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Treat the top gate as a parallel-plate capacitor, neglecting fringing and assuming its voltage change couples to the 2DEG without a dominant trap or series-capacitance correction. Its capacitance per area and induced electron density areIn this top-gate density tuning, a positive gate change increases the electron density in this model, giving $n_s^{\prime }=4.43371\times 10^{15}{\mathrm{m}}^{-2}$. Since $B_j\propto k_F\propto \sqrt{n_s}$,The fields rise by a factor $\sqrt{n_s^{\prime }/n_s}=1.2163$, not by the density ratio itself. Finite quantum capacitance is a small correction here: the one-subband value $e^2{m}^{\ast }/(\pi {\hslash }^2)$ is much larger than the stated geometric capacitance per area.

After field electron emission, let $h$ denote the plate separation of the parallel-plate capacitor, which the question does not specify. The uniform capacitor electric field has magnitude $E_c=|V|/h$, and a nonrelativistic electron of mass $m_e$ has constant acceleration magnitude $a=e{E}_c/m_e$. The sign of its charge affects the direction of the radiation field but not its radiated power.

In the supplied nonrelativistic Liénard-Wiechert field, $\dot{\mathbf{\beta }}=\mathbf{a}/c$. At polar angle $\theta$ to the acceleration, $|\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})|=a\sin \theta$, henceThe outgoing Poynting vector has magnitude $S={ϵ}_{0}c{E}_{\mathrm{rad}}^2$. Its radiated power per solid angle is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ gives the Larmor formula and the requested constant power during the uniform-field part of the flight:The brackets in the given field require the acceleration at retarded time; the radiation reaches the far observer after the propagation delay. This derivation assumes that radiation reaction is a small correction to the acceleration and that the flight is nonrelativistic, for example $e|V|\ll m_e{c}^2$ for release from rest. Entry, emission and impact transients are not described by the uniform-field model. A result solely in terms of $V$ cannot be obtained without a plate spacing or a specified internal field.

Take $x\in [-s/2,s/2]$ along the tilt, with local separation $h(x)=a+2\delta x/s$. Since $s\gg a$ and the tilt is small, each narrow strip is a local parallel-plate capacitor at the same voltage $V$. Its surface charge densities are approximatelyParallel strip capacitances add, so for $|\delta |<a$,Using $\log [(1+z)/(1-z)]=2(z+z^3/3+z^5/5+\cdots )$ givesThe denser charge at the smaller gap explains why tilting increases the capacitance. The local-strip approximation neglects fringing and slope corrections of order $(\delta /s{)}^2$, smaller than the retained $(\delta /a{)}^2$ term.

For a charging parallel-plate capacitor of area $A$, neglect fringing. The field in the gap is $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isAn Amperian loop around the feed wire can bound a surface cutting the wire, which has conduction current $I$, or a bulging surface passing through the dielectric gap, which has no conduction current. Including displacement current makes both surfaces give the same magnetic circulation. It is also required by charge conservation: the divergence of the Ampère-Maxwell equation and Gauss's law yields $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the continuity equation.

Neglect fringing field near the edges of the parallel-plate capacitor. Put the plates at $z=0,d$, with prescribed electric potentials $V_A,V_B$. Translational symmetry parallel to a plane makes the electrostatic potential independent of the coordinates parallel to the plates. Thus Laplace's equation reduces to $d^{2}V/d{z}^2=0$. Its general solution is $V=A+Bz$, and the two boundary conditions determine both constants:The electric field is uniform and points toward decreasing electric potential. No plate voltages are specified, so a numerical electric potential cannot be fixed; choosing one plate as the reference only fixes the arbitrary additive constant.

Use the parallel-plate capacitor approximation, neglecting the fringing field of a capacitor. With plate area $A=\pi (0.10\mathrm{m}{)}^2$, the initial capacitance and opposite plate electric charges areAfter the electric battery is disconnected, the isolated plate electric charges stay fixed. Doubling the separation halves the capacitance. The quasistatic pulling work is the increase of electrostatic energy:The final electric potential difference is thereforeAn electron released at rest gains kinetic energy $e\Delta V_f$, where $e$ is the positive elementary charge. Neglecting radiation and the negligible perturbation caused by this one electron, conservation of energy givesThis is $0.0280c$, so the nonrelativistic kinetic energy is adequate. The circular plates have finite radii; the numerical results are leading parallel-plate capacitor values rather than exact finite-plate capacitances.

The continuity equation for electric charge is $\nabla \cdot J=-{\partial }_t\rho$. Taking the divergence of the magnetostatic Ampère's law would instead require $\nabla \cdot J=0$. The displacement current repairs this conflict: using Gauss's law, $\nabla \cdot ({ϵ}_0{\partial }_{t}E)={\partial }_t\rho$, so the Ampère-Maxwell equation isIn the parallel-plate capacitor, neglect fringing fields and take $E_z=V_0{e}^{-i\omega t}/d$. A circular Ampère loop of radius $s<a$ encloses displacement current $-i\omega {ϵ}_0\pi s^2{V}_0{e}^{-i\omega t}/d$. Therefore the magnetic field between circular capacitor plates has phasor and amplitudeIt is zero on the axis, grows linearly with radius, and is azimuthal by the right-hand rule. This is a quasistatic approximation: in addition to $d\ll a$, the uniform-field model requires $\omega a/c\ll 1$ and excludes the edge region. The answer gives the oscillation amplitude, rather than a time-independent instantaneous magnitude.

In vacuum, Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:Take opposite electric charges $\pm Q$ on the two electrical conductors, with surface charge densities $\pm \sigma =\pm Q/A$. A Gaussian pillbox crossing one isolated large sheet gives $2ES=\sigma S/{ϵ}_0$, so each sheet contributes $\sigma /(2{ϵ}_0)$ on either side. By the superposition principle, the electric fields add between the sheets and cancel outside. Ignoring the fringing field of a capacitor, the gap electric field is $E=\sigma /{ϵ}_0$. The electric potential difference and capacitance of this parallel-plate capacitor are consequentlyThe approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with permittivity $ϵ$ replaces ${ϵ}_0$ by $ϵ$.

Put electric charge $+Q$ on the inner electrical conductor and $-Q$ on the outer one. In the gap, spherical symmetry makes the electric field radial and constant on a spherical Gaussian surface. Gauss's law givesThe inner-to-outer electric potential difference is thereforeThus the capacitance of the spherical capacitor isAs $b\to \infty$ this gives the capacitance of an isolated conducting sphere, $4\pi {ϵ}_{0}a$; for $b-a\ll a$ it approaches the parallel-plate capacitor result with area $4\pi a^2$.

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]

The quoted capacitance is that of the filled parallel-plate capacitor, $C_i=1.0\times 10^{-6}\mathrm{F}$. Disconnecting the source imposes charge conservation on the isolated plates:On removing the dielectric, the capacitance falls by its relative permittivity, so $C_f=C_i/5=2.0\times 10^{-7}\mathrm{F}$. The capacitor energy at fixed electric charge is $U=Q^2/(2C)$. ThereforeandThe increase is supplied by the external work needed to withdraw the dielectric against its attraction into the electric field, in the lossless quasistatic limit. The final electric potential difference is $Q/C_f=12500\mathrm{V}$, not the original charging value.

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 breakdown-limited parallel-plate charge follows from the ideal parallel-plate capacitor relation $Q=CV=({ϵ}_{0}A/d)(Ed)={ϵ}_{0}AE$. Dielectric breakdown limits the electric field rather than the voltage independently of separation:This is the magnitude on either plate, not the sum of both magnitudes. Edge enhancement and fringing are neglected, as in the stated parallel-plate model.

In vacuum the electric and magnetic field energy densities areFor example, charging a capacitor requires $U={\int }_0^Q(q/C)dq=Q^2/(2C)=C{V}^2/2$. A parallel-plate capacitor with $C={ϵ}_{0}A/d$ and $E=V/d$ then gives $U={ϵ}_0{E}^{2}Ad/2$, identifying the electric field energy density. Similarly an inductor stores $L{I}^2/2$; applying this to a long solenoid yields $B^2/(2{\mu }_0)$ times its interior volume.

Electromagnetic energy is transported by the Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$. Poynting's theorem,expresses local energy conservation and transfer to matter. In a vacuum plane wave $E=cB$, so the two energy densities are equal and the energy flux has magnitude $c(u_E+u_B)$. For nondispersive linear media the corresponding reversible expressions are $\mathbf{E}\cdot \mathbf{D}/2$ and $\mathbf{B}\cdot \mathbf{H}/2$; they must not be applied unchanged to arbitrary dispersive or hysteretic materials.

For an ideal air-gap parallel-plate capacitor, neglecting fringing, the electric field between oppositely charged plates is $E=\sigma /{ϵ}_0=Q/({ϵ}_{0}A)$. Dielectric breakdown therefore limits the magnitude of the electric charge on either plate toThe two plates carry $+Q$ and $-Q$; the quoted stored electric charge is the magnitude on one plate, rather than the net electric charge of the whole capacitor. The gap does not enter this ideal bound: $C={ϵ}_{0}A/d$ and $V_{\max }=E_{\mathrm{bd}}d$ make $C{V}_{\max }$ independent of $d$. Edge enhancement of the electric field can lower the practical limit.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

Permittivity describes how an electric field and a material's electric polarization combine into the electric displacement field. In a static, local response with linearity and isotropy, and no electric polarization at zero electric field, the constitutive equation isHere ${ϵ}_0$ is vacuum permittivity, while ${ϵ}_r$ is the relative permittivity, a dimensionless quantity. The SI unit of permittivity is farad per metre, $F/m$: multiplying it by an electric field in $V/m$ gives an electric displacement field in $C/{\mathrm{m}}^2$. A larger positive permittivity lets the same parallel-plate capacitor hold more plate electric charge at the same electric potential difference.

The material contribution is electric polarization $\mathbf{P}$, the electric dipole moment per unit volume. The electric displacement field and electric susceptibility ${\chi }_e$ satisfySubstitution gives the first constitutive equation. The electric field here is the averaged, total field inside the material, not just the field of the external plate electric charges. The continuum approximation averages over many chemical atoms or molecules; locality means that the electric polarization at a position depends on the electric field there. Static equilibrium, fixed material conditions and sufficiently small electric fields are needed for this scalar proportionality. ^[ref].

For an ideal parallel-plate capacitor completely filled by a uniform dielectric, hold the free plate electric charge magnitude $Q$ fixed and neglect the fringing field of a capacitor. Writing $A$ for the plate area, Gauss's law gives $D=Q/A$. ThusThe electric dipoles point along the electric field. Their surface electric charges, described by polarization density, oppose the field from the free plate electric charges. For a hypothetical dielectric with ${ϵ}_r=4$, the internal electric field is one quarter of its vacuum value, while $P=3D/4$. At fixed electric potential difference instead, the internal electric field stays approximately $V/d$ and the plate electric charge grows fourfold. These are different experimental constraints. ^[ref].

Permittivity is a useful material summary, not a universal constant of the material under every condition. Temperature, frequency and electric field strength can change the response. In a linear anisotropic dielectric, permittivity is a tensor, $D_i={\sum }_j{ϵ}_{ij}{E}_j$, so the electric displacement field need not be parallel to the electric field. At nonzero frequency, delayed electric polarization requires complex permittivity and may produce dielectric loss. ^[ref].

An approximation that keeps slow electromagnetic time variation while treating propagation across characteristic size $L$ as rapid compared with the variation time $1/\omega$. The dimensionless parameter $\omega L/c$ measures that separation. Which terms of the Maxwell equations can be dropped still depends on the physical scaling; small propagation delay alone does not justify deleting both displacement current and electromagnetic induction. For a slowly driven parallel-plate capacitor, a nearly uniform gap electric field can coexist with the magnetic field generated by its changing electric flux.

For a static rest-frame electromagnetic stress-energy tensor with no Poynting flux, a Lorentz transformation to a laboratory where it moves along $+x$ gives $u={\gamma }^2(u_0+{\beta }^2{T}_{\ast }^{xx})$ and $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})$. Thus energy flux includes stress as well as rest energy density. In an ideal parallel-plate capacitor, $T_{\ast }^{xx}=+u_0$ for motion perpendicular to the electric field, but $T_{\ast }^{xx}=-u_0$ for motion parallel to it. The former has $S_x=2{\gamma }^{2}v{u}_0$; the latter has zero interior flux, with work at the moving plates transferring energy at the boundaries. The field subsystem need not have an integrated four-momentum independent of simultaneity surface; include material support stresses for a closed system.