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

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.

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

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

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

A driven cylindrical conductor between two grounded coaxial conductors forms two coaxial cylindrical capacitors in parallel. For radii $a,\lambda a,2a$, the capacitance per unit length is the sum $C^{\prime }=2\pi {ϵ}_0[1/\log \lambda +1/\log (2/\lambda )]$. Writing $u=\log \lambda$ and $L=\log 2$, the function $1/u+1/(L-u)$ is strictly convex on $(0,L)$ and symmetric about $L/2$. Hence $C^{\prime }$ has its unique minimum at $\lambda =\sqrt{2}$, with value $8\pi {ϵ}_0/\log 2$. As $\lambda =1+\delta$ approaches the inner conductor, $C^{\prime }\sim 2\pi {ϵ}_0/\delta$, agreeing with the local parallel-plate capacitor approximation.

A dielectric is a material in which an electric field produces electric polarization while transport of electric charge is sufficiently small for the intended use. An insulating solid in a capacitor is an example; real dielectrics can still have dielectric loss and nonzero electrical conductivity.

The important distinction is between rearranging bound electric charge and transporting mobile charge carriers through the specimen. Bound electric charges can move a little within a chemical atom or molecule, or an existing electric dipole can turn. Their averaged electric dipole moments produce electric polarization even if there is negligible steady electric current between the electrodes. A changing electric polarization does give a bound current; the absence of steady conduction does not mean that the material is electrically unresponsive.^[ref]

For uniform electric polarization $\mathbf{P}=P\hat{\mathbf{x}}$ in a rectangular slab, polarization density gives zero bound volume charge density in the interior and opposite bound surface charge densities $-P,+P$ on the left and right faces. The electric charge has been redistributed, not created. More generally, the divergence theorem makes the cancellation explicit:Here $\mathbf{n}$ is the outward normal vector on the material boundary. This accounts for both bound volume charge density and bound surface charge density; omitting either can falsely suggest that an initially neutral dielectric acquired net electric charge.

An electrical insulator between capacitor plates supports a useful combination: appreciable electric polarization, little leakage electric current, and tolerance of the intended electric field. This lets a small parallel-plate capacitor store separated electric charge and recoverable capacitor energy. The comparison in dielectric capacitor response explains why specifying fixed electric charge or fixed electric potential difference changes the answer. High relative permittivity alone does not establish good insulation or a large usable energy density; dielectric loss and dielectric breakdown matter as well.

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].

Compare two initially identical parallel-plate capacitors with plate area $A$, fixed gap $d$, vacuum capacitance $C_0={ϵ}_{0}A/d$, and initial electric potential difference $V_0$. Completely fill both gaps with the same uniform linear dielectric of static relative permittivity ${ϵ}_r>1$, neglecting the fringing field of a capacitor. Keep one connected to an ideal voltage source and isolate the other. The geometric capacitance becomes $C={ϵ}_r{C}_0$ in both cases; the electrical constraint determines what else changes.^[ref]

Write $Q_0=C_0{V}_0$, $E_0=V_0/d$ and $U_0=C_0{V}_0^2/2$. Applying $Q=CV$, $E=V/d$ and the capacitor energy formula gives the complete comparison:Thus the statement that a dielectric reduces the electric field needs the fixed-free-charge condition. With fixed electric potential difference, the voltage source supplies additional free plate electric charge, maintaining $E=V_0/d$ despite the new bound electric charges.

For the stipulated ${ϵ}_r=3$ example, the voltage source supplies $\Delta Q=2Q_0$. At constant $V_0$, its delivered electrical work is $V_0\Delta Q=4U_0$, whereas the increase in capacitor energy is $2U_0$. In an ideal reversible quasistatic insertion, the remaining $2U_0$ is mechanical work delivered by the assembly, as accounted for in fixed-voltage capacitor energy balance. For the isolated assembly, capacitor energy falls from $U_0$ to $U_0/3$, allowing $2U_0/3$ of mechanical work in the same ideal limit. Actual losses reduce recoverable work; energy increases in the connected case do not arise from the dielectric supplying free energy.

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.

In vacuum, displacement current is the term ${\mathbf{J}}_D={ϵ}_0{\partial }_t\mathbf{E}$ in the Ampère-Maxwell equation. It has the units of current density, but need not describe electric charge crossing the region. In Maxwell equations in matter, the corresponding term is ${\partial }_t\mathbf{D}$, with $\mathbf{D}$ the electric displacement field.

Consider a slowly charging parallel-plate capacitor with vacuum gap, plate area $A$, and charge $Q(t)$ on its positive plate. Neglect fringing and assume the charging timescale is long compared with the light-travel time across the apparatus. The nearly uniform gap electric field has magnitude $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isFor $A=0.010{\mathrm{m}}^2$ and charging electric current $I=1.0\mathrm{mA}$, this gives $dE/dt\simeq 1.13\times 10^{10}\mathrm{V}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$ and $J_D=0.10A/{\mathrm{m}}^2$. No conducting electric current crosses the gap. A spanning surface cutting the wire and another passing between the plates nevertheless give the same magnetic circulation when the full Ampère-Maxwell equation is used. ^[ref].

The local consistency check is short. Without displacement current, taking the divergence of $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$ would force $\nabla \cdot \mathbf{J}=0$, allowing no local accumulation of electric charge. With displacement current, the divergence of a curl is zero and Gauss's law giveThis is precisely the charge continuity equation. Electric charge conservation fixes the required divergence of the added term; it does not by itself exclude other divergence-free additions. The specific Maxwell equations also yield the electromagnetic wave equation, linking this consistency correction to propagation at the speed of light.

In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

An electric potential difference compares electric potentials at two points. For a move from $a$ to $b$, define $\Delta \Phi =\Phi (b)-\Phi (a)$; the terminal convention $V_{AB}=\Phi (A)-\Phi (B)$ instead means the potential at $A$ relative to $B$, so the order must always be stated. In electrostatics, the line integral of the electric field givesHere $\Delta U$ is the change in interaction potential energy of the test electric charge $q$ with fixed sources, while $W_{\mathrm{field}}$ is work done by the electric field. For an external agent moving the charge slowly without a change in kinetic energy, its work is $\Delta U$. A positive electric charge gains kinetic energy when accelerated toward lower electric potential; a negative electric charge gains kinetic energy when accelerated toward higher electric potential. Both motions lower $q\Phi$.^[ref]

For example, fix $Q=+2\mathrm{nC}$ at the origin and compare distances $r_a=0.20\mathrm{m}$ and $r_b=0.10\mathrm{m}$. With the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and zero at infinity,Moving $q=-3\mathrm{nC}$ inward therefore givesIf only the electric force does work, conservation of energy makes the gain in kinetic energy $2.697\times 10^{-7}\mathrm{J}$. Direct integration confirms the signs: ${\int }_{r_a}^{r_b}kQdr/r^2=kQ(1/r_a-1/r_b)=-89.9\mathrm{V}$, which multiplied by the negative test electric charge gives positive work. The electric potential difference is unchanged by choosing a different electrical reference potential.

In an electrical circuit, an electric potential difference is the voltage across a component's terminals. In an electrical network, a vertex voltage is specified relative to a reference vertex; its differences across edges drive current flow. For a uniform electric field, the integral reduces to $\Delta \Phi =-\mathbf{E}\cdot ({\mathbf{r}}_b-{\mathbf{r}}_a)$. A field $\mathbf{E}=300V/m\hat{\mathbf{x}}$ across a displacement $0.04\mathrm{m}\hat{\mathbf{x}}$ therefore gives $\Delta \Phi =-12\mathrm{V}$, as in the interior approximation for a parallel-plate capacitor. The volt, measured in joules per coulomb, expresses energy transfer per unit electric charge, which makes it useful for comparing components without specifying the amount of charge transported.

The farad, symbol $\mathrm{F}$, is the SI unit of capacitance:A capacitor of capacitance $1\mathrm{F}$ carries separated electric charge magnitude $1\mathrm{C}$ at terminal electric potential difference $1\mathrm{V}$. A picofarad is $10^{-12}\mathrm{F}$, so the worked parallel-plate capacitor's capacitance $88.54\mathrm{pF}$ is $8.854\times 10^{-11}\mathrm{F}$. ^[ref].

For a quasistatic change of capacitance at fixed terminal electric potential difference, the capacitor energy is $U=C{V}^2/2$ and plate electric charge is $Q=CV$. Electrical work supplied to the electric field by the battery is $VdQ$, so mechanical work done on the capacitor is $d{W}_{\mathrm{mech}}=dU-VdQ=-V^{2}dC/2$. Increasing the separation of a parallel-plate capacitor makes $dC<0$: mechanical work is positive even though electrostatic energy decreases, because electrical energy returns to the source. Conversely, when $dC>0$, the battery supplies $V^{2}dC$: half increases capacitor energy and half is available as mechanical work done by the electric field. Differentiating capacitor energy alone at fixed terminal electric potential difference omits this exchange with the battery.

The nonuniform electric field near the edges of finite electrodes, where electric field lines bend away from an ideal uniform or one-dimensional configuration. A parallel-plate capacitor with separation much smaller than its lateral size has nearly uniform interior field, but its edge field contributes a correction to capacitance.

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]

For a circular air-filled parallel-plate capacitor with voltage phasor $V_0{e}^{-i\omega t}$, uniform gap $d$ and radius $a$, the Ampère-Maxwell equation on a circular loop of radius $s<a$ gives the displayed magnetic field. The enclosed displacement current is proportional to $s^2$, whereas the circulation is $2\pi s{B}_{\varphi }$. Consequently its amplitude grows linearly from zero on the axis. This assumes negligible fringing fields and a quasistatic approximation $\omega a/c\ll 1$.

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.

Neglect end effects, so the electrostatic potential depends only on the cylindrical radius $r$. The radial Laplace equation is $(r{\varphi }^{\prime }{)}^{\prime }=0$, with general solution $A\log r+B$. Imposing the conductor boundary conditions givesThe radial electric field is $E_r=-{\varphi }^{\prime }$. Just inside the middle cylinder it is $-V/[\lambda a\log \lambda ]$, directed towards the inner cylinder; just outside it is $V/[\lambda a\log (2/\lambda )]$, directed towards the outer cylinder. By Gauss's law, its charge per unit length isHence the capacitance per unit length isThe two coaxial cylindrical capacitors contribute in parallel because both connect the middle cylinder to the same grounded potential. This is the coaxial capacitor with grounded inner and outer cylinders construction.

For $\lambda =1+\delta$ with $\delta \to 0^{+}$, $\log \lambda =\delta +O({\delta }^2)$ while $\log (2/\lambda )\to \log 2$. ThereforeThe inner gap has width $a\delta$ and area per unit length approximately $2\pi a$. Its parallel-plate capacitor approximation gives exactly the same leading result; the outer gap contributes only a bounded correction.

For the extremum, set $u=\log \lambda$, $L=\log 2$, so $0<u<L$. ThenThus the unique stationary point is $u=L/2$. Strict convexity and divergence at both endpoints show it is a global minimum:There is no maximum on $1<\lambda <2$.

Treat the film as a parallel-plate capacitor, initially compensated and then isolated electrically. Its capacitance and temperature-induced electric polarization charge areThe corresponding ideal open-circuit signal, with the electrode polarity defined consistently with this charge convention, isArea cancels because both charge and capacitance are proportional to it. Substitution givesThe voltage magnitude is approximately $0.99\mathrm{V}$; interchanging the voltmeter leads reverses its sign. This is an ideal capacitive result, before leakage, mobile-charge screening or loading by the measurement circuit reduces the signal.