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The current associated with electric polarization and magnetization in matter. If the magnetic response is neglected, it is ${\mathbf{J}}_b={\partial }_t\mathbf{P}$. It is distinguished from a free conduction current in the Maxwell equations in matter.

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

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

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

In a linear isotropic nondispersive medium with time-independent permittivity $ϵ$ and magnetic permeability $\mu$, the stored electromagnetic energy per volume is $u=ϵ|\mathbf{E}{|}^2/2+|\mathbf{B}{|}^2/(2\mu )$. For a short derivation, the Maxwell equations in matter and the divergence identity for a cross product give $\nabla \cdot (\mathbf{E}\times \mathbf{H})=-\mathbf{H}\cdot {\partial }_t\mathbf{B}-\mathbf{E}\cdot {\partial }_t\mathbf{D}-{\mathbf{J}}_f\cdot \mathbf{E}$. The constitutive relations turn the first two terms into $-{\partial }_{t}u$, proving the Poynting theorem. Choose $u=0$ when the electric field and magnetic field both vanish. Dispersive or time-dependent material response requires a different accounting of stored energy.

The Maxwell equations are the field equations of classical electromagnetism. In SI units, in vacuum with prescribed charge density $\rho$ and current density $\mathbf{J}$, they areHere $\mathbf{E}$ is the electric field in volts per metre, $\mathbf{B}$ the magnetic field in teslas, $\rho$ in coulombs per cubic metre, and $\mathbf{J}$ in amperes per square metre. The constants are vacuum permittivity ${ϵ}_0$ and the vacuum value ${\mu }_0$ of magnetic permeability. “Vacuum” specifies the field laws without a material response; “source-free” additionally means $\rho =0$ and $\mathbf{J}=0$. Matter described by averaged fields instead requires Maxwell equations in matter and constitutive relations.

The first equation, Gauss's law, makes electric charge a source of electric flux. Gauss's law for magnetism sets the net magnetic flux through a closed surface to zero. The two curl equations, Faraday's law and the Ampère-Maxwell equation, couple changes of the electric field and magnetic field. Appropriate initial conditions and boundary conditions select a physical solution; the two divergence equations constrain the initial fields. ^[ref].

Allow time dependence while keeping the constitutive relations linear, isotropic, nondispersive, and independent of time. The Maxwell equations in matter giveFor the Poynting vector $\mathcal{P}=\mathbf{E}\times \mathbf{H}$, the divergence identity yieldsThe first two terms are exact time derivatives. Taking the zero-field energy density as zero gives the electromagnetic energy densityConsequently the Poynting theorem is the continuity equationIn integral form,The left side is the rate of increase of stored electromagnetic energy plus outward Poynting flux. The right side accounts for work delivered to the free electric current; in an Ohmic medium ${\mathbf{J}}_f\cdot \mathbf{E}$ is the local rate of Joule heating.

The macroscopic Maxwell equations in matter areHere $\rho$ and $\mathbf{J}$ are the free charge and current densities.

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

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

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

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