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The fringing field of a capacitor is the nonuniform electric field near the edges of its electrical conductors, extending outside the gap. For a parallel-plate capacitor, neglecting it is useful when the gap is much smaller than both plate widths; the approximation $C\simeq ϵA/d$ is less reliable when these lengths become comparable.

Finite plates break the translational symmetry of the infinite-plate model. Each conductor in electrostatic equilibrium is an equipotential, so the electric field must meet a smooth conducting surface normally. Near an edge, the equipotentials curve into the surrounding space and the electric field changes both direction and magnitude. The surface charge density adjusts to enforce the conducting boundary conditions; it is not uniform all the way to a sharp edge. The electrostatic boundary conditions at a conductor explain why zero electric field inside the metal does not imply zero electric field just outside it.

The practical issue is where the nearly uniform region ends. A probe near the centre can support $E\simeq V/d$ while a probe close to an edge detects a lateral component and a different magnitude. A total capacitance reading also includes electric flux outside the simple footprint. The fringing field of a capacitor can therefore matter in permittivity measurements even when a central electric field measurement looks nearly ideal. Keysight's three-electrode fixture diverts the fringe contribution away from the measured response.^[ref]

An ideal infinite sheet midway between parallel grounded plates has equal fields of magnitude $\sigma /(2{ϵ}_0)$ on its two sides, directed from positive sheet charge toward the plates. Integrating over the half-separation d/2 gives the displayed electric potential. Each plate carries induced inner-face surface charge density $-\sigma /2$. The potential is triangular between the plates, and the field jumps by $\sigma /{ϵ}_0$ across the sheet. Edge fields are excluded.

Linear charge density is electric charge divided by length along a charged line, or locally $dQ/d\ell$. Its SI unit is coulombs per metre. For a uniformly charged long cylindrical conductor, it equals the total surface charge density integrated around the cylinder's circumference; it need not represent a charge literally supported on a one-dimensional line.

For ideal oppositely charged parallel plates, the normal electric field is $\sigma /{ϵ}_0$. A boost along that normal leaves each transverse plate area unchanged. Invariance of total charge therefore preserves surface charge density and the normal electric field. The plate separation contracts, but the large-plate field does not depend on that separation. Including longitudinal contraction in a voltage argument while holding voltage fixed would impose an unsupported extra assumption.

An annulus with uniform surface charge density $\sigma$, radii $a,b$ and rigid angular velocity $\Omega$ carries ring electric current $dI=\sigma \Omega rdr$. The centre magnetic field of each ring is ${\mu }_{0}dI/(2r)$ by the Biot-Savart law. Integrating gives $\mathbf{B}(0)={\mu }_0\sigma \Omega (b-a)\hat{\mathbf{n}}/2$, equivalent to the title expression with $Q=\pi \sigma (b^2-a^2)$. The oriented normal follows the positive rotation convention, with the sign of $Q$ retained.

A parallel-plate capacitor consists of two parallel electrical conductors with overlapping area $A$ and separation $d$. If the gap is small compared with both plate widths and is filled by a uniform material of positive permittivity $ϵ$, its capacitance is approximatelyThis neglects the fringing field of a capacitor near the plate edges. The interior electric field points from the plate of positive electric charge to the plate of negative electric charge, and is approximately uniform. ^[ref].

For a worked vacuum example, stipulate $A=0.01{\mathrm{m}}^2$, $d=0.001\mathrm{m}$, and vacuum permittivity ${ϵ}_0\approx 8.854\times 10^{-12}F/m$. This rounds the value in Fundamental Physical Constants — Complete Listing by National Institute of Standards and Technology and others. Within the approximation,At terminal electric potential difference $V=10\mathrm{V}$, the electric field magnitude is $E\simeq V/d=10^{4}V/m$, the plate electric charge magnitude is $Q=CV=0.8854\mathrm{nC}$, and the capacitor energy is $U=C{V}^2/2=4.427\mathrm{nJ}$. These numbers use the stipulated rounded vacuum permittivity; finite plates also have corrections from the fringing field of a capacitor.

The formula makes design changes easy to predict: doubling the overlapping area doubles the capacitance, and halving the gap doubles the capacitance, with permittivity fixed and the fringing field of a capacitor neglected. Doubling permittivity also doubles capacitance if the material uniformly fills the gap and remains linear. This lets a compact device store more separated electric charge at the same terminal electric potential difference.

The short derivation uses surface charge density $\sigma =Q/A$. In the interior, Gauss's law gives electric field magnitude $E\simeq \sigma /ϵ=Q/(ϵA)$. The electric potential difference is $V\simeq Ed=Qd/(ϵA)$, so dividing $Q$ by $V$ gives $C\simeq ϵA/d$. A narrow curved capacitor gap has the same leading local approximation when its width is much less than the curvature radius.

Write $e=q>0$ for the elementary charge and $ϵ={ϵ}_0{ϵ}_r$ for the permittivity. Taking the bulk electric potential as zero, the Boltzmann distribution of point ions gives $c_i(x)=c_i^{\infty }\exp [-e{z}_i\varphi (x)/(k_{B}T)]$. Here $c_i$ is a number density, not a molar concentration. Combining this with Poisson's equation gives the planar Poisson-Boltzmann equationMultiplying by ${\varphi }^{\prime }$ and integrating from the neutral bulk, where $\varphi ={\varphi }^{\prime }=0$, yieldsFor the solution occupying $x>0$, with the field on the other side neglected, Gauss's law gives $\sigma =-ϵ{\varphi }^{\prime }(0)$. On the branch that decays to zero without crossing it,Source sign qualification: the unsigned square root printed in the paper gives $|\sigma |$. A negative wall electric potential, such as the next part's $-73\mathrm{mV}$, has ${\varphi }^{\prime }(0)>0$ and hence $\sigma <0$, providing a concrete counterexample to interpreting that printed expression as signed surface charge density. This is the planar Grahame equation, with an explicit sign convention.

Assume the structural surface charge density stays fixed while a divalent chloride is added, without changing $T$ or $ϵ$. Let $v=e{\varphi }_0/(k_{B}T)$ and $\delta c$ be the increment of the bulk divalent concentration. Electroneutral addition also increases bulk chloride by $2\delta c$. Varying the squared Grahame equation at fixed $\sigma$ givesAt a strongly negative wall, coions are small. Dropping their contribution to the denominator and $2e^v$ in the numerator gives the requested leading approximationThe $-3$ includes the change in bulk osmotic concentration of one added divalent counterion and its two monovalent coions. The numerator's increment is a bulk concentration; the denominator contains surface concentrations.^[ref] Extra counterions screen the wall and make its negative electric potential less negative. At fixed bulk calcium, $c_{\mathrm{Ca}}(0)=c_{\mathrm{Ca}}^{\infty }{e}^{-2{\varphi }_0/V_T}$ therefore decreases. Under the stipulated proportionality, synaptic activity decreases as well; this argument isolates electrostatic accumulation rather than assuming a separate specific magnesium effect on the ion channels.

For $\delta c=5\mathrm{mM}$, the printed linear approximation gives $\delta {\varphi }_0=9.53434\mathrm{mV}$. Linearizing the activity too gives $\delta A/A=-2\delta {\varphi }_0/V_T=-0.7422$, or approximately a $74\%$ reduction. Exponentiating this approximate voltage instead gives $A^{\prime }/A=0.4761$. Neither is a controlled finite-increment prediction: the added divalent concentration is 2.5 times its initial value.

A consistent finite-change calculation is possible within the same Poisson-Boltzmann equation. Define, using $\mathrm{mM}$ throughout,Fixed $\sigma$ requires $S_5(u^{\prime })=S_0(2.84129)$. The physical positive root is $u^{\prime }=2.58241$, givingThus the self-consistent finite-change estimate is about a 40.4% reduction. Both estimates rely on fixed surface charge, ideal ionic activities, and unchanged bulk calcium; the limitations of the large local concentrations remain.

Integrate the surface charge density using the surface integral in spherical coordinates:Since $r_0>a$, the elementary integral isSubstitution yields the total induced electric chargeThe equality with the image charge also follows by applying Gauss's law to the exterior electric field: the physical and fictitious descriptions give the same electric flux through a sphere enclosing the conductor but excluding the physical point charge.

Let $\mathbf{n}=\mathbf{r}/a$ be the unit normal from the electrical conductor into the vacuum. The electrostatic boundary conditions at a conductor, together with the vanishing interior electric field, giveWrite $u=\cos \theta$ in spherical coordinates, $D=|\mathbf{r}-{\mathbf{r}}_0|$ on $r=a$, and $z^{\star }=a^2/r_0$. Differentiating the electric potential at fixed angles givesHere the distance identity in the preceding answer was used before simplifying. Hence the induced surface charge density isFor positive $q$ this surface charge density is negative everywhere, with largest magnitude nearest the point charge.

Choose a spherical Gaussian surface of radius $s$ with $a<s<r_0$. Its physical enclosed electric charge is the total induced electric charge $Q_a$ on the inner sphere. The method of images produces the same electric field on that Gaussian surface, so Gauss's law identifies $Q_a$ with the sum of the enclosed image charges. These are the first family with $n\le -1$ and the second family with $n\le 0$:Summing these geometric series givesThe electric field in the empty central cavity vanishes by electrostatic shielding, so it contributes no additional inner-face surface charge density. The formula tends to $-q$ as $r_0\downarrow a$ and to zero as $r_0\uparrow b$, as expected when the point charge approaches one grounded boundary or the other.

Write the constant spherical harmonic $Y_{00}=C$. Its orthonormality gives $4\pi |C{|}^2=1$, independently of its phase convention. Orthogonality to $Y_{00}$ givesOnly the $\ell =m=0$ term contributes to the surface integral of the surface charge density. Multiplying by the surface-area factor $a^2$ givesThus the spherical harmonic calculation reproduces the total induced electric charge:

Let $r_0=|{\mathbf{r}}_0|$ and let ${\Omega }_0=({\theta }_0,{\phi }_0)$ describe its direction in spherical coordinates; write $\Omega =(\theta ,\phi )$. The given Dirichlet Green function between concentric spheres has normalization ${\nabla }^{2}G=-4\pi {\delta }^{(3)}$, so the electric potential is $\varphi =qG/(4\pi {ϵ}_0)$. On the inner sphere the unit normal from metal into the gap is $+{\mathbf{e}}_r$, hence its induced surface charge density isNear this boundary $r_{<}=r$ and $r_{>}=r_0$. The needed radial derivative isIt cancels the factor $2\ell +1$ in the Dirichlet Green function. Therefore the answer in spherical harmonics isThis spherical harmonic expansion converges on the boundary: after the angular sum, the two radial factors decay geometrically with ratios $a/r_0<1$ and $a{r}_0/b^2<1$, while the angular factors grow at most polynomially. This justifies the boundary derivative and the subsequent surface integral.

For this radially layered coaxial cable, defineUse the solution regular at the axis with bounded exterior electric potential, so that the empty interior and exterior have the constant electric potentials fixed by their adjacent conducting cylinders.

There is no free charge in either electrical insulator or on their interface. Gauss's law and continuity of normal electric displacement field give $D_{\rho }=C/\rho$ in both layers. The electric potential difference fixes the constant becauseThe electric potential, including the empty regions bounded by the conducting cylinders, is thereforeBoth inner and outer Dirichlet boundary conditions hold, and the electric potential is continuous at the material interface, whereTaking the negative gradient of the electric potential yieldsThe radial electric field generally jumps at $\rho =c$, but $ϵE_{\rho }(c^{-})={ϵ}^{\prime }{E}_{\rho }(c^{+})$, as required by the absence of free surface charge density there. The conducting cylinders carry opposite linear charge densities $\lambda =2\pi C$ and $-\lambda$.

For an Amperian loop of radius $\rho$, the enclosed electric current is zero, $I$, or zero according as $\rho <a$, $a<\rho <b$, or $\rho >b$. The magnetostatic Ampère law therefore gives $H_{\phi }=I/(2\pi \rho )$ in the gap, and zero outside. Using the two magnetic permeabilities,The tangential magnetic field strength is continuous at the material interface, while the tangential magnetic field need not be. At each ideal conducting surface and at the material interface, the displayed electric field and magnetic field formulas specify one-sided limits; a single interface value is unnecessary when the limiting values differ.

Take zero electric potential at infinity and put $K=(4\pi {\epsilon }_0{)}^{-1}$. By spherical symmetry and Gauss's law, the electric field is radial, with magnitude determined by the total enclosed electric charge. Thus, away from the idealized shell surfaces,Integrating $E_r=-d\Phi /dr$ from infinity, or adding the three spherical-shell potentials, givesThe expressions agree at each interface, so the potential is continuous everywhere, including the origin. The electric field has different interior and exterior limits at each infinitely thin charged surface, with jump equal to surface charge density divided by ${\epsilon }_0$. An ideal sheet has no single classical field value on the sheet itself; the displayed adjacent limits specify the field there. In conducting material of nonzero thickness the internal field is zero.

The electrostatic energy is the integral of ${\epsilon }_0|\mathbf{E}{|}^2/2$. There is no contribution from $r<a$, and spherical integration givesThe dependence on $b$ cancels because the magnitude of enclosed electric charge is $|q|$ on both sides of the middle shell, even though the field direction changes.

Take the divergence of the Ampere-Maxwell law and use Gauss's law $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Since the divergence of a curl is zero, this gives the charge continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{j}=0$. Uniform conductivity and Ohm's law imply $\nabla \cdot \mathbf{j}=\sigma \nabla \cdot \mathbf{E}=\sigma \rho /{ϵ}_0$. ThereforeThe spatially uniform initial density stays uniform in the interior; $\tau$ is the charge relaxation time.

Denote the cylinder radius by $R$. Cylindrical symmetry makes the interior electric field radial. Gauss's law on a coaxial cylinder of radius $r<R$ and length $\ell$ gives $2\pi r\ell E_r=\rho (t)\pi r^2\ell /{ϵ}_0$. ThusThe current does not disappear at the interface: the vacuum conducts no current, so charge accumulates on the cylindrical surface. Initially taking the charge to be the stated volume charge alone, its surface charge density satisfies ${\dot{\rho }}_s=j_r(R^{-},t)$, giving the charge relaxation in an isolated conducting cylinder lawThe total enclosed charge per unit length is $\pi R^2\rho (t)+2\pi R{\rho }_s(t)=\pi R^2{\rho }_0$. Consequently the exterior electric field isindependent of time. The boundary jump obeys $E_r(R^{+})-E_r(R^{-})={\rho }_s/{ϵ}_0$, as it must. Applying Gauss's law to the decaying volume charge alone would incorrectly predict a decaying exterior electric field and violate electric charge conservation. The vector formulas also specify the inward direction if ${\rho }_0<0$.

Take the divergence of the Ampère-Maxwell equation, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{ϵ}_0{\partial }_t\mathbf{E}$. Since the divergence of a curl is zero vanishes, and Gauss's law gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the result isThis is the charge continuity equation.

At an internal point of a homogeneous Ohmic conductor, Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Since $\sigma$ is constant, the charge continuity equation becomes ${\partial }_t\rho =-(\sigma /{ϵ}_0)\rho$. ThereforeThis is charge relaxation. Here $\rho$ is the total charge appearing in the vacuum form of Gauss's law; in a homogeneous dielectric description of free charge the corresponding constant permittivity replaces ${ϵ}_0$.

For a finite isolated body, charge travels to the surface. The bulk formula applies at internal points, not across the interface where electrical conductivity jumps and surface charge density accumulates. Integrating the charge continuity equation over the body, including its surface charge, gives constant total electric charge because no electric current crosses the external boundary. Bulk decay and conservation of electric charge are therefore compatible.

An ideal conductor in electrostatic equilibrium has zero field in its interior and constant potential on its surface. Grounding fixes that potential to zero. The tangential field is continuous and therefore vanishes; the normal field has the jump fixed by the induced surface charge density $\sigma$:Here the normal points upward out of the conductor. The normal field is not generally zero; grounding prescribes the potential, not the induced charge density.

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.

For the conductor in electrostatic equilibrium, the interior electric field is zero and the electrostatic potential is constant. Earthing fixes that constant to zero. Thus the conductor boundary conditions at $x=d$ areThe normal electric field need not vanish; it is supported by a surface charge density. The outward normal from this conductor into the exterior is $\mathbf{n}=-{\mathbf{e}}_x$, so the normal jump condition isThis sign convention will determine the sign of the induced surface charge density below.