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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.

The integral field equation. The Ampere-Maxwell law is $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$. Apply Stokes theorem to any fixed oriented surface bounded by the chosen contour:The added displacement current is required for compatibility with electric charge conservation. Taking a divergence and using $\nabla \cdot \mathbf{D}={\rho }_f$ gives $\nabla \cdot \mathbf{J}+{\partial }_t{\rho }_f=0$.

Field geometry. Choose the upper plate positive and let $z$ increase upward. In the long-plate, slow-discharge approximation, $\mathbf{E}=-(V/d)\hat{\mathbf{z}}$. Conventional wire current runs downward from the positive plate. The magnetic field intensity circulates clockwise as viewed from above. The Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ points radially toward the resistive wire.

Wire and displacement currents. The capacitance and discharge law areBecause $V$ decreases, ${\partial }_t\mathbf{D}$ points upward, opposing the downward wire current. Through a disk of radius $r$ between the plates its magnitude isThe effective downward current enclosed by the circular contour is therefore $I-I_D(r)$. The Ampere-Maxwell law givesThis is the magnitude; with the stated coordinates, $\mathbf{H}=-H\hat{\mathbf{\varphi }}$. The factor $1-r^2/a^2$ expresses cancellation between conduction and displacement currents. The approximation requires distances outside the thin wire and away from detailed electrode-edge fields.

Energy flux and power. Since $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$,Its units are watts per square metre. Integrating over the cylinder's side area $2\pi rd$, the signed outward power is negative:There is no leading axial flux through the cylinder's ends in this ideal field geometry.

Where the energy goes. The Poynting flux in a centrally discharged parallel-plate capacitor carries field energy through the dielectric space into the resistor, where it becomes Joule heating at rate $V^2/R$. Inside radius $r$ the stored electric energy isHenceThe resistor receives the inward flux from outside that cylinder plus the field energy released inside it. As $r$ approaches the wire radius, essentially all its power arrives as radial electromagnetic flux. Over the whole capacitor, $-d(C{V}^2/2)/dt=V^2/R$. The Poynting theorem thus accounts for the complete discharge without treating energy as transported only along the conducting wire. High resistance makes the quasistatic electric-field approximation and neglect of changing magnetic energy consistent.

In a homogeneous Ohmic conductor, $\mathbf{J}=\sigma \mathbf{E}$. Combining this with Maxwell's equations givesFor a good conductor, $\sigma \gg \omega ϵ$, so conduction dominates displacement current. With time dependence $e^{-i\omega t}$ and depth $z>0$, take $E\propto e^{ikz}$. ThenThe field is consequently proportional to $e^{-z/\delta }{e}^{iz/\delta }$: it both attenuates and changes phase with depth. The skin depth is the distance for amplitude to fall by $e^{-1}$; power density falls as $e^{-2z/\delta }$, so its attenuation length is $\delta /2$.

Physically, changing fields induce currents whose magnetic field opposes penetration. Equivalently, after neglecting displacement current, magnetic diffusion obeys ${\partial }_t\mathbf{B}=(\mu \sigma {)}^{-1}{\nabla }^2\mathbf{B}$. Faster changes have less time to diffuse into the metal. Increasing frequency, conductivity or magnetic permeability reduces $\delta$, provided the relevant material parameters remain applicable.

For a thick conductor the tangential surface electrical impedance is $Z_s=(1-i)/(\sigma \delta )$ with the stated time convention and consistent field orientation. Its real part $1/(\sigma \delta )$ represents dissipative resistance per square. Current crowding near the surface raises the AC resistance of a wire when $\delta$ is small compared with its radius; thin strands and laminated magnetic cores reduce suitable eddy-current losses. When $\sigma /\gg \omega ϵ$, the full complex propagation constant must replace the good-conductor approximation. In a magnetic metal the magnetic permeability can itself be nonlinear and frequency dependent.

For circular plates of radius $a$, separation $d\ll a$, and a thin axial resistor $R$, the quasistatic capacitance is $C={\epsilon }_0\pi a^2/d$. During discharge, $I=V/R=-C\dot{V}$. A circular Ampere-Maxwell law contour between the plates encloses wire current $I$ and opposing displacement current $-I{r}^2/a^2$, giving $H=I(1-r^2/a^2)/(2\pi r)$. The electric field has magnitude $V/d$, so the Poynting vector points radially inward with magnitude $V^2(1-r^2/a^2)/(2\pi rdR)$. Its flux through the cylindrical side is the displayed inward power. The field inside that cylinder releases $V^2{r}^2/(R{a}^2)$ per unit time; adding the incoming flux gives the resistor's full Joule heating rate $V^2/R$. End fringing, wire radius and magnetic-energy corrections are neglected.