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Charge conservation says that the total signed electric charge of an isolated collection stays constant. More generally, a collection closed to electric charge exchange has constant total electric charge, even if energy crosses its boundary. This is the charge bookkeeping relevant to a closed system or an isolated system; isolation from energy exchange is not required. Positive and negative electric charge count with opposite signs. Creating or removing an electron and a Positron together is compatible with charge conservation, because their combined electric charge is zero. ^[ref].

Imagine a fixed box initially containing electric charge $Q(0)=10\mu \mathrm{C}$. Stipulate incoming conventional current $I_{\mathrm{in}}=3\mu \mathrm{A}$ and outgoing conventional current $I_{\mathrm{out}}=1\mu \mathrm{A}$ throughout $0\le t\le 2\mathrm{s}$, with no other boundary transport. Since one ampere is one coulomb per second, the stored electric charge grows at $2\mu C/s$:If instead the outgoing electric current also equals $3\mu \mathrm{A}$, the stored electric charge remains $10\mu \mathrm{C}$ while electric charge passes through. These are stipulated transport examples; constant terminal electric currents alone do not guarantee a time-independent interior charge density.

Locally, charge conservation is the charge continuity equation, an instance of the general continuity equation:Here $\rho (\mathbf{r},t)$ is signed volume charge density and $\mathbf{J}(\mathbf{r},t)$ is current density. A positive divergence of current density means net outward transport from a small neighborhood, so its charge density decreases. For a fixed control volume $V$ with closed surface $\partial V$ and outward unit normal $\mathbf{n}$, the equivalent integrated balance isThe surface integral is signed: incoming positive electric charge contributes negatively to outward electric current, and outgoing positive electric charge contributes positively. Integrate the charge continuity equation over fixed $V$ and apply the divergence theorem to obtain this balance. Zero net boundary transport gives constant $Q_V$; the global and local statements describe the same conservation law at different scales. ^[ref].

In a steady transport state, charge density and current density are time-independent, so the charge continuity equation reduces to $\nabla \cdot \mathbf{J}=0$. At an electrical circuit junction, choose each terminal electric current $I_k$ positive outward and assume the terminals account for all boundary transport. ThenKirchhoff's current law, ${\sum }_k{I}_k=0$, follows when accumulation of electric charge at the junction is negligible. During storage, retain the same balance with its nonzero derivative; charge conservation continues to hold. This makes charge conservation useful for checking electrical circuit models and for tracking the charging of a capacitor. ^[ref].

The beautiful consistency with Maxwell equations is that the electric field and magnetic field already enforce this transport balance. For sufficiently smooth fields in vacuum SI, take the divergence of the Ampère-Maxwell equation, use that the divergence of a curl is zero, and substitute Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Constant vacuum permittivity ${ϵ}_0$ givesThe displacement current term supplies the part needed when electric charge accumulates. Historically, Maxwell added that term to extend the steady-current equation; see the existing charge conservation from Maxwell equations for this connection. ^[ref].

For a twice continuously differentiable vector field in Cartesian coordinates, expand the curl and then its divergence. Every mixed partial derivative occurs twice with opposite signs, proving the identity. In electromagnetism this forces the right side of the Ampère-Maxwell equation to be divergence-free, yielding local charge conservation.

Electric charge is an additive, signed property of matter that determines its coupling to the electromagnetic field. Write a particle's electric charge as $q$, or a collection's total electric charge as $Q$; the SI unit is the coulomb, $\mathrm{C}$. A stationary test particle in an electric field experiences electric force $\mathbf{F}=q\mathbf{E}$, while motion adds the magnetic part of the Lorentz force. Electric charge both produces an electric field and determines the response to one. For two stationary point particles in vacuum, Coulomb's law makes like signs repel and opposite signs attract. The choice of which sign to call positive is a convention, used consistently: protons carry $+e$ and electrons carry $-e$, where $e$ is the elementary charge. ^[ref].

Additivity means adding signed electric charges, rather than their magnitudes:for ordinary matter containing $N_p$ protons, $N_e$ electrons, and electrically neutral neutrons. A toy collection with seven protons and ten electrons has $Q=-3e$, however much neutral material accompanies it. Charge neutrality means $Q=0$; it does not mean an absence of charged constituents. Nor does it guarantee an absence of electric force: the spatial arrangement of positive and negative electric charges matters, as illustrated by an electric dipole. This makes a tiny imbalance in an otherwise nearly neutral object physically important. ^[ref].

Charge conservation states that a collection closed to transport of electric charge keeps its total signed $Q$. If $n$ electrons move from object A to object B without other exchange, then $\Delta Q_A=+ne$ and $\Delta Q_B=-ne$; the sum is unchanged. The amount of positive electric charge alone need not be conserved: creation of an electron and a Positron adds opposite electric charges with zero sum. These bookkeeping rules let us distinguish separation or transport of electric charge from creation of net electric charge. ^[ref].

A simple motivating experiment uses two charge-producing paddles, a conducting pail on an insulating support, a grounded outer shield and a high-input-impedance electrometer. Zero the pail, then isolate it. Rub the initially neutral paddles together, keeping their charged surfaces away from other objects. Lower each deeply into the pail without touching: the electrometer gives opposite signs. Inserting both together tests cancellation. For a separate contact test, insert one charged paddle without touching and withdraw it: the reading returns to zero. Repeat after zeroing, this time touching the inner wall before withdrawal: a residual reading shows transfer of electric charge. Zero the pail and reinsert the paddle to test how much electric charge remains on it. PASCO describes these procedures in Basic Electrostatics System manual, demonstrations 1A and 1B, printed pp. 13–16.

The crucial control is withdrawal without contact: a response during insertion can come from electrostatic induction, which redistributes the pail's existing electric charge. In the fixed-geometry approximation, the electrometer measures electric potential difference proportional to enclosed electric charge, with conversion set by effective capacitance; a voltage reading is not already a measurement in coulombs. Opposite readings and their cancellation test additivity and charge conservation within resolution. Leakage or accidental contact spoils that interpretation, and contact need not transfer all the paddle's electric charge. This macroscopic experiment cannot resolve elementary charge steps: that requires the much smaller particles of the oil drop experiment.

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 electron filling an electron hole removes both excitations relative to the filled-band background. Their electric charges cancel, consistent with charge conservation; the remaining electrons are still present. The released energy can leave through photon emission or other channels, including phonons. Formation of an exciton instead retains the excited electron and electron hole as a bound pair; recombination removes the pair. A valence band electron filling a nearby vacancy merely moves the electron hole and does not remove the total vacancy count.

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.

An elementary ionization-chamber charge-loss model makes the inference explicit. Let a positively charged electrode and its electrometer have effective capacitance $C$, electric potential difference $U$, and enclosed gas volume $\mathcal{V}$. Assume constant $C$, a short enough interval that collection conditions remain stable, and no multiplication of charge in the gas. If ionization produces $q$ ion pairs per unit volume per unit time and a fraction $\eta$ is collected, thenHere $e$ is the elementary charge and $I_{\mathrm{leak}}$ is the parasitic electric current through imperfect insulation. Each fully collected ion pair transfers one elementary charge through the measuring circuit; counting its two opposite carriers as two transfers would introduce a factor-of-two error. The relation follows from charge conservation, rather than from a model of the incoming particle's identity.

For an invented example, take $C=2.00\mathrm{pF}$, $\mathcal{V}=2.00\mathrm{L}$, $\eta =1$, a measured fall of $6.00\mathrm{V}{\mathrm{h}}^{-1}$, and a separately determined leakage-equivalent fall of $0.60\mathrm{V}{\mathrm{h}}^{-1}$. The leakage-subtracted fall is $5.40\mathrm{V}{\mathrm{h}}^{-1}$, soThis is about $1.87\times 10^4$ collected ion pairs per second throughout the chamber, not that many primary cosmic rays. Unknown collection losses make $\eta <1$ another uncertainty; changes in capacitance would require differentiating $Q=CU$ with both factors variable.

Gas density is a separate issue. Under a fixed radiation field, fixed gas composition and approximately constant energy deposited per unit gas mass, the volume-based ionization rate scales as $q\propto \rho$. For a vented ionization chamber, the ideal gas law gives the illustrative correctionwhere $T$ is absolute temperature. A pressure-only correction presumes equal temperature. A sealed rigid chamber instead fixes gas number density as it rises: its internal pressure may change with temperature without losing gas. This does not make its insulation, electrometer calibration or collection efficiency insensitive to temperature. Nor can dividing by gas density correct arbitrary radiation from the vessel walls or changing particle energies. Those are distinct effects, not a universal normalization recipe.

Take $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. The four-current, four-potential and four-gradient areThe first two transform as contravariant four-vectors under an inertial-frame Lorentz transformation, within a consistently transformed electromagnetic gauge. The lowered-index gradient is a covector. Their contraction is a Lorentz scalar:Indeed, $J^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{J}^{\nu }$ and ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\alpha }{}_{\mu }{\partial }_{\alpha }$, so the two matrices cancel in the contraction. Local charge conservation in one inertial frame implies local charge conservation in every inertial frame. Integrating over a region gives the charge change equal to minus its outward current flux; global conserved charge requires no flux through the system boundary.

In SI units the Maxwell's equations in vacuum, with sources, areTaking the divergence of the last equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot j=0$, the local conservation of electric charge. Thus sources that violate charge conservation cannot satisfy all the Maxwell's equations. Equivalently, with this conservation law the curl evolution preserves the electric divergence constraint if it holds initially; Faraday's law similarly preserves $\nabla \cdot B=0$.

With $\rho =j=0$, use $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-\Delta E$ and substitute the two curl equations. This gives $-\Delta E=-{\mu }_0{ϵ}_0{\partial }_t^{2}E$. Applying the same calculation to $B$ yieldsA monochromatic plane wave has one angular frequency $\omega$ and constant-phase planes $\mathbf{K}\cdot r=\omega t-\text{constant}$, normal to the wavevector $\mathbf{K}$. Its wavelength is $2\pi /|\mathbf{K}|$.

For a nontrivial traveling wave with $\omega \ne 0$, substitution of the displayed electric field into the divergence equation gives $\mathbf{K}\cdot e_1=\mathbf{K}\cdot e_2=0$. Faraday's law is satisfied byIndeed ${\partial }_{t}B=\mathbf{K}\times E_{\varphi }$ and $\nabla \times E=-\mathbf{K}\times E_{\varphi }$. The magnetic divergence is zero, while Ampère's law reduces to $|\mathbf{K}{|}^2{E}_{\varphi }/\omega =(\omega /c^2)E_{\varphi }$, giving the dispersion relation. The amplitudes may be arbitrary transverse vectors; they need not be mutually orthogonal or have equal lengths. Their combination describes the polarization.

For a charging parallel-plate capacitor of area $A$, neglect fringing. The field in the gap is $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isAn Amperian loop around the feed wire can bound a surface cutting the wire, which has conduction current $I$, or a bulging surface passing through the dielectric gap, which has no conduction current. Including displacement current makes both surfaces give the same magnetic circulation. It is also required by charge conservation: the divergence of the Ampère-Maxwell equation and Gauss's law yields $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the continuity equation.

Take the divergence of the proposed Ampère's law. Since the divergence of a curl vanishes identically, it implies $\nabla \cdot \mathbf{J}=0$. However, local charge conservation requires the continuity equationFor example, electric current entering a charging capacitor accumulates electric charge on its plates, so the charge density is time dependent. The proposed equation cannot describe that process.

From Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Thus $\nabla \cdot ({ϵ}_0{\partial }_t\mathbf{E})={\partial }_t\rho$, which supplies precisely the missing divergence. Adding the displacement current density gives the Ampère-Maxwell equation:Its divergence now yields the correct continuity equation. Through a fixed surface, the associated displacement current is $I_D={ϵ}_{0}d{\Phi }_E/dt$, where ${\Phi }_E$ is the electric flux. In a vacuum capacitor gap, this equals the conduction electric current charging the plates, even though no charges cross the gap. Charge conservation fixes the required divergence of the correction; by itself it does not exclude adding a further divergence-free vector field.

The magnetostatic Ampère's law, in vacuum, isThe path direction and surface normal are related by the right-hand rule. Taking the divergence of the differential law gives $\nabla \cdot \mathbf{J}=0$, since the divergence of a curl vanishes. However, charge conservation requires the continuity equation $\nabla \cdot \mathbf{J}=-\partial \rho /\partial t$, which need not vanish during charging.

Using Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the combination $\mathbf{J}+{ϵ}_0\partial \mathbf{E}/\partial t$ has zero divergence. Replacing $\mathbf{J}$ by this conserved total current density gives the Ampère-Maxwell equation:Its integral form is $\oint \mathbf{B}\cdot d\mathbf{\ell }={\mu }_0[I_{\mathrm{enc}}+{ϵ}_{0}d{\Phi }_E/dt]$. The extra term is the displacement current. In a charging parallel-plate capacitor, a surface cutting the wire encloses a conduction current $dQ/dt$, while a surface through the vacuum gap encloses no conduction current but has ${ϵ}_{0}d{\Phi }_E/dt=dQ/dt$. This modification makes the magnetic circulation independent of which spanning surface is chosen. The vacuum displacement current represents a changing electric field, not literal transport of charge across the gap.

Let the unit normal point from layer 1 to layer 2. The electrostatic electric field boundary condition is $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$, while Gauss's law for the electric displacement field gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Here ${ϵ}_i$ denotes absolute permittivity; replace it by ${ϵ}_0{ϵ}_{ri}$ if dielectric constants are relative.

In the steady conducting state, charge conservation makes the normal current density the same in both layers:Writing $\mathcal{D}={\sigma }_2{d}_1+{\sigma }_1{d}_2$, the steady fields in two leaky dielectric layers areThe signed interfacial free surface charge density is the jump of $D_n$:The total free-plus-bound sheet charge instead follows from the vacuum form of Gauss's law,The bound contribution is ${\sigma }_{\mathrm{total}}-{\sigma }_f$. These are signed quantities for the chosen normal and voltage direction; their magnitudes are obtained by absolute values. They describe the long-time DC state, not the initial capacitive transient. Positive conductivities suffice for the derivation; if both vanish, conductivity alone does not determine the field division, which also depends on the initial free interfacial charge.

The fundamental mode of an ideal coaxial cable is a transverse electromagnetic mode: the electric field is radial between the conductors, the magnetic field is azimuthal, and energy flows along the cable through the dielectric. There is no cutoff for this mode; higher modes, conductor losses and dielectric dispersion can matter outside the ideal transmission-line regime.

Ignoring edge fields in the parallel-strip geometry, $E_x=V/a$ and $H_y=I/b$. Per unit length the charge is $ϵbV/a$ and the linked magnetic flux is $\mu aI/b$. Thus the parallel-plate transmission line hasFor a segment $dz$, charge conservation, equivalently the appropriate integral of the Ampère-Maxwell equation, gives $I(z)-I(z+dz)=Cdz{\partial }_{t}V$. Applying Faraday's law around a longitudinal rectangle between the plates gives $V(z)-V(z+dz)=Ldz{\partial }_{t}I$. These are the lossless telegrapher's equations,Differentiate the second with respect to $z$ and use the first, and conversely for $I$:The signal speed is $1/\sqrt{LC}=1/\sqrt{\mu ϵ}$, equal to $c$ for a vacuum-filled line. It is the speed of the coupled electromagnetic disturbance, rather than the electron drift velocity. Electrons locally adjust their motion as the field arrives; one electron need not travel the cable's length to transmit a signal. The Poynting vector in the dielectric integrates to power $VI$.

In microscopic SI form, Maxwell's equations couple charge density $\rho$ and current density $\mathbf{J}$ to electric and magnetic fields:Gauss's electric law expresses electric flux from charge and reduces to the inverse-square Coulomb field for a point charge. Gauss's magnetic law gives zero net magnetic flux through every closed surface; magnetic field lines do not begin or end on magnetic monopoles in this theory.

Faraday's induction law says changing magnetic flux produces a circulating electric field. For a fixed loop, its circulation is minus the rate of change of enclosed magnetic flux, explaining the sign of induction and Lenz's law. For example, changing a primary coil's electric current changes the flux through a stationary secondary coil and induces a voltage. A moving wire also has the magnetic part of the Lorentz force on its charges; the fixed-loop field equation and the full motional-emf calculation must be distinguished.

The Ampère-Maxwell law relates magnetic circulation to conduction current and displacement current. The latter is essential around a charging capacitor: a surface cutting the lead and another passing through the gap must predict the same circulation. Taking the divergence givesusing Gauss's electric law. Charge conservation therefore fixes the consistency of the time-dependent equations. In matter, the equivalent macroscopic equations use $\mathbf{D}$ and $\mathbf{H}$, free sources and constitutive relations. The Lorentz force density $\rho \mathbf{E}+\mathbf{J}\times \mathbf{B}$ describes exchange of momentum with charges.

In source-free vacuum, take the curl of Faraday's law and use $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. Both fields satisfyFor a plane wave, $\omega =ck$, $\mathbf{k}\cdot \mathbf{E}=\mathbf{k}\cdot \mathbf{B}=0$ and $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. The fields are transverse, mutually perpendicular and in phase. Their energy density and energy flux areThis is Poynting's theorem: field energy is transported and can do work on matter. Electromagnetic radiation is a propagating, energy-carrying solution of the coupled electric and magnetic field equations.

Displacement current makes Ampère's law compatible with charge conservation. The macroscopic Maxwell equations areTaking the divergence of the last equation and using Gauss's law gives $\nabla \cdot {\mathbf{J}}_f+{\partial }_t{\rho }_f=0$. Omitting ${\partial }_t\mathbf{D}$ would wrongly require the conduction current to be divergence free during charging. The displacement-current density ${\partial }_t\mathbf{D}$ has current-density units, but need not represent free charges crossing a region.

The surface independence of Ampère circulation during capacitor charging illustrates the need directly. A surface spanning the circuit can cut the wire or pass between the plates. For ideal parallel plates, $Q={ϵ}_{0}AE$ andThe corrected circulation of $\mathbf{H}$ is independent of that surface choice. Free conduction current stops at a plate; displacement current supplies the gap contribution.

In vacuum the two time-dependent curl equations support waves. Taking a curl of Faraday's law in a source-free region yieldsThe corresponding magnetic equation is identical. For a plane wave, $\mathbf{k}\cdot \mathbf{E}=\mathbf{k}\cdot \mathbf{B}=0$ and $\mathbf{B}=\mathbf{k}\times \mathbf{E}/\omega$. The fields carry energy through the Poynting flux. Dropping displacement current everywhere would remove these freely propagating solutions.^[ref]

In a metal, conduction and displacement compete. For a uniform ohmic material with ${\mathbf{J}}_f=\sigma \mathbf{E}$, constant $ϵ,\mu$, and transverse waves,Use $e^{i\mathbf{k}\cdot \mathbf{r}-i\omega t}$. The dispersion relation and conduction-to-displacement current ratio areWhen $\sigma \gg \omega ϵ$, $k\simeq (1+i)/\delta$ with $\delta =\sqrt{2/(\mu \sigma \omega )}$: fields are screened within a skin depth and energy is dissipated by Joule heating. This quasistatic limit neglects displacement current locally compared with conduction; it does not invalidate the full Maxwell law. At sufficiently high frequency, conductivity and permittivity are frequency dependent, so a constant-$\sigma$ approximation must be replaced by a material model.

In an unmagnetized cold plasma, electron inertia gives a cutoff. With immobile ions and negligible collisions, $m_e\dot{\mathbf{v}}=-e\mathbf{E}$ and $\mathbf{J}=-n_{e}e\mathbf{v}$ give $\sigma (\omega )=i{n}_e{e}^2/(m_e\omega )$. For the transverse wave cutoff in a cold unmagnetized plasma, combining this carrier current with vacuum displacement current givesTransverse waves propagate above ${\omega }_p$ and are evanescent below it. The group velocity is $c\sqrt{1-{\omega }_p^2/{\omega }^2}<c$ even though the phase velocity exceeds $c$. The cold model also has longitudinal charge-density oscillations at ${\omega }_p$, distinct from transverse light waves. Collisions add damping, and pressure or magnetic fields modify these simple dispersions. To avoid carrier-current double counting in effective permittivity, once the carrier response is included in ${ϵ}_{\mathrm{eff}}$ it must not be counted a second time as a separate conductivity. Displacement current therefore connects circuit continuity, vacuum radiation, conductor screening and plasma propagation within one consistent field theory.^[ref]

Maxwell's displacement current is the electric-field contribution needed to make Ampère's law consistent with charge conservation. In macroscopic form,The displacement current density ${\mathbf{J}}_d={\partial }_t\mathbf{D}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Since $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$, it contains the vacuum field term and, in matter, the polarization-current contribution. It need not represent free charge moving through a material.

Taking the divergence gives $0=\nabla \cdot {\mathbf{J}}_f+{\partial }_t(\nabla \cdot \mathbf{D})$. Gauss's law, $\nabla \cdot \mathbf{D}={\rho }_f$, then yields the continuity equation ${\partial }_t{\rho }_f+\nabla \cdot {\mathbf{J}}_f=0$. Without the displacement term, Ampère's original law would incorrectly require every current distribution to be divergence-free even while charge accumulates.

A charging parallel-plate capacitor makes this concrete. Ignoring fringing, $D=Q/A$, so the displacement current across its dielectric gap is $I_d=A{\partial }_{t}D=dQ/dt$, equal to the conduction electric current in its wire. The integral lawtherefore gives the same answer whether the spanning surface cuts the wire or passes through the gap. Together with Faraday's law, this field term also permits self-consistent electromagnetic waves in vacuum, with speed $c=1/\sqrt{{\mu }_0{ϵ}_0}$.

Integral conservation. Electric charge cannot be created by ordinary transport. For a fixed volume $V$, outward current reduces the enclosed charge:Here ${\rho }_e$ has units $\mathrm{C}{\mathrm{m}}^{-3}$ and $\mathbf{J}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Applying the divergence theorem to an arbitrary volume gives the local continuity equationIt is local charge conservation: a net outflow must be accompanied by depletion. In a steady state, $\nabla \cdot \mathbf{J}=0$. Integrating over a small circuit junction then gives Kirchhoff's current law. During a transient the junction can accumulate charge and the incoming and outgoing conduction currents need not yet match.

Consistency with Maxwell's equations. In vacuum, take the divergence of the Ampère-Maxwell equation:Gauss's law, $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, reproduces the continuity equation. The displacement current term makes these equations consistent for time-dependent charge distributions, such as a charging capacitor. It is not a stream of free charges across the insulating gap. In matter, $\nabla \cdot \mathbf{D}={\rho }_f$ and $\nabla \times \mathbf{H}={\mathbf{J}}_f+{\partial }_t\mathbf{D}$ give the analogous conservation law for free charge; total charge also includes bound charge.

For charge sharing between distant connected spheres, use the standard approximation that the spheres are sufficiently far apart for mutual electrostatic effects to be negligible, and that the wire adds negligible capacitance. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}r$. ThusThe initial charges are $Q_i=C_i{V}_i$. Connecting the spheres allows charge to flow until their potentials are equal. Charge conservation givesThe initial and final electrostatic energies are $U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$ and $U_f=\frac{1}{2}(C_1+C_2)V_f^2$. Their difference becomes Joule heating in the resistive wire:Numerically,The resistance determines how fast this happens, not the final energy difference. Sufficient damping suppresses sustained charge-current oscillations; the calculation also neglects energy radiated away. At finite sphere separation the capacitance matrix would be needed, and the radii and initial potentials alone would not specify an exact correction.

The quoted capacitance is that of the filled parallel-plate capacitor, $C_i=1.0\times 10^{-6}\mathrm{F}$. Disconnecting the source imposes charge conservation on the isolated plates:On removing the dielectric, the capacitance falls by its relative permittivity, so $C_f=C_i/5=2.0\times 10^{-7}\mathrm{F}$. The capacitor energy at fixed electric charge is $U=Q^2/(2C)$. ThereforeandThe increase is supplied by the external work needed to withdraw the dielectric against its attraction into the electric field, in the lossless quasistatic limit. The final electric potential difference is $Q/C_f=12500\mathrm{V}$, not the original charging value.