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For cold, collisionless, unmagnetized electrons with stationary heavy ions, $-m_e{\omega }^2\mathbf{x}=-e\mathbf{E}$. Therefore $\mathbf{P}=-ne\mathbf{x}=-n{e}^2\mathbf{E}/(m_e{\omega }^2)$ andThis is the cold collisionless plasma dielectric response. For a propagating transverse wave, $\mathbf{B}=\mathbf{k}\times \mathbf{E}/\omega$, so $E/B=c$ in vacuum and $E/B=c/n_p$ in plasma, where $n_p=\sqrt{{ϵ}_r}$. Equivalently $E/H=Z_0$ and $Z_0/n_p$ for $\mu ={\mu }_0$.

Converting the density gives $n=10^{18}{\mathrm{m}}^{-3}$, so ${\omega }_p=5.64\times 10^{10}{\mathrm{s}}^{-1}$ and $f_p=8.98\mathrm{GHz}$. At $10\mathrm{GHz}$,At an interface without independently imposed surface charge or sheet current, tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{B},\mathbf{D}$ are continuous, with the electron response included in the effective permittivity. Normal $E$ need not be continuous; the induced electron charge is represented by that response. In a convention treating it explicitly as a surface charge, the normal vacuum-displacement jump equals that charge instead.

For the downward transmitted wave, transversality gives $k_x{E}_x+k_z{E}_z=0$. The measured field therefore has $k_x/(-k_z)=3$, so $\tan {\theta }_t=3$ and $\sin {\theta }_t=3/\sqrt{10}$. Tangential phase matching givesFor inferring incident power from a transmitted plasma field, choose ${\mathbf{E}}_i=E_i(\cos {\theta }_i,0,\sin {\theta }_i)$ and ${\mathbf{E}}_r=E_r(\cos {\theta }_i,0,-\sin {\theta }_i)$. Their magnetic fields have opposite $y$ directions. The two tangential conditions areHence $E_i\simeq 1.2464\mathrm{V}{\mathrm{m}}^{-1}$ and $E_r\simeq -0.1458\mathrm{V}{\mathrm{m}}^{-1}$. Since these are peak phasor amplitudes,This is incident power per area perpendicular to its propagation. Per area of the interface the incoming normal flux is $⟨S_i⟩\cos {\theta }_i\simeq 1.87\times 10^{-3}\mathrm{W}{\mathrm{m}}^{-2}$. The reflected sign is a phase reversal, not negative reflected energy.

At $1\mathrm{GHz}<f_p$, the relative permittivity is negative and there is no propagating transverse bulk wave. The field decays evanescently into the plasma and the ideal collisionless half-space reflects all incoming normal power. Collisions would allow absorption; a complex refractive index here does not by itself imply dissipative heating.

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.

Find the potential from $\mathbf{E}=-\nabla \varphi$ and $\nabla \cdot (ϵ\nabla \varphi )=-{\rho }_f$. One must specify boundary data: a potential on each Dirichlet boundary, a normal displacement on each Neumann boundary, or an appropriate mixture. Conductors are equipotentials; at dielectric interfaces $\varphi$ is continuous and the normal displacement jumps by the specified free surface charge. With no singular dipole sheet, tangential electric field is continuous.

To establish uniqueness of Poisson equation, subtract two solutions with the same charge and boundary data, writing their difference as $u$. Integration by parts givessince $\nabla \cdot (ϵ\nabla u)=0$. The right side vanishes for identical Dirichlet data, identical Neumann data, or their usual mixed combination. Interface terms cancel under the homogeneous difference conditions. With positive permittivity, $\nabla u=0$: the field is unique, and the potential is unique if any suitable reference potential is fixed. For a pure Neumann problem an additive constant remains, and the prescribed outward displacement flux must equal the enclosed free charge. This theorem justifies image-charge solutions once their boundary data have been checked.

For a smooth isolated conducting sphere, Gauss's law gives the surface electric field $E_s=Q/(4\pi {ϵ}_0{a}^2)$ and electric potential $V=Q/(4\pi {ϵ}_{0}a)=a{E}_s$. At the specified dielectric strength, $V_{\max }=3.0\times 10^3\mathrm{V}$. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}a$, so its capacitor energy isThis is the ideal breakdown-limited energy of a conducting sphere, taking air's permittivity approximately equal to vacuum permittivity and neglecting nearby objects and surface irregularities. The vacuum permittivity used numerically is $8.8542\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$.^[ref]

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.

For a linear nondispersive medium with real absolute permittivity ${ϵ}_a=ϵ{ϵ}_0$ and permeability ${\mu }_0$, instantaneous field energy density and energy flux arewith local balance ${\partial }_{t}u+\nabla \cdot \mathbf{N}=-\mathbf{J}\cdot \mathbf{E}$. energy stored locally and energy passing through a surface are distinct: integrating the balance relates their rates and the power given to charges.

Between the coaxial conductors, cylindrical Gauss and Ampère laws giveThe inner line charge is $\lambda =2\pi {ϵ}_{a}V/\ln (b/a)$, so the capacitance per length is $C=2\pi {ϵ}_a/\ln (b/a)$. Integrating magnetic energy over the annulus gives $\int B^2(2\pi rdr)/(2{\mu }_0)={\mu }_0{I}^2\ln (b/a)/(4\pi )=L{I}^2/2$, hence $L={\mu }_0\ln (b/a)/(2\pi )$. ThusThese are per-length parameters for ideal conductors, excluding internal magnetic penetration.

Now let the voltage and current amplitudes be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular dielectric. The two nontrivial curl equations reduce toConsequently both $k^2={\omega }^2{\mu }_0{ϵ}_a$ and $V_0=Z_0{I}_0$ are required for a forward wave. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy Maxwell's equations. This TEM travelling-wave voltage-current relation and the dispersion constraint identify the transverse electromagnetic mode and its speed $v_p=1/\sqrt{LC}$.

For the travelling wave, $⟨N_z⟩=\mathrm{Re}(V_0{I}_0^{\ast })/[4\pi r^2\ln (b/a)]$. Integrating over the dielectric annulus yieldsThis is the familiar circuit power, carried here through the dielectric rather than through the ideal metal. In root mean square notation it is $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ for the forward lossless wave; the corresponding static fields give $VI$ without a time-average factor. The average line energy is $(C|V_0{|}^2+L|I_0{|}^2)/4=C|V_0{|}^2/2$, whose product with wave speed agrees with the power.

Finite conductor resistance requires a small longitudinal electric field to drive current. Its cross product with the azimuthal magnetic field produces a radial Poynting component into each conductor: inward radially at the inner conductor and outward radially into the outer conductor. This Poynting flux into lossy coaxial conductors explains why forward power decreases along the cable as energy becomes heat in the metal. The mode is then only approximately TEM; strictly zero longitudinal electric field would be inconsistent with resistive current.

An electrostatic field calculation begins by specifying the charge information, material response and boundary conditions. In vacuum, prescribed volume and surface electric charge densities can be superposed directly using Coulomb's law. For a volume distribution,Differentiating this electric potential gives the electric field. This is useful for known source distributions, but in the presence of conductors the induced surface electric charge is usually an unknown rather than an independently specified input.

Gauss's law is especially effective when symmetry fixes the direction and magnitude of the electric field on a suitable closed surface. A spherical charge distribution gives a radial field, an infinitely long uniform line gives a cylindrical field, and a uniform infinite sheet gives a planar field. In an arbitrary geometry, the flux law remains true but does not alone determine the pointwise electric field. A multipole expansion is another useful approximation: far from a localized charge distribution, successive monopole, dipole and higher terms separate the effects of total charge and its spatial structure.

For a boundary-value formulation, use the electrostatic equationsThey imply Poisson equation ${\nabla }^2\varphi =-\rho /{ϵ}_0$, or Laplace equation where $\rho =0$. A conductor in electrostatic equilibrium is an equipotential with zero internal electric field. Its exterior normal electric field fixes the surface electric charge through $E_n=\sigma /{ϵ}_0$. One must specify whether each conductor is grounded, maintained at a prescribed voltage, or isolated with prescribed total electric charge.

Uniqueness of Poisson equation justifies methods that construct a convenient candidate solution. With fixed boundary electric potential and suitable behaviour at infinity, a solution satisfying the field equation is the solution. With normal-derivative boundary conditions, the source and flux must be compatible, and the electric potential is determined only up to an additive constant.

For example, a point charge $q$ a distance $h$ above a grounded infinite plane has the same upper-half-space electric potential as $q$ and a mirrored image charge $-q$ below it. Their potentials cancel on the plane. The image is a mathematical source in the excluded region, not a second physical charge. Separation of variables similarly constructs solutions suited to rectangular, cylindrical or spherical boundaries; matching coefficients enforces the prescribed boundary data.

In dielectrics, the distinction between total and free charge is essential. In a linear isotropic material,Only where permittivity is constant may this be reduced to ${\nabla }^2\varphi =-{\rho }_f/ϵ$. At an ordinary interface, $\varphi$ and the tangential electric field are continuous, while the jump in normal electric displacement field is the free surface charge. Applying Gauss's law to total charge with ${ϵ}_0$, or its displacement-field form to free charge, gives consistent descriptions; mixing their sources and permittivities does not.

A useful interface example is the image charge at a dielectric interface. If $q$ lies in medium 1, the source-side potential is represented bywhere $r_1$ is distance to the real charge and $r_2$ to its mirrored position. The other-side potential can be written ${\varphi }_2=q_t/(4\pi {ϵ}_2{r}_1)$, where $q_t=2{ϵ}_{2}q/({ϵ}_1+{ϵ}_2)$. Continuity of $\varphi$ and of normal $\mathbf{D}$ fixes both coefficients. The equal-permittivity limit gives no image; the conductor limit ${ϵ}_2\to \infty$ gives $q^{\prime }\to -q$.

For a sphere of radius $a$ and relative permittivity ${ϵ}_r$ in a uniform external electric field ${\mathbf{E}}_0$, matching the spherical solutions gives a uniform internal electric field and an external induced electric dipole moment:This illustrates how a material boundary modifies an otherwise simple applied electric field.

For complex boundaries or spatially varying materials, finite difference methods or the finite element method solve the same boundary value problem numerically. A sound calculation checks its boundary conditions, charge or flux balance and convergence with mesh refinement. The governing equations, source classification and uniqueness argument remain the organizing principles even when an analytic formula is unavailable.

Field equations. In a homogeneous source-free nonmagnetic medium, $\mathbf{D}=ϵ{\epsilon }_0\mathbf{E}$ and $\mathbf{B}={\mu }_0\mathbf{H}$. Maxwell's equations becomeThe first equation follows from $\nabla \cdot \mathbf{D}=0$ for spatially constant nonzero $ϵ$. When the effective relative permittivity is negative below, its charge response is already included in $\mathbf{D}$; it must not also be counted as an extra free source.

Amplitude relations and bulk dispersion. Work with complex phasors proportional to $e^{ikx-Kz-i\omega t}$. For $\mathbf{H}=(0,H_y,0)$, the Ampere-Maxwell law givesso the amplitudes must obeyThey are not arbitrary independent constants: the dispersion relation alone would not make arbitrary amplitudes satisfy the equations. These relations give $ik{E}_x-K{E}_z=0$, so the electric divergence vanishes; the magnetic divergence vanishes because there is no $y$ dependence. The only nonzero component of Faraday's law of induction isSubstitution then givesThis proves that all four field equations hold for the specified transverse magnetic polarization when both the amplitude relations and the dispersion relation are imposed.

Interface matching. Let region 1 be vacuum above the interface, with ${ϵ}_1=1$, $K={\beta }_1>0$; region 2 has ${ϵ}_2=ϵ$, $K={\beta }_2<0$. These signs make the evanescent fields decay as $z\to \pm \infty$. In the absence of free surface current, tangential $H_y$ is continuous, giving a common interface amplitude. Tangential $E_x$ is also continuous, henceNormal $D_z$ is continuous as well: both sides have $D_z=-k{H}_y/\omega$. This does not require $E_z$ itself to be continuous; the material can have polarization surface charge. The normal magnetic field is zero on both sides.

Existence and surface dispersion. The opposite signs of ${\beta }_1$ and ${\beta }_2$ require $ϵ<0$. The two bulk equations areUse ${\beta }_2^2={ϵ}^2{\beta }_1^2$ to eliminate the decay constants. For the nontrivial interface branch,A positive ${\beta }_1^2$ together with real propagating $k$ gives $ϵ<-1$. A negative permittivity between $-1$ and zero is insufficient. This lossless planar surface-plasmon dispersion describes a surface plasmon polariton; $ϵ=1$ would instead be identical media and supplies no decaying bound mode.

Frequency range and long wavelengths. With $ϵ=1-{\omega }_p^2/{\omega }^2$, the existence condition becomesThe upper value is a limiting frequency approached as $k\to \infty$, not an attained maximum at finite $k$. Substitution into the dispersion relation givesFor $\omega \ll {\omega }_p$, choosing propagation in the positive $x$ direction,It approaches the vacuum light line from larger $k$. The vacuum decay constant is then ${\beta }_1\simeq {\omega }^2/(c{\omega }_p)$, so confinement becomes weak at low frequency. This calculation assumes the stated real, lossless effective permittivity; dissipation would require complex propagation and decay constants.

Effective permittivity. Use peak phasors with time dependence $e^{-i\omega t}$. The Ampere-Maxwell law isThe conductive contribution to complex permittivity is thereforeHere ${ϵ}^{\prime }$ is the source's effective relative permittivity, not just the real part of a complex response. The positive imaginary term is the loss sign appropriate to this time convention.

The penetrating electric wave. Combine this equation with Faraday's law of induction. In a uniform source-free medium the transverse phasor satisfiesFor a good conductor of electrical conductivity $\sigma$, $\sigma \gg \omega ϵ{\epsilon }_0$, so $k^2\simeq i{\mu }_0\sigma \omega$. The root which decays into $z>0$ isChoose the surface phasor amplitude $E_0$ real. Taking the real part of ${\tilde{E}}_x=E_0{e}^{ikz}$ givesThe skin depth $\delta$ is an amplitude-decay distance in metres.

The magnetic wave and phase. The curl of an $x$-polarized field varying in $z$ is in the $y$ direction. Since $\nabla \times \tilde{\mathbf{E}}=i\omega {\mu }_0\tilde{\mathbf{H}}$,ThusThe phase sign follows the specified convention. Equivalently the surface impedance of a good conductor is $E_x/H_y=(1-i)/(\sigma \delta )$.

How the fields differ from those in vacuum. The two metal fields have the same exponential envelope and spatial period $2\pi \delta$, but their phase arguments differ by $\pi /4$.

An individual travelling wave in the lossless exterior has no attenuation, real wave impedance and in-phase electric and magnetic fields. The exterior total field also includes the reflected wave and may show interference, so it need not share that simple phase relation everywhere. At the interface the tangential fields match; $E_0$ is the field just inside the metal, not necessarily the incident amplitude. A good conductor reflects strongly and its wave impedance is much smaller than the vacuum impedance.

Average inward energy flux. For peak phasors, the Poynting vector averaged over a cycle is $⟨\mathbf{S}⟩=\mathrm{Re}(\tilde{\mathbf{E}}\times {\tilde{\mathbf{H}}}^{\ast })/2$. The fields giveConsequently the constant in the question isThe flux is in the $+z$ direction and has units $\mathrm{W}{\mathrm{m}}^{-2}$.

Local energy conservation. The Poynting flux and local heating in a skin layer satisfyThe decrease becomes Joule heating in each layer. The Poynting theorem includes this transfer from field energy to the metal; averaged electromagnetic energy storage has zero time derivative in the periodic state, so no energy is lost from the complete balance.

Distributed line parameters. The geometry is the wide parallel-plate transmission line described in the question, with absolute permittivity $\epsilon =ϵ{\epsilon }_0$. A length $dx$ has charge $CVdx$ and field $E=V/d$, givingThe opposed plate currents have sheet density $I/a$ and confine the leading magnetic field $H=I/a$ between the plates. The magnetic energy per unit length is ${\mu }_0{H}^{2}ad/2=L{I}^2/2$, soHere $C$ and $L$ are per unit length, in $\mathrm{F}{\mathrm{m}}^{-1}$ and $\mathrm{H}{\mathrm{m}}^{-1}$, not lumped component values. Internal conductor fields and edge corrections are neglected.

Voltage and current evolution. The current entering a short length minus that leaving it is the rate of increase of its plate charge:The voltage drop across that length is its inductive drop, $V(x)-V(x+dx)=Ldx{\partial }_{t}I$. Dividing by $dx$ gives the lossless Telegrapher's equations,

Propagation speed. Differentiate these equations once more and eliminate the other variable:The wave equation speed is

Travelling-wave impedance. For a forward wave $V=F(t-x/v)$, take $I=V/Z_0$. Substitution into ${\partial }_{x}I+C{\partial }_{t}V=0$ gives $1/(v{Z}_0)=C$, hence the characteristic impedanceThe backward-wave current has the opposite sign relative to its voltage; the fixed ratio above applies to one travelling direction.

Matching the antenna. The voltage reflection coefficient is $\Gamma =(Z_{\mathrm{load}}-Z_0)/(Z_{\mathrm{load}}+Z_0)$. No reflection requires $Z_0=5\Omega$. With $a=0.020\mathrm{m}$, $ϵ=1.5$, and $\sqrt{{\mu }_0/{\epsilon }_0}\simeq 376.7\Omega$,The resulting $d/a\simeq 0.016$ is consistent with the wide-plate approximation.

Both transverse forces. The matched transverse electromagnetic mode has electric attraction and magnetic repulsion. For time-averaged power $P=100\mathrm{kW}$,The force on one plate excludes its own field. Thus the average electric attraction and magnetic repulsion per unit length are respectivelyThe electric and magnetic pressure cancellation in a matched TEM line follows directly from $V_{\mathrm{rms}}=Z_0{I}_{\mathrm{rms}}$:The equality also holds instantaneously for a single travelling wave. Quoting either $0.628\mathrm{N}{\mathrm{m}}^{-1}$ alone would omit the opposing contribution. This is the ideal electromagnetic force in the stated lossless uniform-dielectric model; a standing wave would not generally have the same cancellation.

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 dielectric responds to an electric field mainly by displacing or orienting bound charges, rather than by sustaining a steady conduction current. The polarization density $\mathbf{P}$ is electric dipole moment per unit volume. It creates bound chargeSeparating these charges from free charge motivatesFor an isotropic linear dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, soThe electric susceptibility ${\chi }_e$ describes material polarization and is not the magnetic susceptibility. More generally it is a tensor and can depend on frequency, field strength and temperature.

Electronic polarization displaces an electron cloud relative to its nucleus; ionic polarization displaces positive and negative ions; orientational polarization partially aligns permanent molecular dipoles against thermal disorder. For independent permanent moments of magnitude $p$, weak-field orientational response gives ${\chi }_e\simeq n{p}^2/(3{ϵ}_0{k}_{B}T)$ when local-field corrections and interactions are negligible. This explains a common inverse-temperature contribution without asserting that every dielectric follows that law.

At a static interface, $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$ and the tangential electric field is continuous. Bound surface charge is already included through $\mathbf{P}$. For a homogeneous filled capacitor, $C=ϵA/d$, and a reversible linear dielectric stores energy density $u=\mathbf{E}\cdot \mathbf{D}/2$. At fixed voltage, insertion raises the stored field energy but the voltage source also supplies energy; force arguments must include that source. At fixed charge the capacitor energy instead decreases on insertion.

At frequencies too high for a polarization mechanism to follow the field, its contribution weakens. A complex permittivity represents dissipative dielectric loss as well as phase lag; the sign of its imaginary part depends on the time convention. Strong fields can cause nonlinear response or dielectric breakdown, beyond the reversible linear model.

Electromagnetic waves drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent refractive index is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.

Use phasors proportional to $e^{i(kz-\omega t)}$, with $k=k^{\prime }+i{k}^{\prime \prime }$ and $k^{\prime \prime }\ge 0$ for a passive medium. In a homogeneous, isotropic, linear material, let ${ϵ}_b$ represent the permittivity associated with bound-charge electric polarization, $\mu$ the magnetic permeability and $\sigma (\omega )$ the mobile-charge conductivity. The macroscopic Maxwell equations giveFor a transverse plane wave, taking another curl yieldsThis division of the response avoids counting mobile electrons both in ${ϵ}_b$ and in $\sigma$. Define the complex refractive index $\tilde{n}=ck/\omega =n+i\kappa$. For a nonmagnetic medium, ${\tilde{n}}^2={ϵ}_{\mathrm{eff}}/{ϵ}_0$. Its real part determines the phase variation, while the positive optical extinction coefficient $\kappa$ gives amplitude attenuation $e^{-\omega \kappa z/c}$ and intensity attenuation $e^{-2\omega \kappa z/c}$. The amplitude penetration length is

A simple microscopic explanation comes from the Drude model. Let electrons of charge $-e$, mass $m$ and number density $n_e$ move against an effectively stationary positive background. Collisions relax their mean velocity at rate $\nu$:ConsequentlyWith a vacuum-like bound-charge background, this givesIts imaginary part is positive for positive $\omega ,\nu$, consistent with dissipation under the chosen phasor convention. Real materials can require an effective electron mass, a nontrivial bound-charge response, or additional absorption associated with the electronic band structure.

A plasma is a conducting ionized medium with mobile electrons and ions, approximately neutral on large scales because of plasma screening. For the electron-frequency transverse waves considered here, the heavy ions can often be treated as stationary. In a dilute weakly collisional plasma, $\nu \ll \omega$, soFor $\omega >{\omega }_p$, $k$ is real in this collisionless model and the wave propagates. Its phase velocity is $v_{\mathrm{ph}}=c/n>c$, while its group velocity is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega <{\omega }_p$, the refractive-index square is negative andThe transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless plasma reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a plasma cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; plasmas also support longitudinal collective modes.

In a metal, positive ions usually form a crystal lattice. Mobile conduction electrons occupy states described by the electronic band structure and normally form a degenerate Fermi gas. The conduction-electron density is usually much higher than in a dilute gas plasma. Thus its plasma frequency is commonly in the optical or ultraviolet range, rather than the lower radio-frequency values possible in dilute plasmas. Electron scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every plasma is collisionless and every metal collisional: the controlling ratio for either material is $\omega /\nu$, together with its carrier density and bound-charge response.

At sufficiently low frequency, $\omega \ll \nu$, the metal's conductivity is approximately real, ${\sigma }_0=n_e{e}^2/(m\nu )$. In a good conductor the conduction current dominates the displacement current, ${\sigma }_0\gg \omega {ϵ}_b$. The wave equation then givesFor example, a tangential electric field varies asThe current $\mathbf{J}={\sigma }_0\mathbf{E}$ is therefore concentrated within a few skin depths of the surface: this is the skin effect. As frequency increases within this regime, $\delta \propto {\omega }^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.