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The effective electric charge associated with electric polarization is bound charge. For polarization density $\mathbf{P}$, its volume charge density is ${\rho }_{\mathrm{b}}=-\nabla \cdot \mathbf{P}$. At a boundary with unpolarized surroundings, its surface charge density is ${\sigma }_{\mathrm{b}}=\mathbf{P}\cdot \mathbf{n}$, where $\mathbf{n}$ is the outward unit normal vector.

The electric field produced by the bound charge of a polarized body is its depolarization field. For a uniformly polarized ellipsoid in vacuum, the internal depolarization field is uniform and opposes the relevant component of the electric polarization: $E_{\mathrm{dep},i}=-N_i{P}_i/{ϵ}_0$. Its dimensionless geometric factors obey $N_x+N_y+N_z=1$.

A state with a stationary electric charge distribution and no net drift of mobile charge driven by the electric field. A conductor in electrostatic equilibrium has zero interior electric field, while bound charges in an insulating material need not move to eliminate its interior field.

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.

A bound charge in a harmonic restoring potential has induced electric dipole moment $d=(q^2/k)E$. Completing the square proves this for every static field within the strictly harmonic model. First-order nondegenerate perturbation theory gives the same coefficient because the position operator connects the unperturbed ground state only to the first excited state. This exact response coefficient does not make its two-state wavefunction approximation exact.

A bound charge obeying $m\ddot{u}+m\gamma \dot{u}+m{\omega }_0^{2}u=qE$ gives a resonant electric susceptibility. For carrier density $n$ and the $e^{-i\omega t}$ convention,Positive damping gives $\mathrm{Im}{ϵ}_r\ge 0$ for $\omega >0$, expressing passive absorption. Independent resonances add to the electric susceptibility. Setting ${\omega }_0=0$ gives the Drude frequency response.

Both Thomson scattering and Rayleigh scattering can be understood as radiation from a dipole driven by the incident electric field. Their different frequency dependences come from the response of a free charge compared with a bound charge.

For a free particle of charge $q$ and mass $m$, a harmonic field gives $m\ddot{\mathbf{x}}=q\mathbf{E}$, hence $\mathbf{p}=q\mathbf{x}=-q^2\mathbf{E}/(m{\omega }^2)$. The radiated field contains two time derivatives of $\mathbf{p}$, cancelling the response's ${\omega }^{-2}$. The Thomson cross section is consequently independent of frequency in the classical low-recoil regime:For incident linear polarization with unit polarization $\mathbf{ϵ}$, summing over outgoing polarizations gives $d\sigma /d\Omega =r_q^2[1-|\hat{\mathbf{n}}\cdot \mathbf{ϵ}{|}^2]$. Radiation vanishes along the acceleration and is strongest perpendicular to it. Averaging an incident unpolarized light beam giveswhere $\Theta$ is the scattering angle. Scattering through $90^{\circ }$ is fully linearly polarized perpendicular to the scattering plane for this ideal isotropic dipole model. The electron case dominates over equally charged heavier particles because the cross section scales as $m^{-2}$. recoil corrections become necessary when photon energy is no longer small compared with $m{c}^2$.

For a bound charge with natural frequency ${\omega }_0$ and damping rate ${\gamma }_d$,Far below resonance, ${\alpha }_{\mathrm{e}}$ is approximately constant, soThis is Rayleigh scattering. It requires a scatterer small compared with the wavelength and, for the simple expression, an isotropic linear dipole response. Atoms, molecules and small dielectric particles provide examples. The stronger scattering of shorter visible wavelengths explains blue skylight; absorption and the nonuniform solar spectrum prevent treating that simple scaling as a complete color calculation.^[ref]

For isotropic electric scatterers the normalized angular and polarization patterns are the same as in Thomson scattering. The frequency scaling distinguishes their limits: a free charge has a displacement proportional to ${\omega }^{-2}$, whereas a tightly bound charge has an approximately frequency-independent displacement. Near resonance, far outside the small-particle limit, or with anisotropic polarizability, the simple Rayleigh formula must be modified. In particular, the electric-only pattern must not silently be applied to a perfectly conducting sphere whose magnetic dipole response is also retained.

A scattering measurement should subtract stray-light and detector backgrounds, control scatterer density and optical thickness, and calibrate detector response and incident power when comparing wavelengths. A polarization analyzer at $90^{\circ }$ tests the dipole pattern; multiple scattering or unresolved nonisotropic contributions can reduce its polarization.

Apply Faraday's law to a shrinking rectangular loop crossing the interface. For bounded fields the magnetic-flux term vanishes with its area, giving ${\mathbf{E}}_{t,+}={\mathbf{E}}_{t,-}$. A shrinking Gaussian pillbox similarly gives $D_{n,+}-D_{n,-}={\sigma }_{\mathrm{free}}$. Thus normal $D$ is continuous when no free surface charge is present, even though a bound polarization charge may remain. These are the assumptions needed for the stated dielectric-interface boundary conditions.

Write $ϵ={ϵ}_r{ϵ}_0$, $r_{+}=\sqrt{(x-a{)}^2+y^2+z^2}$ and $r_{-}=\sqrt{(x+a{)}^2+y^2+z^2}$. Use each region's own permittivity in its Coulomb potential:At $x=0$, continuity of potential and normal displacement givesSolving,This permittivity convention for transmitted image charges matters: if the vacuum Coulomb prefactor is used even for ${\varphi }_{-}$, its combined effective source is $2Q/(1+{ϵ}_r)$ instead. The physical potential is the same. The two numbered leaves identify how these sources satisfy the equations in their respective regions.

The real charge feels only the image contribution after removing its own singular self-field. The image at $(-a,0,0)$ is separated from it by $2a$, soFor the bound charge, the outward normal of the dielectric is $+\hat{x}$ and $\mathbf{P}={ϵ}_0({ϵ}_r-1){\mathbf{E}}_{-}$. Differentiating ${\varphi }_{-}$ at $x=0$ gives $E_{-,x}=-a(Q+Q^{\prime })/[4\pi ϵ(a^2+y^2+z^2{)}^{3/2}]$. ThereforeIts integral over the plane is $Q^{\prime \prime }$, consistent with the external image representation. For ${ϵ}_r=1$ both the image force and induced surface charge vanish.

Polarization density $\mathbf{P}$ is electric dipole moment per unit volume. Displacing positive and negative bound charges leaves the bulk nearly neutral locally but can produce uncompensated charge where polarization varies or terminates. For a dielectric occupying a volume with outward unit normal $\mathbf{n}$,The divergence theorem makes their total sum zero for a body described entirely by bound dipoles. Uniform polarization gives no interior volume charge but generally gives opposite surface charges on different faces.

In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}=ϵ\mathbf{E}$. Gauss's law becomes $\nabla \cdot \mathbf{D}={\rho }_f$, separating free from bound charge. Across a static interface, tangential electric field is continuous, while the normal displacement jumps by the free surface charge. Polarization charges alter the internal field, as in the dielectric cylinder calculation. Molecular distortion and orientation contribute differently, and response can depend on temperature and frequency. Bound charge is a redistribution of material charge, not an additional free-charge reservoir.

An applied electric field displaces bound charges or reorients permanent dipoles, producing electric polarization $\mathbf{P}$, the dipole moment per unit volume. It gives bound charges ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\sigma }_b=\mathbf{P}\cdot \mathbf{n}$. Definefor a linear isotropic dielectric. Then $\nabla \cdot \mathbf{D}={\rho }_f$ separates free charge from the bound-charge response. Electronic, ionic and orientational electric polarization respond on different timescales; real permittivity is frequency dependent, and its dissipative part describes dielectric loss.

A uniform dielectric filling a parallel-plate capacitor raises its capacitance to $ϵA/d$. At fixed charge this reduces field energy $Q^2/(2C)$; at fixed voltage the stored energy $C{V}^2/2$ increases because the voltage source supplies charge and energy. Insulation and energy storage are useful applications, limited by loss and dielectric breakdown. Static interface conditions are continuous tangential $\mathbf{E}$ and a jump in normal $\mathbf{D}$ equal to free surface charge.

Free charge is charge assigned to externally supplied or mobile carriers in a macroscopic description. Bound charge describes the net charge associated with a material's polarization density $\mathbf{P}$. Its bulk and surface densities arewhere the normal points out of the polarized material. Uniform bulk polarization density has no bulk bound charge, but it can still have surface bound charge.

The electric field is sourced by the total electric charge. Introducing the electric displacement field separates the free source in the macroscopic field equation:This does not mean that bound charges are physically absent from $\mathbf{D}$: they are represented through $\mathbf{P}$. In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}=ϵ\mathbf{E}$, with $ϵ={ϵ}_0(1+{\chi }_e)$.

Across a stationary ordinary dielectric interface, electrostatics gives continuity of tangential electric field, while a free sheet charge givesThe normal electric field need not be continuous. For a fully filled parallel-plate capacitor at fixed free plate charge, polarization reduces the internal electric field by ${ϵ}_r$ and increases capacitance by the same factor. If a battery instead holds the voltage fixed, extra free plate charge flows from the battery; the constraint must be stated before comparing stored energies.^[ref]

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.

A sphere with constant electric polarization $\mathbf{P}=P\hat{\mathbf{z}}$ has zero volume bound charge and surface bound charge density ${\sigma }_b=P\cos \theta$. Integrating $\mathbf{r}{\sigma }_{b}dA$ gives the displayed electric dipole moment. The surface distribution is purely degree one in spherical harmonics, so its external static electric potential is exactly dipolar. Its interior field is finite and is not that of a point singularity.