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For a parallel-plate capacitor of area $A$ and separation $d$, filled by a spatially varying positive scalar permittivity, neglect edge fields. At fixed terminal voltage, the physical electric potential minimizes $\int \epsilon |\nabla V{|}^{2}d\mathcal{V}/2$. A linear trial potential gives the arithmetic upper bound. At fixed plate charge, minimizing $\int |\mathbf{D}{|}^2/(2\epsilon )d\mathcal{V}$ over divergence-free electric displacement fields gives the harmonic lower bound by using a uniform trial displacement. The averages are volume averages. Layers parallel to the field attain the arithmetic bound; layers perpendicular to the field attain the harmonic bound. Equal volumes of vacuum and relative permittivity $ϵ$ therefore give $2ϵC_0/(1+ϵ)\le C\le (1+ϵ)C_0/2$, where $C_0={\epsilon }_{0}A/d$.

For harmonic electric fields with time convention $e^{i\omega t}$, write $\mathbf{D}={ϵ}^{\ast }(\omega )\mathbf{E}$ for their complex amplitudes, where $\omega$ is angular frequency. The real part ${ϵ}^{\prime }$ describes the in-phase electric displacement field; the minus imaginary part ${ϵ}^{\prime \prime }$ describes the loss component. The sign reverses with the opposite time convention. Complex permittivity can also be a tensor in a linear anisotropic dielectric.

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.

The time derivative of the electric displacement field enters the macroscopic Ampère-Maxwell equation alongside free electric current density. In vacuum, ${\mathbf{J}}_D={ϵ}_0{\partial }_t\mathbf{E}$. It is not transport of free charges through the vacuum gap of a capacitor; its surface integral nevertheless supplies the term needed for compatibility with the continuity equation.

For a lossless nonmagnetic uniaxial optical medium, the electric displacement field of a plane electromagnetic wave is perpendicular to its wavevector. The magnetic field is perpendicular to the plane containing the electric field and wavevector, while the Poynting vector is perpendicular to the electric field in that same plane. Thus the angle between $\mathbf{E}$ and $\mathbf{D}$ equals the walk-off angle between energy flow and phase propagation. At normal incidence on a parallel slab of thickness $L$, a lateral separation $\Delta$ of the ordinary and extraordinary exiting rays gives ${\alpha }_e=\arctan (|\Delta |/L)$, while ${\alpha }_o=0$. The output rays can be parallel even though their internal energy paths differ.

The dimensionless quantity relating induced electric polarization to the internal electric field through $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ in the static scalar model. It describes the material contribution; relative permittivity includes the vacuum term as well, ${ϵ}_r=1+{\chi }_e$. Zero electric susceptibility gives ${ϵ}_r=1$, not zero permittivity.

The scalar ${\chi }_e$ assumes a local response with linearity and isotropy, settled electric polarization, fixed material conditions, and no residual electric polarization at zero electric field. It relates $\mathbf{P}$ to the averaged total internal electric field, which already includes the field of the bound electric charges. It does not relate $\mathbf{P}$ to the field the external electrodes would produce in vacuum. The distinction is essential for a self-consistent calculation.^[ref]

Substitute the scalar relation into the definition of electric displacement field:Comparison with $\mathbf{D}=ϵ\mathbf{E}$ proves $ϵ={ϵ}_0(1+{\chi }_e)$ and ${ϵ}_r=1+{\chi }_e$. Both $\mathbf{D}$ and $\mathbf{P}$ have units $C/{\mathrm{m}}^2$, so ${\chi }_e$ is a dimensionless quantity. In a stipulated static model with ${ϵ}_r=3$, ${\chi }_e=2$: at a prescribed internal electric field, the material contribution $P$ is twice ${ϵ}_{0}E$. This is a hypothetical response, not a measured material constant. The electric susceptibility is a convenient bridge from microscopic dielectric polarization to a macroscopic capacitance measurement, but it is not a measure of electrical conductivity.

The electrostatic boundary conditions at a conductor relate the one-sided total electric fields at a smooth metal–vacuum interface. Choose a unit normal $\mathbf{n}$ from metal into the adjacent vacuum, let ${\mathbf{E}}_{\mathrm{metal}}$ and ${\mathbf{E}}_{\mathrm{vac}}$ be the two limiting electric fields, and let $\sigma$ be the signed surface charge density. A thin Gaussian pillbox with cap area $A$ has outward cap normals $+\mathbf{n}$ on the vacuum side and $-\mathbf{n}$ on the metal side. As its height tends to zero, its side electric flux tends to zero for bounded local electric fields, leavingThe minus sign comes from the second outward cap normal, not from the sign of the electric charge. A shrinking rectangular loop across the interface and the zero closed line integral in electrostatics give continuity of the tangential component:For a conductor in electrostatic equilibrium, ${\mathbf{E}}_{\mathrm{metal}}=0$, henceThus positive surface charge density gives an electric field away from metal and negative surface charge density gives an electric field toward metal. The derivative uses the same chosen unit normal, from the vacuum side. The ideal surface layer has no unique electric field value on the layer itself; these are limiting values. Ordinary surface electric charge preserves continuity of electric potential; an ideal electric dipole sheet would require a different model. ^[ref].

Here “exterior” means just outside the metal, including the vacuum within a cavity in a conductor. On a spherical inner wall of radius $a$, its unit normal from metal into the cavity in a conductor is $-\hat{\mathbf{r}}$. A central positive point charge $q$ gives ${\mathbf{E}}_{\mathrm{vac}}=q\hat{\mathbf{r}}/(4\pi {ϵ}_0{a}^2)$ there, soThe outward normal of the cavity in a conductor as a volume is instead $+\hat{\mathbf{r}}$; confusing these two unit normals reverses the inferred sign. For a stipulated spherical enclosure with inner radius $a$, outer radius $b$, no external sources and total isolated metal electric charge $Q_c$, spherical symmetry and Gauss's law give the analytical modelIts outer surface charge density is $(Q_c+q)/(4\pi b^2)$. Grounding to $V_c=0$ with zero electric potential at infinity instead gives $Q_c=-q$ and zero exterior electric field, while preserving the same nonzero electric field inside the charged cavity in a conductor.

As a numerical sign check, stipulate $q=1\mathrm{nC}$, $a=0.10\mathrm{m}$, $b=0.20\mathrm{m}$ and $1/(4\pi {ϵ}_0)\simeq 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. The inner electric field is approximately $899N/C$ along $+\hat{\mathbf{r}}$, but ${\mathbf{E}}_{\mathrm{vac}}\cdot \mathbf{n}=-899N/C$ and ${\sigma }_{\mathrm{inner}}\simeq -7.96nC/{\mathrm{m}}^2$. An isolated neutral enclosure has $+1\mathrm{nC}$ on the outer wall, approximately $+1.99nC/{\mathrm{m}}^2$ and $225N/C$ just outside. Multiplying each surface charge density by its spherical surface area recovers the signed wall electric charges.

The local result uses the total electric field, including all induced and imposed sources. The factor $\sigma /{ϵ}_0$ follows from the zero metal-side electric field; a symmetric isolated sheet in vacuum instead has magnitude $|\sigma |/(2{ϵ}_0)$ on each side. ^[ref]. At sharp corners, limiting electric fields can be singular and the smooth local pillbox argument needs care. In material outside the metal, use the normal free-charge jump of the electric displacement field and the material's permittivity, as in electrostatic interface boundary conditions. ^[ref].

The extraordinary wave's electric displacement field lies in that plane and is perpendicular to the wavevector; its electric field need not be. For angle $\theta$ between the wavevector and optic axis,Thus $n_{\mathrm{ex}}(0)=n_o$ and $n_{\mathrm{ex}}(\pi /2)=n_e$.^[ref]

A short derivation starts from the existing plane wave in an anisotropic dielectric equation. Put $\mathbf{k}=(k_x,0,k_z)$ and $k_0=\omega /c$, with angular frequency $\omega$ and vacuum speed of light $c$. The $y$ component gives the ordinary wave relation $k_x^2+k_z^2=n_o^2{k}_0^2$. The determinant of the $x,z$ system givesSubstituting $k_x=n_{\mathrm{ex}}{k}_0\sin \theta$ and $k_z=n_{\mathrm{ex}}{k}_0\cos \theta$ yields the displayed directional refractive index.

Southern Methodist University's teaching lab, Introduction to the Parallel Plate Capacitor Lab, makes the response tangible with a capacitor assembled from aluminium foil, paper and taped wire contacts. Its instructions specify approximately $10\mathrm{cm}\times 10\mathrm{cm}$ foil rectangles, aligned with the paper completely separating them. A multimeter in capacitance mode supplies the readout; a ruler measures plate dimensions and a micrometer measures specimen thickness. The lab replaces the paper with other sheets, including cardstock, plastic wrap and wax paper. It asks students to keep foil area and alignment consistent, measure thickness at several positions, estimate thickness uncertainty, and wait for the capacitance reading to stabilize. These are published preparation and measurement instructions, not measured results reported here.^[ref]

The inference uses each specimen's own thickness $t$, not just a comparison of raw meter readings. With corrected capacitance $C_s$ and overlapping area $A$, the ideal estimate is ${ϵ}_r=C_{s}t/({ϵ}_{0}A)$. A thicker sheet can give a smaller capacitance despite larger relative permittivity. This is why changing material while letting the plate gap change does not isolate electric susceptibility by itself.

Useful additional controls are to keep leads in the same positions, use capacitance measurement calibration, repeat one sheet after the others to check drift, and repeat the assembly to assess contact reproducibility. An air-gap reference must have known spacers maintaining the same separation; removing a supporting sheet and letting the foil touch is not such a reference. Record the instrument's test frequency and measurement mode when available. A capacitance display alone does not establish negligible dielectric loss or a strictly static permittivity. These are proposed controls extending the published teaching instructions.

Air gaps produce a particularly revealing systematic error. Suppose a uniform sheet of thickness $t$ and relative permittivity ${ϵ}_r$ is separated from the electrodes by air layers of total thickness $g$. Approximate air permittivity by vacuum permittivity, neglect the fringing field of a capacitor and use the same area in each layer. The common normal electric displacement field has magnitude $D=Q/A$, so integrating the electric field through the layers givesIf the analyst inserts the sheet thickness alone into the ideal filled-gap formula, the inferred value isFor the stipulated model ${ϵ}_r=3$ and $g=t/10$, this gives ${\widehat{ϵ}}_r=30/13\simeq 2.308$, not three. These values are calculated examples, not measured sheet properties. If total plate separation $d=t+g$ is used instead, the effective estimate is $(t+g)/(g+t/{ϵ}_r)$; it still mixes the sheet and the air. Thus measuring a thickness accurately cannot compensate for an unrecognized layer.

The fringing field of a capacitor and additive lead capacitance further limit a foil measurement. Keep lateral plate dimensions large relative to separation, and use a guard electrode of a capacitor in a suitable fixture for more demanding work.^[ref] A family of thickness measurements can help distinguish material response from an additive offset, but air gaps can vary with the assembly. A large, repeatable reading supports a capacitance change; extracting a material electric susceptibility requires the geometry, gaps, instrument response and dielectric loss assumptions to survive these checks. No measured relative permittivity is claimed for any of the lab's sheets.

Birefringence produced by subwavelength structure rather than solely by intrinsic crystalline anisotropy. For ideal dielectric lamellae, continuity of tangential electric field gives ${ϵ}_{\parallel }=f{ϵ}_1+(1-f){ϵ}_2$, while continuity of normal electric displacement field gives ${ϵ}_{\perp }^{-1}=f/{ϵ}_1+(1-f)/{ϵ}_2$. The unequal effective permittivities give different principal refractive indices.

In macroscopic electromagnetism, free charge is the part of the charge density represented explicitly as ${\rho }_f$, separately from bound charge associated with the electric polarization of matter. The electric displacement field satisfies $\nabla \cdot \mathbf{D}={\rho }_f$. The word free does not imply zero force or zero energy: it distinguishes this source from the polarization response in the chosen material description.

For a point electric charge $q$ in a homogeneous medium of permittivity ${ϵ}_1$ beside a plane interface with permittivity ${ϵ}_2$, its potential on the source side is reproduced by a mirrored image charge $q^{\prime }=q({ϵ}_1-{ϵ}_2)/({ϵ}_1+{ϵ}_2)$. The interface has no free surface charge; continuity of tangential electric field and normal electric displacement field fixes this coefficient.

For electrically isolated electrodes with unchanged free charge, the increment of normal electric displacement field is approximately zero. A constitutive relation such as $D_3=d_{31}({\sigma }_1+{\sigma }_2)+ϵE_3$ then gives an induced field that balances stress-induced displacement. Finite leakage and input capacitance modify the voltage response.

A long coaxial cylindrical capacitor of length $L$ and radii $a,b$ has a dielectric-filled overlap of length $x$ and relative permittivity ${ϵ}_r$. Neglecting end and interface fringing, its filled and vacuum portions are in parallel because both share the same two conductors. The electric potential difference $V$ produces $E_r=V/[r\ln (b/a)]$ in either portion; the electric displacement field and electrode charge differ by the local permittivity. Adding their capacitances gives the title formula. The dielectric boundary is axial while the interior electric field is radial and tangential to that interface, consistent with the electrostatic interface boundary conditions.

Use the constitutive relations $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{H}=\mathbf{B}/\mu$, where $\mathbf{D}$ is the electric displacement field, $\mathbf{H}$ the magnetic field strength, $ϵ$ the permittivity, and $\mu$ the magnetic permeability. Let ${\rho }_f$ be the charge density of free charge and ${\mathbf{J}}_f$ its current density. In this homogeneous medium, Gauss's law and the magnetostatic Ampère law are, respectively,Here $W$ is a volume with closed boundary $\partial W$, and $S$ is an oriented spanning surface whose boundary orientation follows the right-hand rule. Thus the electric field form is $\nabla \cdot \mathbf{E}={\rho }_f/ϵ$ and ${\oint }_{\partial W}\mathbf{E}\cdot \mathbf{n}dS=Q_f/ϵ$; the magnetic field form is $\nabla \times \mathbf{B}=\mu {\mathbf{J}}_f$ and ${\oint }_{\partial S}\mathbf{B}\cdot d\mathbf{\ell }=\mu I_f$.

The remaining static Maxwell equations areThese express Gauss's law for magnetism and the static limit of Faraday's law. The volume and circulation forms follow from the divergence theorem and Stokes theorem. The electric charge and electric current on the right of the macroscopic laws are free sources; material electric polarization and magnetization are already included in $\mathbf{D}$ and $\mathbf{H}$.

For this radially layered coaxial cable, defineUse the solution regular at the axis with bounded exterior electric potential, so that the empty interior and exterior have the constant electric potentials fixed by their adjacent conducting cylinders.

There is no free charge in either electrical insulator or on their interface. Gauss's law and continuity of normal electric displacement field give $D_{\rho }=C/\rho$ in both layers. The electric potential difference fixes the constant becauseThe electric potential, including the empty regions bounded by the conducting cylinders, is thereforeBoth inner and outer Dirichlet boundary conditions hold, and the electric potential is continuous at the material interface, whereTaking the negative gradient of the electric potential yieldsThe radial electric field generally jumps at $\rho =c$, but $ϵE_{\rho }(c^{-})={ϵ}^{\prime }{E}_{\rho }(c^{+})$, as required by the absence of free surface charge density there. The conducting cylinders carry opposite linear charge densities $\lambda =2\pi C$ and $-\lambda$.

For an Amperian loop of radius $\rho$, the enclosed electric current is zero, $I$, or zero according as $\rho <a$, $a<\rho <b$, or $\rho >b$. The magnetostatic Ampère law therefore gives $H_{\phi }=I/(2\pi \rho )$ in the gap, and zero outside. Using the two magnetic permeabilities,The tangential magnetic field strength is continuous at the material interface, while the tangential magnetic field need not be. At each ideal conducting surface and at the material interface, the displayed electric field and magnetic field formulas specify one-sided limits; a single interface value is unnecessary when the limiting values differ.

The electric polarization $P$ is electric dipole moment per unit volume. A polarized body acts as the bound volume and surface charge densityIndeed, summing the scalar potential of its dipoles and integrating by parts givesThis is exactly the potential of ${\rho }_{\mathrm{b}}$ and ${\sigma }_{\mathrm{b}}$.

For the central free charge, Gauss's law for the electric displacement field givesThus on the sphere,

For a time-independent field, Faraday's law gives $\nabla \times \mathbf{E}=0$. Integrating around a thin rectangular loop crossing $x=0$ and shrinking its height gives continuity of the tangential component:A thin Gaussian pillbox and $\nabla \cdot \mathbf{D}=0$ give continuity of the normal electric displacement field:These are the electrostatic cases of the dielectric-interface boundary conditions.

The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume. It producesThe electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ is useful becauseso it separates free charge from the bound charge represented by $\mathbf{P}$. In a linear isotropic dielectric, $\mathbf{D}=ϵ\mathbf{E}$.

Spherical symmetry and Gauss's law for $\mathbf{D}$ give, on both sides of the interface,ThusWith $\mathbf{n}={\mathbf{e}}_r$ pointing from medium one to medium two, the net bound sheet charge at $r=R$ is the polarization jumpIf one assigns surface charge separately to the inner body and outer medium, their contributions are ${\mathbf{P}}_1\cdot \mathbf{n}$ and $-{\mathbf{P}}_2\cdot \mathbf{n}$; their sum is the displayed physical interface charge.

Birefringence is polarization-dependent propagation in an optically anisotropic medium. Its permittivity tensor relates $\mathbf{D}$ to $\mathbf{E}$. In a lossless uniaxial dielectric, take the optic axis along $z$ and the principal relative permittivities as $(n_o^2,n_o^2,n_e^2)$. For a wave vector making angle $\vartheta$ with that axis, the ordinary mode has index $n_o$, while the extraordinary phase-index angular dependence isThe ordinary electric field is perpendicular to the plane containing the wave vector and optic axis. The extraordinary electric displacement field lies in that plane and is perpendicular to the wave vector, but its electric field need not be. Its energy-flow direction can therefore differ from the wave-normal direction: birefringent walk-off. Along the optic axis the two indices coincide. Refraction generally splits an incident beam into two differently polarized rays, producing double images, as in calcite.

A waveplate instead uses the relative phase of two eigenpolarizations. With the axes chosen for the propagation direction, the retardance is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate gives $|\Delta |=\pi /2$ modulo $2\pi$ and converts equal-amplitude orthogonal components between linear and circular polarization. A half-wave plate gives $|\Delta |=\pi$ and rotates a linear polarization. This is a phase effect; unequal absorption is a separate property.^[ref]