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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 $C_0={\epsilon }_{0}A/d$.

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.

For an isolated smooth conducting sphere of radius $a$, Gauss's law gives surface electric field $E_s=V/a$. Combining its capacitance $4\pi {ϵ}_{0}a$ with capacitor energy $C{V}^2/2$ gives the displayed bound when the surrounding medium has dielectric strength $E_{\mathrm{bd}}$ and approximately vacuum permittivity. Nearby conductors, surface roughness and discharge geometry can lower the practical limit.

The capacitance of a linear two-terminal capacitor is the constant relating its separated electric charge to its terminal electric potential difference:Here $Q$ is the electric charge on one electrical conductor, the other carries $-Q$, and $V$ is their electric potential difference with the same ordering. Reversing the terminal ordering reverses both $Q$ and $V$ and leaves capacitance unchanged. An ideal passive capacitor has $C\ge 0$; an ordinary pair of separated electrical conductors with positive permittivity has $C>0$. At $V=0$, use $Q=CV$ rather than the undefined ratio $0/0$.

For example, electric charge magnitude $Q=0.8854\mathrm{nC}$ at terminal electric potential difference $V=10\mathrm{V}$ gives capacitance $C=88.54\mathrm{pF}$. At $20\mathrm{V}$, the same linear capacitor has $Q=1.7708\mathrm{nC}$ and the same capacitance. The plots in capacitor energy show that capacitance is the slope of the electric charge versus terminal electric potential difference graph.

For fixed geometry and linear material response, capacitance depends on dimensions and permittivity, rather than the source's terminal electric potential difference. This is why capacitance is a useful property of a component: once it is known, $Q=CV$ predicts the separated electric charge for each operating terminal electric potential difference. The SI unit is the farad, $1\mathrm{F}=1C/V$. ^[ref].

For combinations of capacitors, the same terminal definition gives an equivalent capacitance. Two identical capacitors of capacitance $C$ connected in parallel share the same terminal electric potential difference, and their electric charges add, giving $C_{\mathrm{eq}}=2C$. Connected in series with an initially neutral isolated junction, they carry equal plate electric charge magnitudes and their terminal electric potential differences add, giving $C_{\mathrm{eq}}=C/2$. ^[ref].

Characterize the instrument and fixture before inferring permittivity: use the prescribed open and short measurements to compensate residual parallel response and series electrical impedance, and check a known reference capacitor or the required load standard. Keep cables and electrode geometry reproducible. An open fixture measurement is distinct from the empty-gap reference capacitor; blindly subtracting its entire capacitance would remove the desired reference signal. Compensation cannot determine an incorrectly measured specimen thickness or eliminate all specimen-dependent fringing field of a capacitor errors. ^[ref].

An isolated conducting sphere of radius $R$ in a medium of permittivity $ϵ$ has surface electrostatic potential $V=Q/(4\pi ϵR)$ relative to infinity. Thus its capacitance is $C=Q/V=4\pi ϵR$.

A 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 energy is the electrostatic energy stored when electric charge is separated. For a linear capacitor of fixed positive capacitance $C$, plate electric charge magnitude $Q$, and terminal electric potential difference $V=Q/C$, it isThe zero is the uncharged state. Thus the capacitor energy is positive for nonzero $Q$, even though the total electric charge of the two plates is zero. ^[ref].

For the parallel-plate capacitor example, capacitance $C=8.854\times 10^{-11}\mathrm{F}$ and terminal electric potential difference $V=10\mathrm{V}$ giveThe plate electric charge magnitude is $Q=0.8854\mathrm{nC}$. Doubling terminal electric potential difference with capacitance unchanged doubles the plate electric charge magnitude and quadruples the capacitor energy.

The electric charge is a linear function of terminal electric potential difference, while capacitor energy is a quadratic function. The second curve in each plot uses twice the capacitance: at a given terminal electric potential difference, both plate electric charge magnitude and capacitor energy double. These relations let us calculate how much energy an electrical circuit can receive when a capacitor discharges.

A brief charging argument explains the factor $1/2$. If the plate electric charge magnitude has reached $q$, the terminal electric potential difference is $q/C$. Transferring a further amount $dq$ of electric charge quasistatically requires work $dW=(q/C)dq$. With the geometry and permittivity fixed, the integral isUsing $Q=CV$ gives the other two forms. This integral accounts for the gradually rising terminal electric potential difference; it is the work stored in the capacitor's electric field, rather than a claim that all energy drawn from a fixed-voltage battery during charging ends up stored.

The field picture gives the same result. In the vacuum gap of a parallel-plate capacitor, the electric field has energy density $u={ϵ}_0{E}^2/2$. Multiplying by the gap volume $Ad$ gives the electrostatic energyFor the worked example, $u=4.427\times 10^{-4}J/{\mathrm{m}}^3$ and $Ad=10^{-5}{\mathrm{m}}^3$ reproduce $U=4.427\mathrm{nJ}$. The electrostatic energy resides in the electric field throughout the gap, despite overall charge neutrality.

When capacitance changes, specify the constraint. At fixed plate electric charge magnitude $Q$, increasing capacitance decreases capacitor energy $Q^2/(2C)$; doubling capacitance halves it. At fixed terminal electric potential difference $V$, increasing capacitance increases capacitor energy $C{V}^2/2$; doubling capacitance doubles it. In the latter case a connected battery supplies additional electric charge and energy, and there may also be mechanical work. Use the fixed-voltage capacitor energy balance to account for all transfers before inferring a force from changing capacitor energy.

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 $-Q$.

The region around a p-n junction with reduced mobile charge carrier density.

An illustrative abrupt-junction depletion approximation assumes uniform, fully ionized donor impurities and acceptor impurities, negligible mobile space charge in the depleted layer, and permittivity ${\epsilon }_s$. For area $A$, doping number densities $N_A,N_D$ and built-in electric potential difference $V_{\mathrm{bi}}$, it givesHere $q>0$ is the elementary charge and $V$ is positive for forward bias. Increasing both doping densities reduces $W$ in this estimate. That explains the preparation strategy for a tunnel diode; it does not make this sharp-depletion approximation a precise microscopic model of a degenerate semiconductor.

The motivating observation is that inserting an insulating material between fixed capacitor plates increases its capacitance. ^[ref]. To investigate this response, use two flat parallel metal electrodes with measured overlapping area $A$ and separation $d$, a uniform specimen filling the gap, and an instrument that measures electrical impedance. Record the capacitance with and without the specimen at the same small applied electric potential difference, frequency and temperature, after capacitance measurement calibration. An increase in corrected capacitance indicates a larger permittivity, provided the geometry is unchanged. A guard electrode reduces the contribution of the fringing field of a capacitor. ^[ref].

For negligible dielectric loss in the parallel-plate capacitor approximation,An air reference approximates $C_0$; it is not an exact vacuum reference. For the following synthetic example, stipulate $A=0.010{\mathrm{m}}^2$, $d=1.00\mathrm{mm}$ and the rounded vacuum permittivity ${ϵ}_0\approx 8.854\times 10^{-12}F/m$. Then $C_0=88.54\mathrm{pF}$ in this model, and relative permittivity four gives $C=354.16\mathrm{pF}$. A residual parallel capacitance of $20\mathrm{pF}$ would make an uncorrected sample/reference ratio $374.16/108.54\approx 3.447$, rather than four. This additive error does not cancel in a ratio.

At nonzero frequency $f$, measure both magnitude and phase of the electric current relative to the applied electric potential difference. Define $Y=I/V=1/Z$, where $Z$ is electrical impedance. For the stated complex permittivity convention, the ideal uniform specimen givesThus a parallel capacitor and electrical conductance describe the readout at that frequency; a single scalar capacitance cannot describe both components. The diagram's assumed dielectric relaxation gives $C_p=265.62\mathrm{pF}$ and $C_0{ϵ}_r^{\prime \prime }=88.54\mathrm{pF}$ at $1\mathrm{kHz}$. These are model predictions, not observations.

Repeat the reference measurement after removing the specimen and vary the small applied electric potential difference to test drift and linearity. Record specimen thickness, electrode area, contact conditions, frequency and temperature. Air gaps act as additional layers in series and can lower inferred relative permittivity; roughness, uncertain thickness and the fringing field of a capacitor limit the geometric formula. At sufficiently high frequency, the fixture itself changes the electromagnetic field, so low-frequency guarding and a lumped capacitor model require reassessment. ^[ref].

Dielectric strength characterizes the electric field at dielectric breakdown under specified test conditions. In an ideal uniform gap, it sets a voltage scale $V_{\mathrm{bd}}\simeq E_{\mathrm{bd}}d$.^[ref] The parallel-plate capacitor model consequently gives the stored energy density scaleThis is a model scale, not an operating rating: fringing field of a capacitor make local electric fields differ from $V/d$, while leakage and dielectric loss spend some supplied energy as heat. Comparing usable capacitor energy therefore requires insulation and breakdown performance as well as permittivity.

A quasi-static chemical potential outside a fixed-shape droplet solves the same Laplace equation as an electrostatic potential outside an equipotential conductor. If $\mu ={\mu }_s$ on the droplet and zero on the surrounding reservoirs, its outward diffusive current iswhere $C$ is the corresponding electrostatic capacitance in a medium of permittivity $ϵ$. The arbitrary $ϵ$ cancels against the capacitance's proportionality to $ϵ$. Thus electrostatic geometry directly determines evaporation rates.

The dimensionless quantity relating induced electric polarization to the internal electric field through $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ in the static scalar model. It describes the material contribution; relative permittivity includes the vacuum term as well, ${ϵ}_r=1+{\chi }_e$. Zero electric susceptibility gives ${ϵ}_r=1$, not zero permittivity.

The scalar ${\chi }_e$ assumes a local response with linearity and isotropy, settled electric polarization, fixed material conditions, and no residual electric polarization at zero electric field. It relates $\mathbf{P}$ to the averaged total internal electric field, which already includes the field of the bound electric charges. It does not relate $\mathbf{P}$ to the field the external electrodes would produce in vacuum. The distinction is essential for a self-consistent calculation.^[ref]

Substitute the scalar relation into the definition of electric displacement field:Comparison with $\mathbf{D}=ϵ\mathbf{E}$ proves $ϵ={ϵ}_0(1+{\chi }_e)$ and ${ϵ}_r=1+{\chi }_e$. Both $\mathbf{D}$ and $\mathbf{P}$ have units $C/{\mathrm{m}}^2$, so ${\chi }_e$ is a dimensionless quantity. In a stipulated static model with ${ϵ}_r=3$, ${\chi }_e=2$: at a prescribed internal electric field, the material contribution $P$ is twice ${ϵ}_{0}E$. This is a hypothetical response, not a measured material constant. The electric susceptibility is a convenient bridge from microscopic dielectric polarization to a macroscopic capacitance measurement, but it is not a measure of electrical conductivity.

In a linear isotropic nondispersive medium with time-independent permittivity $ϵ$ and magnetic permeability $\mu$, the stored electromagnetic energy per volume is $u=ϵ|\mathbf{E}{|}^2/2+|\mathbf{B}{|}^2/(2\mu )$. For a short derivation, the Maxwell equations in matter and the divergence identity for a cross product give $\nabla \cdot (\mathbf{E}\times \mathbf{H})=-\mathbf{H}\cdot {\partial }_t\mathbf{B}-\mathbf{E}\cdot {\partial }_t\mathbf{D}-{\mathbf{J}}_f\cdot \mathbf{E}$. The constitutive relations turn the first two terms into $-{\partial }_{t}u$, proving the Poynting theorem. Choose $u=0$ when the electric field and magnetic field both vanish. Dispersive or time-dependent material response requires a different accounting of stored energy.

The electrostatic boundary conditions at a conductor relate the one-sided total electric fields at a smooth metal–vacuum interface. Choose a unit normal $\mathbf{n}$ from metal into the adjacent vacuum, let ${\mathbf{E}}_{\mathrm{metal}}$ and ${\mathbf{E}}_{\mathrm{vac}}$ be the two limiting electric fields, and let $\sigma$ be the signed surface charge density. A thin Gaussian pillbox with cap area $A$ has outward cap normals $+\mathbf{n}$ on the vacuum side and $-\mathbf{n}$ on the metal side. As its height tends to zero, its side electric flux tends to zero for bounded local electric fields, leavingThe minus sign comes from the second outward cap normal, not from the sign of the electric charge. A shrinking rectangular loop across the interface and the zero closed line integral in electrostatics give continuity of the tangential component:For a conductor in electrostatic equilibrium, ${\mathbf{E}}_{\mathrm{metal}}=0$, henceThus positive surface charge density gives an electric field away from metal and negative surface charge density gives an electric field toward metal. The derivative uses the same chosen unit normal, from the vacuum side. The ideal surface layer has no unique electric field value on the layer itself; these are limiting values. Ordinary surface electric charge preserves continuity of electric potential; an ideal electric dipole sheet would require a different model. ^[ref].

Here “exterior” means just outside the metal, including the vacuum within a cavity in a conductor. On a spherical inner wall of radius $a$, its unit normal from metal into the cavity in a conductor is $-\hat{\mathbf{r}}$. A central positive point charge $q$ gives ${\mathbf{E}}_{\mathrm{vac}}=q\hat{\mathbf{r}}/(4\pi {ϵ}_0{a}^2)$ there, soThe outward normal of the cavity in a conductor as a volume is instead $+\hat{\mathbf{r}}$; confusing these two unit normals reverses the inferred sign. For a stipulated spherical enclosure with inner radius $a$, outer radius $b$, no external sources and total isolated metal electric charge $Q_c$, spherical symmetry and Gauss's law give the analytical modelIts outer surface charge density is $(Q_c+q)/(4\pi b^2)$. Grounding to $V_c=0$ with zero electric potential at infinity instead gives $Q_c=-q$ and zero exterior electric field, while preserving the same nonzero electric field inside the charged cavity in a conductor.

As a numerical sign check, stipulate $q=1\mathrm{nC}$, $a=0.10\mathrm{m}$, $b=0.20\mathrm{m}$ and $1/(4\pi {ϵ}_0)\simeq 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. The inner electric field is approximately $899N/C$ along $+\hat{\mathbf{r}}$, but ${\mathbf{E}}_{\mathrm{vac}}\cdot \mathbf{n}=-899N/C$ and ${\sigma }_{\mathrm{inner}}\simeq -7.96nC/{\mathrm{m}}^2$. An isolated neutral enclosure has $+1\mathrm{nC}$ on the outer wall, approximately $+1.99nC/{\mathrm{m}}^2$ and $225N/C$ just outside. Multiplying each surface charge density by its spherical surface area recovers the signed wall electric charges.

The local result uses the total electric field, including all induced and imposed sources. The factor $\sigma /{ϵ}_0$ follows from the zero metal-side electric field; a symmetric isolated sheet in vacuum instead has magnitude $|\sigma |/(2{ϵ}_0)$ on each side. ^[ref]. At sharp corners, limiting electric fields can be singular and the smooth local pillbox argument needs care. In material outside the metal, use the normal free-charge jump of the electric displacement field and the material's permittivity, as in electrostatic interface boundary conditions. ^[ref].

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.

Southern Methodist University's teaching lab, Introduction to the Parallel Plate Capacitor Lab, makes the response tangible with a capacitor assembled from aluminium foil, paper and taped wire contacts. Its instructions specify approximately $10\mathrm{cm}\times 10\mathrm{cm}$ foil rectangles, aligned with the paper completely separating them. A multimeter in capacitance mode supplies the readout; a ruler measures plate dimensions and a micrometer measures specimen thickness. The lab replaces the paper with other sheets, including cardstock, plastic wrap and wax paper. It asks students to keep foil area and alignment consistent, measure thickness at several positions, estimate thickness uncertainty, and wait for the capacitance reading to stabilize. These are published preparation and measurement instructions, not measured results reported here.^[ref]

The inference uses each specimen's own thickness $t$, not just a comparison of raw meter readings. With corrected capacitance $C_s$ and overlapping area $A$, the ideal estimate is ${ϵ}_r=C_{s}t/({ϵ}_{0}A)$. A thicker sheet can give a smaller capacitance despite larger relative permittivity. This is why changing material while letting the plate gap change does not isolate electric susceptibility by itself.

Useful additional controls are to keep leads in the same positions, use capacitance measurement calibration, repeat one sheet after the others to check drift, and repeat the assembly to assess contact reproducibility. An air-gap reference must have known spacers maintaining the same separation; removing a supporting sheet and letting the foil touch is not such a reference. Record the instrument's test frequency and measurement mode when available. A capacitance display alone does not establish negligible dielectric loss or a strictly static permittivity. These are proposed controls extending the published teaching instructions.

Air gaps produce a particularly revealing systematic error. Suppose a uniform sheet of thickness $t$ and relative permittivity ${ϵ}_r$ is separated from the electrodes by air layers of total thickness $g$. Approximate air permittivity by vacuum permittivity, neglect the fringing field of a capacitor and use the same area in each layer. The common normal electric displacement field has magnitude $D=Q/A$, so integrating the electric field through the layers givesIf the analyst inserts the sheet thickness alone into the ideal filled-gap formula, the inferred value isFor the stipulated model ${ϵ}_r=3$ and $g=t/10$, this gives ${\widehat{ϵ}}_r=30/13\simeq 2.308$, not three. These values are calculated examples, not measured sheet properties. If total plate separation $d=t+g$ is used instead, the effective estimate is $(t+g)/(g+t/{ϵ}_r)$; it still mixes the sheet and the air. Thus measuring a thickness accurately cannot compensate for an unrecognized layer.

The fringing field of a capacitor and additive lead capacitance further limit a foil measurement. Keep lateral plate dimensions large relative to separation, and use a guard electrode of a capacitor in a suitable fixture for more demanding work.^[ref] A family of thickness measurements can help distinguish material response from an additive offset, but air gaps can vary with the assembly. A large, repeatable reading supports a capacitance change; extracting a material electric susceptibility requires the geometry, gaps, instrument response and dielectric loss assumptions to survive these checks. No measured relative permittivity is claimed for any of the lab's sheets.

Birefringence produced by subwavelength structure rather than solely by intrinsic crystalline anisotropy. For ideal dielectric lamellae, continuity of tangential electric field gives ${ϵ}_{\parallel }=f{ϵ}_1+(1-f){ϵ}_2$, while continuity of normal electric displacement field gives ${ϵ}_{\perp }^{-1}=f/{ϵ}_1+(1-f)/{ϵ}_2$. The unequal effective permittivities give different principal refractive indices.

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]