d\phi /dr$, the electric potential for $d>2$, normalized to vanish at infinity, isThus the four-dimensional electric potential falls as $1/r$, and the five-dimensional one as $1/r^2$. If a convention counts only spatial dimensions as $D$, replace $d$ by $D$ in the first formula. For $d=2$, the integral instead gives $\phi =-q\log (r/r_0)/(2\pi {\epsilon }_2)$; for $d=1$, it gives $\phi =-q|x|/(2{\epsilon }_1)$ up to a constant. Neither case has a finite zero at infinity.

For a Newtonian gravitational potential ${\Phi }_g$, define the flux-normalized coupling ${\kappa }_d$ by ${\nabla }^2{\Phi }_g={\kappa }_{d}m{\delta }^{(d)}(\mathbf{x})$ and acceleration $-\nabla {\Phi }_g$. Exactly the same flux integration gives the attractive potentialIn $d=3$, ${\kappa }_3=4\pi G_4$ reproduces $-G_{4}m/r$. If $G_D$ is defined by the $D$-dimensional Einstein-Hilbert action $S=(16\pi G_D{)}^{-1}\int \sqrt{-g}R$, the weak static Einstein field equations give ${\kappa }_d=8\pi G_D(D-3)/(D-2)$ for $D>3$. ThereforeIndeed, $R_{00}\simeq {\nabla }^2{\Phi }_g$ and $T_{00}-g_{00}T/(D-2)=(D-3)\rho /(D-2)$ in signature $(-+\cdots +)$. A Poisson-defined Newtonian theory in two spatial dimensions has a logarithmic potential, but pure three-dimensional Einstein gravity has no analogous local Newtonian point-mass force; the relativistic normalization must not be extrapolated to that exceptional dimension.

For an unwarped product with a fixed extra-dimensional metric, normalize the Einstein-Hilbert action by $\frac{1}{2}{M}_{\ast }^{n+2}\int d^{4+n}x\sqrt{-g}R$. Integrating the part containing the four-dimensional curvature over the internal space gives $\frac{1}{2}{M}_{\ast }^{n+2}\mathrm{vol}(S_R^n)\int d^{4}x\sqrt{-g_4}{R}_4$. Thus, using the reduced Planck mass,This is a volume relation for compactification; a curved sphere additionally requires a mechanism supporting and stabilizing its background. It is not by itself a proof that a sphere times flat spacetime solves a vacuum gravitational theory.

With $M_{\ast }=10^3\mathrm{GeV}$, $M_{\mathrm{Pl}}=2.435\times 10^{18}\mathrm{GeV}$ and $1{\mathrm{GeV}}^{-1}=1.97327\times 10^{-16}\mathrm{m}$, the spherical-volume factors $A_1=2\pi$, $A_2=4\pi$, $A_6=16{\pi }^3/15$ giveThe first estimate is astronomical; the second is roughly a tenth of a millimetre. If both Planck scales are instead defined by $S=M_D^{D-2}\int \sqrt{-g}R/(16\pi )$, the same algebraic volume relation uses the unreduced $M_{\mathrm{planck}}=1.221\times 10^{19}\mathrm{GeV}$. Holding that convention's $M_{\ast }=1\mathrm{TeV}$ gives radii larger by $(8\pi {)}^{1/n}$: approximately $4.69\times 10^{12}\mathrm{m}$, $6.80\times 10^{-4}\mathrm{m}$, $2.54\times 10^{-14}\mathrm{m}$. These are different conventions for what is held fixed at one TeV, not different scaling laws.

Finally, compactify the fifth coordinate on a circle of circumference $L=2\pi R$. For a source and observer at the same compact coordinate, the method of images turns a five-dimensional static potential $C/{\rho }^2$ intoOne can derive this image sum without assuming a summation formula. The Fourier transform of $1/(r^2+y^2)$ is $\pi e^{-r|k|}/r$; Poisson summation then gives $\pi [1+2{\sum }_{\ell \ge 1}{e}^{-2\pi \ell r/L}]/(Lr)$, which is the displayed hyperbolic cotangent. ThereforeThe short-distance limit is $C/r^2$; the long-distance limit is the four-dimensional $1/r$ potential. These exponential corrections are the nonzero Kaluza-Klein modes, with masses $2\pi |\ell |/L$. For electrostatics $C=q/(4{\pi }^2{\epsilon }_4)$, so the effective permittivity is ${\epsilon }_3=L{\epsilon }_4$. For a Poisson-normalized gravitational potential, $C=-{\kappa }_{4}m/(4{\pi }^2)$ gives ${\kappa }_3^{\mathrm{eff}}={\kappa }_4/L$. A massless scalar associated with the circle radius also contributes in an unstabilized gravitational compactification; recovering pure four-dimensional Einstein gravity with $G_4=G_5/L$ requires its stabilization or removal. The $1/r$ crossover itself is independent of this tensor-versus-scalar normalization issue.

Use the Gaussian units convention implicit in the $4\pi$ coefficient. For a uniform dielectric and a counterion-only electrolyte, the Boltzmann distribution and Poisson-Boltzmann equation areThe reference electric potential is $\varphi (a)=0$, so $n_0=n(a)$; it is not a prescribed nonzero bulk concentration at infinity. A salt reservoir with additional ion species would require their charge densities as well. With $u=\log (r/a)$, the cylindrical Laplacian is $r^{-2}{d}^2/d{u}^2$. Since $\beta q\varphi =\Phi +2u$ and $r^2=a^2{e}^{2u}$, the factors of $e^{2u}$ cancel:Here primes mean differentiation with respect to $u$, and the surface condition is $\Phi (0)=0$ in this coordinate.

Multiply by ${\Phi }^{\prime }$ to obtain a first integral:The two specified far-field limits set $K=0$. For a solution tending upwards to infinity, take the positive square root. Writing $b=\sqrt{C/2}$ gives $(e^{\Phi /2}{)}^{\prime }=b$, hence the cylindrical counterion-only Poisson-Boltzmann profileFor $b>0$, $\Phi \to \infty$ and ${\Phi }^{\prime }=2b/(1+bu)\to 0$. The corresponding number density is

Apply Gauss's law to a coaxial surface per unit cylinder length. Define the positive accumulated cylindrical screening charge byThe signed outward electric field is $E_r=2[-\lambda +Q(r)]/(ϵr)$, so its magnitude is $2|\lambda -Q(r)|/(ϵr)$. In the branch here, $Q(r)<\lambda$, the field is inward andIn this display, ${\varphi }^{\prime }$ is a radial derivative. The quantities $Q$ are electric charges per unit axial length, not total charges of an infinite cylinder.

At the surface, $a{\varphi }^{\prime }(a)=2\lambda /ϵ$. Define the charge-$q$ Bjerrum length ${\ell }_B=\beta q^2/ϵ$ and the Manning parameter $\eta =\lambda {\ell }_B/q$. Matching the derivative of the solution givesThus the condensed branch exists only for $\eta >1$, and its normalization isThe sign condition $b>0$ is essential: simply squaring $\eta -1$ would produce a spurious subcritical density. Direct integration using $rn(r)dr=n_0{a}^{2}du/(1+bu{)}^2$ yieldsEquivalently, $E(r)=2{\lambda }_c[1+b/(1+b\log (r/a))]/(ϵr)$: the remaining far-field line-charge magnitude is ${\lambda }_c$. The Manning condensed fraction is $1-1/\eta$.

Below threshold, no positive-density profile can satisfy those same far-field conditions. Indeed ${\Phi }^{\prime \prime }\le 0$, so a negative initial slope $2(\eta -1)$ cannot increase to zero. The physical interpretation is the infinite-dilution limit of cylindrical counterions: start with a neutral finite cylindrical cell and send its outer radius to infinity. For $\eta \le 1$, counterions escape to arbitrarily large radii, leaving $n_0\to 0$ and no finite condensed charge. The bare-cylinder Boltzmann distribution $n\propto r^{-2\eta }$ also shows why its radial normalization $\int r^{1-2\eta }dr$ diverges in this regime. In the local zero-density limit, $\Phi =2(\eta -1)u$; this does not obey the condensed branch's imposed limits. At $\eta =1$ the branch has zero amplitude and $\Phi =0$, rather than $\Phi \to \infty$.

Consequently the intended counterion condensation result, interpreted as a local infinite-dilution density, isEvery finite neutral cell still contains enough counterions to neutralize the cylinder; the missing residual charge in the local limiting profile is carried to infinity. Taking the infinite-cell limit before the charge integral differs from integrating the entire finite cell first. In SI units, the same reference-charge Bjerrum length is $q^2/(4\pi {ϵ}_{\mathrm{abs}}{k}_{B}T)$; the electrostatic prefactors must be changed consistently.