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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.

A cavity in a conductor is a void surrounded by an electrical conductor; completely enclosed is the essential qualification in exact electrostatic shielding arguments. At electrostatic equilibrium, the surrounding connected metal is an equipotential. An empty closed cavity in a conductor has constant electric potential and zero electric field, even if the metal is isolated and charged. An occupied closed cavity in a conductor generally has a nonzero electric field; its inner-wall total electric charge cancels the total electric charge of the internal sources. These are distinct statements, illustrated by Faraday's ice pail experiment.^[ref]

Conductive-paper electric potential mapping uses a voltmeter to locate equipotentials on a resistive sheet, then infers the in-plane electric field from the measured electric potential differences.

A conductor in electrostatic equilibrium is an electrical conductor at rest whose mobile charge carriers have finished redistributing, with no maintained bulk electric current. In the ideal macroscopic model,A residual bulk electric field would drive further electric charge redistribution. The electric potential difference between any two points joined through the metal is $-\int \mathbf{E}\cdot d\mathbf{l}=0$, so each connected piece is an equipotential. Gauss's law then gives zero net bulk charge density. This is a statement about net macroscopic electric charge, not the absence of positive atomic nuclei and negative electrons. Excess electric charge is represented by surface charge density; atomic surface structure is outside this continuum model. ^[ref].

The word inside must distinguish metal from a cavity in a conductor. A completely enclosed, charge-free cavity in a conductor has constant boundary electric potential $V_c$. The constant solution of Laplace's equation and Uniqueness of the Dirichlet problem give $\Phi =V_c$ and $\mathbf{E}=0$ throughout the cavity in a conductor, regardless of its shape or external electric charges. Grounding is unnecessary for this inward electrostatic shielding; an isolated enclosure can change its overall electric potential without producing an internal electric field. Zero enclosed net electric charge alone would only give zero electric flux. A positive and negative pair inside a cavity in a conductor can have zero total electric charge and a nonzero electric field. ^[ref].

If fixed electric charges inside a cavity in a conductor sum to $q$, a Gaussian surface lying entirely in the metal and enclosing that cavity in a conductor has zero electric flux. Thus the total induced inner-wall electric charge is $-q$. For one such cavity in a conductor and an isolated enclosure with total metal electric charge $Q_c$, the outer-wall electric charge is $Q_c+q$. In particular, an initially neutral isolated enclosure generally has an exterior electric field when $q\ne 0$. For fixed internal sources, external changes only shift the constant electric potential inside an isolated enclosure, leaving the internal electric field unchanged. ^[ref].

A grounded enclosure fixes $V_c=0$ and can acquire metal electric charge $-q$; with no exterior sources and $\Phi \to 0$ at infinity, Uniqueness of the Dirichlet problem gives zero exterior electric field. With external sources present, the exterior electric field need not vanish, but it is independent of the internal sources at fixed grounded boundary electric potential. ^[ref].

Faraday's ice pail experiment makes the distinction between redistribution and transfer visible. In his letter of 4 February 1843, Faraday described an insulated pewter pail connected by wire to a gold-leaf electrometer, and a brass ball suspended on a long dry silk thread to reduce the hand's influence. He discharged the pail and charged the ball away from it. Lowering the ball without contact produced leaf divergence, which grew during entry and became approximately independent of further depth; withdrawing it collapsed the leaves. Contact with the bottom transferred electric charge, and the withdrawn ball was discharged. The readout was leaf divergence, not a direct map of the internal electric field. These observations distinguish electrostatic induction from contact transfer of electric charge. ^[ref].

For a modern version of Faraday's ice pail experiment, connect an insulated inner metal pail to the signal input of a high-input-impedance electrometer, with its reference lead and an outer shield grounded. Zero and momentarily ground the pail, then isolate the signal side. Lower a charged paddle deeply without touching and record the electric potential difference; withdraw it and check return to zero. Repeat with contact, then withdrawal: a residual reading indicates transferred electric charge. Reverse the paddle's sign and repeat. The electrometer reads volts; inferring electric charge requires the effective capacitance of pail, cable and input, so a stable reading at fixed geometry is a relative measure, not automatically a value in coulombs. Keep supports clean, nearby charged objects distant, and the operator grounded; humidity can cause leakage. ^[ref].

A complementary electrostatic shielding demonstration uses a charged hollow metal enclosure, a small metal sampling ball on an insulating handle, and an electrometer. Discharge the sampling ball between trials; touch it to the exterior, withdraw it, and transfer its collected electric charge to the electrometer. Repeat by touching the inner wall without touching the exterior during withdrawal. The exterior sample produces a response, whereas the inner sample gives a null within instrument sensitivity. This compares accessible surface responses rather than assuming a perfectly spherical enclosure. ^[ref]. Useful additional controls are an uncharged enclosure baseline, a repeat exterior sample to verify sensitivity, and repeated trials at different depths. An access hole, charged handle, leakage or disturbed surface charge density limits the inference; a null bounds detectable transfer, rather than proving an exactly zero electric field. An open pail's depth plateau tests good enclosure empirically. Neither demonstration establishes shielding of arbitrary time-dependent electromagnetic fields.

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]

Faraday's ice pail experiment distinguishes redistribution of electric charge by electrostatic induction from transfer of electric charge by contact. Lower a charged body into an initially neutral isolated metal pail: an externally connected electrometer responds even though the body never touches the metal. Withdraw the body without contact and the response disappears. Repeat with conductive contact before withdrawal and a residual response can remain. The striking point is that an exterior response does not establish that the pail has gained net electric charge.

Apparatus and readout. A modern implementation uses two concentric wire-mesh cylinders: a 10 cm diameter, 15 cm high inner pail on insulating rods and a grounded outer shield. Connect the electrometer's signal lead to the inner pail and its reference lead to the shield. Momentarily ground and zero the signal assembly, then isolate it; the shield remains grounded. A charged disk on an insulating handle goes into the lower half without touching. The electrometer measures electric potential difference in volts, with sign determined by the lead connections. It does not directly measure electric charge in coulombs.^[ref]

A charge ledger before contact. To separate exact reasoning from the open apparatus, first idealize the pail as a closed cavity in a conductor with one internal body of total electric charge $q$. Let the isolated metal have total electric charge $Q_0$. The metal's electric field is zero at electrostatic equilibrium.^[ref] A Gaussian surface enclosing the cavity in a conductor but lying wholly in the metal therefore has zero electric flux. Gauss's law and charge conservation giveFor $Q_0=0$ and $q>0$, the inner-wall total is $-q$ and the outer-wall total is $+q$: the metal remains neutral. The exterior electric field can therefore be nonzero without any transfer from the body. The connected electrometer and signal cable belong to the same isolated conducting assembly and share its exterior electric charge. The zero-electric field assertion concerns metal, not the occupied cavity in a conductor.

This ledger fixes totals, not the local surface charge density. Moving an off-center body changes the inner-wall distribution while its integral remains $-q$. With fixed outer geometry, fixed $Q_0+q$ and unchanged exterior sources, Uniqueness of the Dirichlet problem together with the fixed total outer electric charge determines the same exterior solution: the outside cannot reveal the body's position or detailed internal distribution. For example, internal bodies with stipulated electric charges $+2q$ and $-q$ give the same total exterior response as a single body with $+q$. Internal bodies with $+q$ and $-q$ can give a null exterior response while retaining a nonzero internal electric field. This is why a null reading cannot establish an empty cavity in a conductor.

Withdrawal is the control. Record a baseline, insert without contact, withdraw, and check return to baseline. After a fresh zero, repeat with the charged disk touching the inner wall, then withdraw and record the residual response. Finally zero the pail and reinsert the disk without contact to test remaining electric charge. Keep the operator grounded, discharge unwanted handle electric charge, and keep the unused charged disk distant. Dirty supports and humidity promote leakage. These controls separate transfer from electrostatic induction and stray coupling.^[ref]

In the ideal contact test, the internal body is itself an electrical conductor and makes a conducting connection to the wall. After charge relaxation, body and enclosure become a single equipotential; the remaining charge-free void has zero electric field. The body then carries no excess electric charge on surfaces facing that void. Withdrawal without subsequent charging leaves total metal electric charge $Q_0+q$. A coated paddle can retain electric charge on insulating patches, so merely touching it does not guarantee complete transfer. A residual reading establishes some transfer under the controls; the disk retest checks the stronger claim of complete transfer. The ideal argument assumes no leakage and no relevant contact-generated charging.

Grounding changes the ledger. With no exterior sources and zero electric potential at infinity, a closed grounded enclosure has zero exterior electric field, inner-wall electric charge $-q$ and zero outer-wall electric charge.^[ref] Consequently, disconnecting ground before withdrawal retains total metal electric charge $-q$, which moves to the exterior after withdrawal. Keeping ground connected during withdrawal instead restores zero electric charge. A lasting opposite-sign response can therefore arise through grounding without body–wall contact.

For the real open pail, no Gaussian surface lying wholly in metal closes over the mouth. Increasing insertion depth and seeking a stable response tests whether enclosure is a good approximation; mesh openings, rim proximity and nearby objects limit it. In a fixed-geometry approximation, the positive insertion response obeys $\Delta V\simeq q/C_{\mathrm{eff}}$, where $C_{\mathrm{eff}}$ is the effective capacitance of the pail, shield, cable and electrometer input. Changed geometry changes that conversion. Wait for charge relaxation and instrument settling at each step, while keeping the observation short compared with leakage. The experiment probes settled electrostatics; it does not establish shielding against arbitrary changing electromagnetic fields.

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.

Two broad perfect conductors of width $w$, separated by gap $d\ll w$, carry opposite steady electric currents $I$ and connect through a terminating resistor $R$. Away from edges, Ampère's circuital law and the superposition principle give gap magnetic field magnitude ${\mu }_{0}I/w$. Each conductor is an equipotential, and Ohm's law sets their electric potential difference to $IR$. Planar symmetry and charge-free Gauss's law then give uniform gap electric field magnitude $IR/d$. Injection parallel to the upper current makes the electric and magnetic Lorentz forces opposite. Their cancellation selects speed $Rw/({\mu }_{0}d)$ independent of nonzero electric charge and of operating current; nonrelativistic kinetic energy is $m[Rw/({\mu }_{0}d){]}^2/2$.

Take the wire at $(a,0)$ to carry $+\lambda$ and that at $(-a,0)$ to carry $-\lambda$. Ifthen, up to one additive constant,Thus an equipotential with $k=2\pi {\epsilon }_0\varphi /\lambda$ satisfies $r_{-}/r_{+}=e^k$. For $k\ne 0$, completing the square gives the Apollonius circleIts centre is $(a\coth k,0)$ and its radius is $a/|\sinh k|$. Positive and negative values give nested circles around the positive and negative wires. The electric field is orthogonal to these circles and points from the positive wire toward the negative wire. For $\varphi =0$, the limiting equipotential is the straight line $x=0$.

Direct superposition givesIn the limit $a\to 0$ with $\lambda a=p$, the potential and field become those of a two-dimensional electric dipole:and

Use the standard infinite-plane boundary condition: the sheet is an equipotential, with no imposed uniform background electric field and the electric potential tending to zero far from the localized charges. The images of $+Q$ at $z=d$ and $-2Q$ at $z=2d$ are $-Q$ at $z=-d$ and $+2Q$ at $z=-2d$.

At the point $z=0^{+}$ directly below the real charges, horizontal components vanish. The first real-image pair contributes $-2k_{e}Q/d^2$ along $z$, while the second contributes $+2k_e(2Q)/(2d{)}^2=+k_{e}Q/d^2$. ThereforeFor $Q>0$, the electric field points downwards into the sheet. The expression is the one-sided vacuum limit; the electric field inside the conducting material is zero. A prescribed uniform background field would be an additional boundary condition and would change this result.

Spherical symmetry and the vacuum Gauss's law imply $\mathbf{E}=A\hat{\mathbf{r}}/r^2$ in the gap, where $A$ is constant. Integrating the electric field between the two equipotentials givesConsequentlyThe inner charge is $Q=4\pi {ϵ}_{0}A$, which reproduces the spherical capacitor capacitance $4\pi {ϵ}_{0}ab/(b-a)$. The radial direction is outward for positive $V_1$ and inward for negative $V_1$; the formula represents the signed field, not just its magnitude.

The integral Gauss law in vacuum is ${\oint }_{\partial V}\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{ϵ}_0$. A single isolated point charge has spherical symmetry, so its electric field is radial and constant in magnitude on a centred sphere. The flux is $4\pi r^2{E}_r$, giving Coulomb's lawIn electrostatic equilibrium, a conductor is an equipotential: a tangential electric field would move its mobile charges. The electric field inside the conducting material is zero. A thin pillbox straddling its surface therefore gives $E_{n}A=\sigma A/{ϵ}_0$, henceHere $\sigma$ is the charge density on the particular face and $\hat{\mathbf{n}}$ points into vacuum. For an isolated thin plate with total charge divided equally between two faces, each face instead has half that total areal density. The local boundary result must not be confused with the $\sigma /(2{ϵ}_0)$ field of an isolated nonconducting infinite charged sheet.

In SI units, Maxwell's equations in vacuum, allowing charge density $\rho$ and electric current density $\mathbf{J}$, areThe source-free specialization sets $\rho =0$ and $\mathbf{J}=0$; it cannot describe the charged objects at their locations. Applying the divergence theorem to the first equation gives Gauss's law:For an isolated point charge $q_1$, spherical symmetry makes the electric field radial and constant in magnitude on a concentric sphere. Therefore $4\pi r^2{E}_r=q_1/{ϵ}_0$. Multiplying by a second point charge gives Coulomb's law,The force points away from the first charge for like signs and toward it for opposite signs.

For the spherical capacitor, a Gaussian sphere between the shells encloses $q$, so $E_r=q/(4\pi {ϵ}_0{r}^2)$ for $a<r<b$. Integrating the electric field inward from the outer shell yieldsAn arbitrary constant $V(b)$ remains unless a reference is specified. The capacitance is determined by the potential difference:For the subsequent charging operation, take the initial outer shell to be earthed, or interpret the stated $V$ as the inner-to-outer voltage with the outer reference zero. Isolating the inner shell fixes its electric charge at $q=CV$. Removing the outer shell leaves an isolated conducting sphere. Its electric field and electric potential, now with zero at infinity, areConsequentlyThese changes of electric potential are compatible with constant electric charge. Without the initial reference convention the absolute stated $V$ does not determine $q=CV$: for example, if the outer shell were initially isolated and neutral, the absolute inner electric potential would instead be $q/(4\pi {ϵ}_{0}a)$.

Finally take the connected spheres to be sufficiently far apart that the long wire's capacitance and their mutual influence can be neglected. Their common electric potential requiresso $q_a=q/3$ and $q_{2a}=2q/3$. The magnitudes of the surface electric fields areThis is the usual widely separated-sphere approximation. A wire enforces an equipotential, but alone does not make this ratio exact for arbitrary arrangements. As a concrete counterexample, if the second shell enclosed the first concentrically and they were connected, all net electric charge would reside on the exterior and the inner sphere's surface electric field would be zero.

In a charge-free region of homogeneous permittivity, Gauss's law and $\mathbf{E}=-\nabla V$ give Laplace's equation, ${\nabla }^{2}V=0$. Prescribed surface electric potentials specify a Dirichlet problem. Its Uniqueness of the Dirichlet problem permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same boundary conditions is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. Green's first identity gives $\int |\nabla u{|}^2{d}^{3}x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal electric field instead determine a potential only up to an additive constant when the usual compatibility condition holds.

A cylindrical Gaussian surface around an infinite line charge gives $E_r=\lambda /(2\pi {ϵ}_{0}r)$. The electric potential difference is thereforeAn isolated infinite line charge has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to 0$ there. Define $r_b^2=(x-b{)}^2+y^2$ and $r_c^2=(x-c{)}^2+y^2$. superposition givesOn $x^2+y^2=bc$,Thus the circle, and its cylindrical extension, is an equipotential. This is an Apollonius circle: a constant ratio of distances gives a circle rather than a constant distance from either charge.

For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The cylinder-plane image charges are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and Uniqueness of the Dirichlet problem applies. A Gaussian surface enclosing the physical cylinder confirms that its charge per unit length is indeed $\lambda$. Its voltage isThe physical field energy per unit length at fixed cylinder charge is $U^{\prime }=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:The electrostatic attraction per unit length of a cylinder above a plane is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.

The method of images solves certain electrostatics problems by replacing a conductor by fictitious electric charges outside the region where the solution is required. The replacement must satisfy the actual boundary values of the electric potential as well as the prescribed sources. The uniqueness of Poisson equation then establishes that the constructed electric potential equals the physical one in that region. The image locations are mathematical devices; they are not real charges inside the metal.

For a point charge $q$ at $(0,0,a)$ above an infinite grounded conducting plane $z=0$, put an image charge $-q$ at $(0,0,-a)$. In a cylindrical coordinate system with coordinates $(s,z)$ the potential in the physical region isThis vanishes at $z=0$, has the required singularity at the real charge, and tends to zero at infinity. With electric field $\mathbf{E}=-\nabla \varphi$, the induced surface charge density follows from the normal field immediately above the conductor:The force on $q$ is the field of the image evaluated at the real charge:The electrostatic energy relative to infinite separation is $U=\frac{1}{2}q{\varphi }_{\mathrm{image}}=-q^2/(16\pi {ϵ}_{0}a)$. The factor $1/2$ accounts for the fact that the induced charge changes as the real charge is assembled; treating the image as an independently fixed physical charge would give the wrong energy.

A second standard geometry is a grounded conducting sphere of radius $b$ with a charge $q$ a distance $d>b$ from its centre. Its exterior field is generated by an image $q^{\prime }=-qb/d$ at distance $b^2/d$ from the centre on the line towards the real charge. On the sphere the distances to the real and image charges are in the ratio $d/b$, making their potentials cancel. For an isolated neutral sphere, an additional central image $+qb/d$ makes the total induced charge zero and leaves the surface an equipotential. The construction is powerful for suitable symmetric boundaries, but arbitrary boundaries do not in general admit a finite image system.

Keep $+x$ along the upper-sheet current and $+z$ upward, putting the lower sheet at $z=0$ and upper sheet at $z=d$. Between them, the upper sheet's lower-side magnetic field is $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The lower sheet carries current in $-x$, so its upper-side magnetic field is also $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The superposition principle therefore givesEach perfect conductor is an equipotential. The circuit's resistor carries current from upper to lower, so Ohm's law fixes their electric potential difference:The broad parallel faces and their constant potentials make the electric field perpendicular to them and independent of the coordinates along them, away from edges. There is no volume electric charge in the gap. Apply Gauss's law to a small pillbox wholly within the gap, with equal face areas $A$ at heights $z_1,z_2$. Its side flux is zero, soThus $E_z$ is independent of height as well. Since $V_{\mathrm{upper}}-V_{\mathrm{lower}}=-{\int }_0^d{E}_{z}dz$,It points from the higher-potential upper sheet to the lower sheet. Gauss's law requires the planar symmetry argument: zero enclosed charge alone would not prove a uniform electric field in a general geometry. These combined fields characterize parallel current sheets with a resistive termination.

For the focal length of a rotating liquid mirror, consider steady rigid rotation at angular frequency $\Omega =2\pi \nu$. The free surface is an equipotential of $gz-{\Omega }^2{r}^2/2$. HenceA paraboloidal mirror has $z-z_0=r^2/(4f)$, givingThe numerical estimate uses the conventional standard acceleration $g_n=9.80665\mathrm{m}{\mathrm{s}}^{-2}$, rather than an asserted local measurement or a recovered handbook entry.^[ref] The mercury density cancels; the ideal surface shape follows from gravity and rotation.