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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 $c=1/\sqrt{{\mu }_0{ϵ}_0}$.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

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.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

Equilibrium radiation and cavity frequencies. A blackbody absorbs all incident radiation at every frequency and direction. In thermal equilibrium its emission spectrum depends only on its temperature. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes black-body radiation to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.

In a rectangular electromagnetic cavity, the wave equation and perfect conductor wall boundary conditions select component wavenumbers $k_x=\pi l/L_x$, $k_y=\pi m/L_y$, $k_z=\pi n/L_z$. Combining the vacuum dispersion relation $\omega =c|\mathbf{k}|$ and $\nu =\omega /(2\pi )$ givesFor example, electric-field components can have the spatial dependenceFor three positive indices, Gauss's law requires $k_x{A}_x+k_y{A}_y+k_z{A}_z=0$, leaving two independent polarizations of an electromagnetic wave.

Index qualification. The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one polarization of an electromagnetic wave. For instance, with $n=0$, $l,m\ge 1$, only $E_z$ above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional power law.

Three-dimensional mode count. A lattice cell in positive $\mathbf{k}$ space has volume ${\pi }^3/V$, where $V=L_x{L}_y{L}_z$. In a continuum approximation, a shell of radius $k$ and thickness $dk$ contains its positive octant, multiplied by two polarizations:Using $k=2\pi \nu /c$ givesFor a finite cavity the exact density of states is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.

Mean photon number. A mode is a quantum harmonic oscillator with thermal excitation energies $sh\nu$, $s=0,1,2,\dots$. The zero-point energy does not enter its thermal occupation. Put $q=e^{-h\nu /(k_{B}T)}$. Its partition function and mean occupation areThis is the Bose-Einstein distribution with zero photon chemical potential: equilibrium absorption and emission do not conserve photon number.

Thermal energy. Multiply the density of states by the mean excitation energy $h\nu ⟨s⟩$ and integrate:The substitution $x=h\nu /(k_{B}T)$ isolates all the dependence on cavity dimensions and temperature:Expanding $1/(e^x-1)={\sum }_{r=1}^{\infty }{e}^{-rx}$ gives an integral ${\sum }_{r}6/r^4=6\zeta (4)={\pi }^4/15$. HenceThis is the thermal energy only, excluding the zero-point contribution. The three-dimensional continuum regime requires $k_{B}T$ large compared with the geometric level spacings in all three directions.

Spectral sketch. The spectral energy density per unit ordinary frequency is Planck's law,At small frequency it rises as $8\pi k_{B}T{\nu }^2/c^3$, while at large frequency it decays exponentially. Raising temperature increases the spectrum at every fixed positive frequency, moves the peak to larger frequency in proportion to $T$, and increases the integrated area in proportion to $T^4$. A plot per unit wavelength has a different peak position because the integration variable changes.

Thin-cavity radiation. Let $A=L_x{L}_y$. The relevant low-temperature window isNonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The $n=0$ family has one polarization of an electromagnetic wave. Its allowed in-plane wavenumbers form a lattice of cell area ${\pi }^2/A$; a quarter of a circular shell therefore containsThus the dimensional crossover of thermal cavity radiation givesThe integral is $2\zeta (3)$ by the same geometric-series argument. A model with a different number of independent polarizations of an electromagnetic wave changes the prefactor but not the $T^3$ scaling. For a cavity with fixed finite $L_x,L_y$, cooling still further eventually freezes its lowest in-plane mode as well: the strict $T\to 0$ thermal energy is exponentially suppressed. The requested area law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.

The original PDF calls ${ϵ}_r$ a relative magnetic permeability while also specifying a nonmagnetic dielectric. These descriptions are inconsistent if read literally. The physically consistent interpretation of the notation is relative permittivity ${ϵ}_r=2$ and relative magnetic permeability ${\mu }_r=1$.

Let $a=3\mathrm{mm}$ and let $b=6\mathrm{mm}$ be the inner radius of the outer conductor. For line charge density $\lambda$, Gauss's law gives $E_r=\lambda /(2\pi ϵr)$, henceThe magnetic field in the dielectric is $H_{\phi }=I/(2\pi r)$. Its energy per unit length givesUsing $b/a=2$,These are the lossless coaxial line parameters, assuming negligible field penetration into the conductors. If the quoted radii instead referred to an outer diameter or if ${\mu }_r=2$ were intended, the data would need clarification; the stated dielectric and nonmagnetic assumptions support the interpretation above.

An electrical insulator has no freely conducting bulk charge in the ideal static description, but an applied electric field can displace bound positive and negative charge relative to one another. The resulting electric dipole moments per unit volume define the polarization density $\mathbf{P}$. Distortion of electronic charge clouds, relative displacement of ions, and partial alignment of permanent molecular electric dipoles can all contribute. Thermal agitation opposes the last mechanism.

For a weak field in a linear isotropic dielectric, the constitutive law isHere ${\chi }_e$ is the electric susceptibility, $ϵ$ is the permittivity, and $\mathbf{D}$ is the electric displacement field. In an anisotropic material, ${\chi }_e$ is a tensor and $\mathbf{P}$ need not be parallel to $\mathbf{E}$. These response laws describe the macroscopic field, not necessarily the microscopic field acting on one molecule.

Spatial variation and surface termination of the polarization density give bound charge and bound current through the bound charge densitieswhere $\hat{\mathbf{n}}$ is outward from the material. Thus uniform polarization density has no bulk bound charge, but generally has opposite bound charges on opposing surfaces. Gauss's law can be rewritten asWith no free surface charge, the normal component of $\mathbf{D}$ is continuous across an interface; the tangential electric field is continuous in electrostatics.

For example, fill a parallel-plate capacitor of plate area $A$ and gap $d$ with a uniform dielectric, neglecting fringing. At fixed free plate charge $Q$, $D=Q/A$, soThe induced bound surface charges oppose the original field, reducing the electric field and potential difference at fixed $Q$. At fixed applied potential difference, the field remains $V/d$ and the battery instead supplies more free charge. The electric displacement field separates these free-charge and bound-charge roles.

The closed surface integral sums the outward electric flux: $\mathbf{E}$ is the electric field, $d\mathbf{S}$ its outward vector area, $Q$ the total enclosed charge and ${ϵ}_0$ the vacuum permittivity. Gauss's law relates that flux to enclosed charge for any closed surface. High symmetry can make the field's magnitude constant on a suitable Gaussian surface and fix its direction; without symmetry, flux alone does not determine the local field.

For an isolated conducting sphere, charges reside at its surface and the exterior field is radial. ThusThe conductor has zero interior field. Integrating exterior field energy givesLet the initial charge per drop be $q=4\pi {ϵ}_{0}a{V}_1$. After disconnecting the charging supply, charge is conserved. Volume conservation gives final radius $A=2^{1/3}a$, total charge $2q$ andInitial mutual energy is negligible at the stipulated large separation. HenceThe electrostatic energy increase on equal-drop coalescence is positive. It can be supplied by the decrease in surface energy during coalescence and/or by external mechanical work bringing like charges together. Remaining energy can become motion and heat. If the charge is too large, surface-energy release alone need not suffice to make merging spontaneous; conservation does not assert that it will. Keeping the drops connected to a voltage supply would instead be a different charge-balance problem.

Gauss's law statesThe surface is closed and oriented outward, and $Q_{\mathrm{enc}}$ includes all enclosed charge. The integral law follows from the local form through the divergence theorem. It is universally valid in classical electromagnetism, but calculating the field directly from the flux requires enough symmetry.

For spherical charge symmetry, $E(r)=Q(<r)/(4\pi {ϵ}_0{r}^2)$. A uniform infinite line gives $E=\lambda /(2\pi {ϵ}_{0}r)$ from a coaxial cylinder. An infinite sheet gives $E=\sigma /(2{ϵ}_0)$ on either side from a pillbox, with the direction set by charge sign. The normal field jump is $\sigma /{ϵ}_0$; the electrostatic field inside a conductor is zero. Zero enclosed charge implies zero net flux, not generally zero field. ^[ref]

In electrostatic equilibrium a conductor has $\mathbf{E}=0$ throughout its bulk and constant electric potential on each connected component. Otherwise free charge would continue moving. Gauss's law then excludes bulk excess charge; it resides on surfaces. Just outside a conductor the tangential field vanishes and the normal jump isfor a vacuum interface with outward normal $\hat{\mathbf{n}}$. Surface charge rearranges in an external field to enforce these boundary conditions. Greater curvature often concentrates charge, but the actual distribution is a global boundary-value problem.

An empty closed cavity inside a conductor has constant potential and zero field, even in an external electrostatic field. If a charge $q$ is placed in the cavity, a Gaussian surface in the metal shows that the cavity wall carries total induced charge $-q$; its detailed distribution depends on geometry. An initially neutral isolated conductor then has $+q$ on its exterior surface. A grounded conductor can exchange charge with Earth, so this isolation condition no longer applies.

A useful example is a charge $q$ at height $d$ above an infinite grounded plane. In the physical half-space, the method of images replaces the induced charges by an image $-q$ at height $-d$, giving the attractive force $q^2/(16\pi {ϵ}_0{d}^2)$. The image is a mathematical construction, not an actual charge behind the conductor. Equipotential surfaces and charge conservation determine conductor fields together.

Spherical symmetry makes the electric field radial and constant in magnitude on each concentric Gaussian surface. The enclosed charge isGauss's law therefore givesThe direction is outward for $a>0$ and inward for $a<0$; the field vanishes at the centre. No exterior boundary condition changes this spherically symmetric interior result.

An inverse-square point-source field has constant flux through enclosing spheres because their area grows as $r^2$. In electrostatics, Gauss's law is ${\oint }_S\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{\epsilon }_0$, or $\nabla \cdot \mathbf{E}={\rho }_e/{\epsilon }_0$. For a spherical charge distribution, rotational symmetry makes the exterior field radial and constant in magnitude on a concentric sphere, giving $\mathbf{E}=Q\hat{r}/(4\pi {\epsilon }_0{r}^2)$. Symmetry is needed to take the field outside the surface integral; Gauss's law alone does not make an arbitrary charge distribution spherical.

For gravity the corresponding relations are ${\oint }_S\mathbf{g}\cdot d\mathbf{S}=-4\pi G{M}_{\mathrm{enc}}$ and $\nabla \cdot \mathbf{g}=-4\pi G{\rho }_m$. A spherical mass has exterior field $\mathbf{g}=-GM\hat{r}/r^2$. Inside a uniform sphere, $M_{\mathrm{enc}}(r)=4\pi {\rho }_m{r}^3/3$, so $\boxed{\mathbf{g}=-(4\pi G{\rho }_m/3)r\hat{r}}$. Inside an empty spherical shell the enclosed mass is zero and symmetry gives zero field. These examples distinguish an inverse-square exterior field from a field inside distributed matter.

The law also treats other symmetries, but their field-distance dependence differs: an infinitely long line charge has cylindrical rather than spherical spreading and a $1/r$ field. Flux conservation and geometry must both be used, rather than assuming every Gaussian surface leads to an inverse-square answer.

In SI units, with microscopic charge density ${\rho }_e$ and current density $\mathbf{J}$, Maxwell's equations can be writtenThe first relates electric flux to charge. The second states that the net magnetic flux through a closed surface is zero. Faraday's law relates electric circulation to changing magnetic flux. The Ampère-Maxwell equation includes both conduction current and displacement current; the latter is essential to time-dependent fields and local charge conservation.

For fixed contours and spanning surfaces, the integral curl equations are $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$ and $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enc}}+{\mu }_0{\epsilon }_{0}d{\Phi }_E/dt$. For a moving material contour the first must include the motional force term; its total emf is $\oint (\mathbf{E}+{\mathbf{v}}_{\mathrm{loop}}\times \mathbf{B})\cdot d\mathbf{l}$. This prevents confusing a changing circuit area with a partial time derivative of a static magnetic field.

Taking the divergence of the Ampère-Maxwell equation and using electric Gauss's law gives ${\partial }_t{\rho }_e+\nabla \cdot \mathbf{J}=0$. In a source-free vacuum, combining the two curl equations gives ${\nabla }^2\mathbf{E}-{\mu }_0{\epsilon }_0{\partial }_t^2\mathbf{E}=0$, and similarly for $\mathbf{B}$, predicting electromagnetic waves of speed $c=1/\sqrt{{\mu }_0{\epsilon }_0}$. The Lorentz force law then specifies how these fields act on a charge. Macroscopic matter descriptions may instead use $\mathbf{D}$ and $\mathbf{H}$ with free sources and constitutive relations; those should not be silently mixed with the microscopic-source equations above.

In electrostatics, Gauss's law isThe surface normal points outward. Spherical symmetry makes the field radial and constant in magnitude on a concentric Gaussian surface.

For a uniformly charged sphere, enclosed charge is $4\pi {\rho }_0{r}^3/3$ inside and $Q=4\pi {\rho }_0{R}^3/3$ outside. Thus the radial field isSet potential zero at infinity. Outside, $V=Q/(4\pi {ϵ}_{0}r)$. Inside, integrate $dV/dr=-E_r$ from the known boundary value, obtainingThe field rises linearly from zero to its maximum at $R$, then falls as $r^{-2}$. The potential falls quadratically from ${\rho }_0{R}^2/(2{ϵ}_0)$ to its surface value and then as $r^{-1}$. Both are continuous at the surface; there is no separate sheet of surface charge.

For the electric field of a hydrogen ground-state electron cloud, its enclosed charge isWriting $u=2s/a_0$ and using $\int u^2{e}^{-u}du=-e^{-u}(u^2+2u+2)$ givesThe total electron-cloud charge tends to $-e$, as required. Dividing by $4\pi {ϵ}_0{r}^2$ yieldsThe expression is negative and therefore points inward. Its apparently singular terms cancel at small radius, leaving $E_{e,r}\simeq -er/(3\pi {ϵ}_0{a}_0^3)$.

A distant external charge polarizes the neutral atom. Its electric field displaces the positive and negative charge distributions relative to one another, giving an induced dipole $\mathbf{p}=\alpha {\mathbf{E}}_{\mathrm{ext}}$ with static electric polarizability $\alpha >0$. The field is stronger on the nearer side, so the opposite-charge attraction exceeds the like-charge repulsion. This is charge-induced dipole attraction, not a net monopole charge on the atom.

Quantitatively, for external charge $Q=\pm e$ at separation $r$, the polarization energy is $U=-\alpha E_{\mathrm{ext}}^2/2$, henceHere negative $F_r$ means decreasing separation. Reversing the external charge reverses the induced dipole but leaves this leading long-range force attractive and of the same magnitude. The factor one half accounts for the energy cost of inducing the dipole.

The supplied charge density describes the unperturbed atom. Adding its electron field to the nucleus field leaves $E_{\mathrm{atom}}=e{e}^{-2r/a_0}(r^{-2}+2/(a_{0}r)+2/a_0^2)/(4\pi {ϵ}_0)$, an exponentially small residual because the cloud has an infinite tail. A completely frozen density would give only that tiny sign-dependent force. Allowing polarization is essential to obtain the dominant attractive power-law force for either sign of external charge when $r\gg a_0$.

In SI units, the microscopic Maxwell's equations for total charge density $\rho$ and current density $\mathbf{J}$ areThe first is Gauss's law: electric charge sources electric flux. The second is Gauss's law for magnetism, expressing the absence of magnetic monopole sources in this theory. Faraday's law relates changing magnetic flux to circulating electric field. The Ampere-Maxwell law relates magnetic circulation to both material current and changing electric flux.

For fixed loops and surfaces, the integral versions areSurface and loop orientations are related by the right-hand rule. These forms follow from the divergence theorem and Stokes' theorem. In material media, equivalent macroscopic equations can separate free and bound sources using $\mathbf{D}$ and $\mathbf{H}$ together with constitutive relations; the displayed forms use total sources.

The displacement-current term is essential for charge conservation. Taking the divergence of the last differential equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot \mathbf{J}=0$. It also makes magnetic circulation consistent for surfaces spanning the same loop through either a charging capacitor's wire or its dielectric gap.

In source-free vacuum the curl equations combine to give ${\nabla }^2\mathbf{E}={\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$, with the same equation for $\mathbf{B}$. ThusA plane wave is transverse, with $\mathbf{E}$, $\mathbf{B}$ and propagation direction mutually perpendicular and $E=cB$. The fields transport energy with density $u={ϵ}_0{E}^2/2+B^2/(2{\mu }_0)$ and Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$. Together with the Lorentz force law, the equations connect charges, currents, forces, induction and electromagnetic radiation.

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 charge per length ${\lambda }_q$, Gauss's law gives $E_r={\lambda }_q/(2\pi {ϵ}_{0}r)$ between the conductors. Integration gives $V={\lambda }_q\ln (b/a)/(2\pi {ϵ}_0)$. Similarly Ampère's law gives $B_{\varphi }={\mu }_{0}I/(2\pi r)$; integrating the magnetic energy per length and setting it equal to $L{I}^2/2$ givesHere $C,L$ are per unit length. The displayed $L$ is the external TEM inductance, appropriate to the ideal surface-current/high-frequency model. Internal magnetic energy in conductors with appreciable current penetration adds internal inductance; it cannot be determined fully from an outer conductor with unspecified wall geometry.

The distributed series resistance and inductance, with shunt capacitance and negligible leakage, give the telegrapher's equationsDifferentiate the first with respect to $x$ and substitute the second to obtainFor $V\propto e^{ikx-i\omega t}$ the forward attenuating branch has $k^2={\omega }^{2}LC+i\omega RC$. When $R\ll \omega L$,The positive imaginary part gives amplitude $e^{-x/{\ell }_{\mathrm{amp}}}$. Power decays twice as fast in its exponent, so its attenuation length is ${\ell }_{\mathrm{amp}}/2$. Finite conductivity changes the speed only at higher order in this constant-parameter weak-loss expansion, apart from any separately retained internal inductance or dielectric response.