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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].

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.

At fixed electric potential difference $V$, a change of capacitance has electrical work input $VdQ=V^{2}dC$ and field-energy increase $dU=V^{2}dC/2$. The remaining half is mechanical work done by the field, so the force in the direction of increasing overlap is $F_x=V^2{C}^{\prime }(x)/2$. Equivalently the effective mechanical potential is $-C{V}^2/2$. For a partially filled coaxial capacitor, this gives $F_x=\pi {ϵ}_0({ϵ}_r-1)V^2/\ln (b/a)$, directed inward when ${ϵ}_r>1$. The fringing field of a capacitor supplies the local field gradient and transmits this force even when its small correction to total capacitance is neglected.

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 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]

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.

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]

A guard electrode of a capacitor surrounds an electrically isolated sensing electrode and is held at the same electric potential. The small insulating slot separates their readout paths. The guard extends the nearly planar equipotential, moving the main outer edge away from the sensing region; it does not abolish the fringing field of a capacitor everywhere. In a capacitance fixture, fringe electric current is routed through the guard instead of the sensing readout. Keysight describes this three-electrode method in Materials Measurement: Dielectric Materials (Solution, 16451B fixture).^[ref]

Measuring the sensing and guard electric currents together would include the diverted response again. Grounding an arbitrary surrounding electrode also does not reproduce a guard unless its electric potential matches the sensing electrode. Finite guard width, the insulating slot, plate misalignment and specimen air gaps limit the approximation. Use capacitance measurement calibration and the geometry appropriate to the sensing region when inferring permittivity. The source also warns that guarding changes the electromagnetic field unfavourably at high frequency, so the low-frequency electrostatic picture has a restricted range.^[ref]

The Field Mapper Kit (PK-9023), Product Manual 012-04346D by PASCO scientific (p. 3, Parallel plate capacitor) proposes a direct investigation of the plate-length-to-gap ratio using conductive-paper electric potential mapping. Its finite straight electrodes represent a cross-section of the plates. The apparatus is a gridded resistive sheet on a corkboard, silver-ink electrodes, pushpin electrical contacts, a $5$–$20\mathrm{V}$ direct supply capable of $25\mathrm{mA}$, and a voltmeter with input electrical resistance at least $10\mathrm{M\Omega }$. The manual specifies waiting 20 minutes for the ink to reach its maximum electrical conductivity, then checking each electrode: its internal voltage drop should be no more than $1\%$ of the applied electric potential difference.^[ref]

For the readout, keep one voltmeter lead on the reference electrode and scan the other across the sheet. Mark points having the same reading to trace an equipotential. A second method holds the two probe tips at a fixed separation and pivots them to find the largest electric potential difference; a suggested separation is one centimetre. Avoid printed grid marks when making contact.^[ref]

The inference follows from the measured electric potential, rather than assuming straight electric field lines. With probe separation $\Delta s$ and direction $\hat{\mathbf{t}}$, a local finite difference givesAn arrow toward rising electric potential follows $+\nabla \varphi$, so it must be reversed to show $\mathbf{E}$. To investigate the fringing field of a capacitor, compare contours well inside the gap with contours near and beyond the ends, then repeat at a changed gap while keeping the sheet boundary and plate length fixed. Curving contours and an outside gradient establish nonuniformity in this planar setup. Compare $-E_{y}d/V$ at recorded positions to distinguish simple voltage rescaling from geometric changes; no measured values are supplied here.

Additional useful controls are repeat reference readings to detect drift, reversal of supply polarity, and repeating selected readings with reduced probe spacing. Keep contact quality and sheet edges fixed during the aspect-ratio comparison. Finite probe separation averages the gradient; reducing it improves spatial resolution but magnifies the effect of voltage uncertainty. Voltmeter loading, imperfect ink electrodes, and inhomogeneous sheet electrical conductivity can mimic distortions. These proposed checks supplement the manufacturer's electrode check, rather than claiming a completed experiment.

The sheet carries a steady electric current. For uniform electrical conductivity, Ohm's law and electric charge conservation give the two-dimensional Laplace equation away from electrodes, as explained in conductive-paper electric potential mapping. Uncontacted sheet edges have zero normal electric current, whereas the diagram model fixes zero outer electric potential. These distinct boundary conditions must be matched before a quantitative comparison. The apparatus reveals a planar electric potential map; it does not directly measure the free-space three-dimensional capacitance of metal plates.

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.

Use the parallel-plate capacitor approximation, neglecting the fringing field of a capacitor. With plate area $A=\pi (0.10\mathrm{m}{)}^2$, the initial capacitance and opposite plate electric charges areAfter the electric battery is disconnected, the isolated plate electric charges stay fixed. Doubling the separation halves the capacitance. The quasistatic pulling work is the increase of electrostatic energy:The final electric potential difference is thereforeAn electron released at rest gains kinetic energy $e\Delta V_f$, where $e$ is the positive elementary charge. Neglecting radiation and the negligible perturbation caused by this one electron, conservation of energy givesThis is $0.0280c$, so the nonrelativistic kinetic energy is adequate. The circular plates have finite radii; the numerical results are leading parallel-plate capacitor values rather than exact finite-plate capacitances.

For this space-charge-filled parallel-plate capacitor, place the first plate at $x=0$ and the second at $x=d=x_o$, ignoring fringing field of a capacitor. With vacuum permittivity between the plates and fixed charge density $\rho$, Poisson's equation and the prescribed boundary electric potentials giveIntegrating twice and imposing both values yieldsThe electric field is minus the gradient of the electric potential, henceThus the electric potential is a parabola, concave downward for $\rho >0$ and upward for $\rho <0$. The signed electric field is a straight line. Its magnitude has a V-shaped corner if its zerolies inside the gap; otherwise its magnitude is a single straight segment. An interior zero requires $|V_1-V_2|<|\rho |d^2/(2{ϵ}_0)$. At that point the electric potential has its interior extremum. For $\rho =0$, these reduce to a linear electric potential and a uniform electric field.

In vacuum, Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:Take opposite electric charges $\pm Q$ on the two electrical conductors, with surface charge densities $\pm \sigma =\pm Q/A$. A Gaussian pillbox crossing one isolated large sheet gives $2ES=\sigma S/{ϵ}_0$, so each sheet contributes $\sigma /(2{ϵ}_0)$ on either side. By the superposition principle, the electric fields add between the sheets and cancel outside. Ignoring the fringing field of a capacitor, the gap electric field is $E=\sigma /{ϵ}_0$. The electric potential difference and capacitance of this parallel-plate capacitor are consequentlyThe approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with permittivity $ϵ$ replaces ${ϵ}_0$ by $ϵ$.

Use opposite electric charges $\pm Q$ over a common length $L$, and write the inner linear charge density as $\lambda =Q/L$. Away from the ends, axisymmetry and Gauss's law on a coaxial cylindrical Gaussian surface of length $\ell$ giveThe electric potential difference and capacitance of the coaxial cylindrical capacitor areThis neglects the fringing field of a capacitor at the ends, as justified when $L\gg b$.

Choose the plate polarity so that the gap electric field points along $+z$ when $V>0$. Away from the plate edges, the quasistatic electromagnetic approximation gives $E_z=V_0\sin \omega t/d$. A circular Ampere-Maxwell law contour of radius $r<a$ then givesIt is displacement current, rather than a conduction current across the air, that supplies the enclosed current term. The direction follows the right-hand rule for increasing positive electric flux; it reverses every half cycle. In the ideal extension that sets the gap electric field to zero outside the plate footprint, $B_{\varphi }={\mu }_0{ϵ}_0\omega V_0{a}^2\cos \omega t/(2dr)$ for $r>a$. That extension is an edge approximation, not an exact exterior solution for finite plates. Neglect fringing field of a capacitor and retardation, requiring $d\ll a$ and $\omega a/c\ll 1$.

In electrostatics, the Maxwell equations in matter areHere ${\rho }_f$ is free charge density and $\mathbf{P}$ is polarization density. For a homogeneous isotropic linear dielectric, the constitutive equation is $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. Integrating the first equation over a thin pillbox straddling an interface, with unit normal $\mathbf{n}$ from medium 1 to medium 2, gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Integrating the second around a vanishingly narrow rectangular loop gives $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$. These are the electrostatic interface boundary conditions. When there is no free surface electric charge, the normal electric displacement field and tangential electric field are continuous; the normal electric field generally is not.

Neglect end fringing field of a capacitor. In either axial segment of the coaxial cylindrical capacitor, a cylindrical Gaussian surface gives $D_r={\lambda }_f/(2\pi r)$, where ${\lambda }_f$ is free electric charge per length. Integrating the radial electric field between the conductors givesThe liquid and air segments share the same conductor electric potential difference, so their capacitances add in parallel. Thus the partially filled coaxial capacitor hasHere the question's $ϵ$ is relative permittivity, and air's relative permittivity is approximated by one.

Write $x$ for the rise above the external level. The wide reservoir keeps that level approximately fixed. At constant applied voltage, one must include work done by the source: the appropriate electrical contribution to the mechanical potential is $-C{V}^2/2$. The upward dielectric insertion force at fixed voltage is consequentlyThe excess hydrostatic pressure $\rho gx$ acts over the annular area $A=\pi (b^2-a^2)$. Alternatively its gravitational potential energy is $\rho gA{x}^2/2$, since liquid initially at the reservoir level must be raised through all heights from zero to $x$. Balancing the derivatives givesThis assumes $h+x<L$, a communicating liquid reservoir, negligible capillarity, and a sufficiently long capacitor that end effects are small. Differentiating gives the dielectric-rise voltmeter's sensitivity at a bias $V$:To obtain its numerical value in $\mu \mathrm{m}{\mathrm{V}}^{-1}$, multiply the right-hand side in SI units by $10^6$. A convenient parametric scale, with $\beta =b/a$, isFor a concrete estimate, choose $a=1\mathrm{mm}$, $b=2\mathrm{mm}$, mass density $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$, relative permittivity $ϵ=2$ and operating voltage $V=100\mathrm{V}$. These are illustrative choices, not supplied apparatus data. They giveThus this millimetre-scale example requires fine optical displacement measurement for volt-level changes. The parametric formula states exactly how the estimate changes with the chosen geometry, liquid and bias.

The displacement is quadratic in voltage, cannot distinguish its sign, and has zero first-order sensitivity at zero bias. Smaller gaps and larger relative permittivity increase sensitivity, but capillarity, leakage, dielectric breakdown, temperature-dependent liquid properties, and finite available rise then limit performance. A nonconducting liquid avoids a sustained conduction load, although charging still draws electric charge and leakage is possible. A high voltage can give a readable displacement; a low-voltage instrument would require fine displacement measurement and careful calibration.

The electric field is approximately radial in the long interior, but near an electrode end its lines spread into a fringing field of a capacitor. The electric field magnitude falls on moving outward through that end region. The dielectric's induced electric dipoles experience a net force toward the stronger field, pulling the material into the capacitor. For a weakly perturbing polarizable element this is expressed by $\mathbf{F}=\frac{1}{2}\alpha \nabla E^2$, with positive electric polarizability $\alpha$.

In the sketch the field lines run from the positive electrode to the negative one, curving outward near their ends. The material is attracted toward the interior where the field is stronger. A perfectly uniform interior field by itself would not give a longitudinal dipole-gradient force; the end region is essential to the local explanation. Neglecting its small contribution to the total capacitance does not justify omitting the finite insertion force that it transmits. The energy argument of part (d) gives that force without needing the detailed fringe solution.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

Permittivity describes how an electric field and a material's electric polarization combine into the electric displacement field. In a static, local response with linearity and isotropy, and no electric polarization at zero electric field, the constitutive equation isHere ${ϵ}_0$ is vacuum permittivity, while ${ϵ}_r$ is the relative permittivity, a dimensionless quantity. The SI unit of permittivity is farad per metre, $F/m$: multiplying it by an electric field in $V/m$ gives an electric displacement field in $C/{\mathrm{m}}^2$. A larger positive permittivity lets the same parallel-plate capacitor hold more plate electric charge at the same electric potential difference.

The material contribution is electric polarization $\mathbf{P}$, the electric dipole moment per unit volume. The electric displacement field and electric susceptibility ${\chi }_e$ satisfySubstitution gives the first constitutive equation. The electric field here is the averaged, total field inside the material, not just the field of the external plate electric charges. The continuum approximation averages over many chemical atoms or molecules; locality means that the electric polarization at a position depends on the electric field there. Static equilibrium, fixed material conditions and sufficiently small electric fields are needed for this scalar proportionality. ^[ref].

For an ideal parallel-plate capacitor completely filled by a uniform dielectric, hold the free plate electric charge magnitude $Q$ fixed and neglect the fringing field of a capacitor. Writing $A$ for the plate area, Gauss's law gives $D=Q/A$. ThusThe electric dipoles point along the electric field. Their surface electric charges, described by polarization density, oppose the field from the free plate electric charges. For a hypothetical dielectric with ${ϵ}_r=4$, the internal electric field is one quarter of its vacuum value, while $P=3D/4$. At fixed electric potential difference instead, the internal electric field stays approximately $V/d$ and the plate electric charge grows fourfold. These are different experimental constraints. ^[ref].

Permittivity is a useful material summary, not a universal constant of the material under every condition. Temperature, frequency and electric field strength can change the response. In a linear anisotropic dielectric, permittivity is a tensor, $D_i={\sum }_j{ϵ}_{ij}{E}_j$, so the electric displacement field need not be parallel to the electric field. At nonzero frequency, delayed electric polarization requires complex permittivity and may produce dielectric loss. ^[ref].