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In an unmagnetized plasma, an electron displacement driven by $\mathbf{E}{e}^{-i\omega t}$ obeys $-m_e{\omega }^2{\mathbf{x}}_e=-e\mathbf{E}$. Its dipole moment is $-e{\mathbf{x}}_e$, giving electric polarization $\mathbf{P}=-n{e}^2\mathbf{E}/(m_e{\omega }^2)$. With $\mathbf{D}={\epsilon }_0\mathbf{E}+\mathbf{P}$, this yields the displayed relative permittivity and ${\omega }_p^2=n{e}^2/(m_e{\epsilon }_0)$. Ions contribute a further negative term smaller by $Z{m}_e/m_i$ for one ion species of charge $Ze$. Neglecting it requires heavy ions at electron-wave frequencies; collisions, background magnetization and pressure dispersion are excluded. The transverse electromagnetic wave relation is $c^2{k}^2={\omega }^2-{\omega }_p^2$.

A dielectric is a material in which an electric field produces electric polarization while transport of electric charge is sufficiently small for the intended use. An insulating solid in a capacitor is an example; real dielectrics can still have dielectric loss and nonzero electrical conductivity.

The important distinction is between rearranging bound electric charge and transporting mobile charge carriers through the specimen. Bound electric charges can move a little within a chemical atom or molecule, or an existing electric dipole can turn. Their averaged electric dipole moments produce electric polarization even if there is negligible steady electric current between the electrodes. A changing electric polarization does give a bound current; the absence of steady conduction does not mean that the material is electrically unresponsive.^[ref]

For uniform electric polarization $\mathbf{P}=P\hat{\mathbf{x}}$ in a rectangular slab, polarization density gives zero bound volume charge density in the interior and opposite bound surface charge densities $-P,+P$ on the left and right faces. The electric charge has been redistributed, not created. More generally, the divergence theorem makes the cancellation explicit:Here $\mathbf{n}$ is the outward normal vector on the material boundary. This accounts for both bound volume charge density and bound surface charge density; omitting either can falsely suggest that an initially neutral dielectric acquired net electric charge.

An electrical insulator between capacitor plates supports a useful combination: appreciable electric polarization, little leakage electric current, and tolerance of the intended electric field. This lets a small parallel-plate capacitor store separated electric charge and recoverable capacitor energy. The comparison in dielectric capacitor response explains why specifying fixed electric charge or fixed electric potential difference changes the answer. High relative permittivity alone does not establish good insulation or a large usable energy density; dielectric loss and dielectric breakdown matter as well.

The motivating observation is that inserting an insulating material between fixed capacitor plates increases its capacitance. ^[ref]. To investigate this response, use two flat parallel metal electrodes with measured overlapping area $A$ and separation $d$, a uniform specimen filling the gap, and an instrument that measures electrical impedance. Record the capacitance with and without the specimen at the same small applied electric potential difference, frequency and temperature, after capacitance measurement calibration. An increase in corrected capacitance indicates a larger permittivity, provided the geometry is unchanged. A guard electrode reduces the contribution of the fringing field of a capacitor. ^[ref].

For negligible dielectric loss in the parallel-plate capacitor approximation,An air reference approximates $C_0$; it is not an exact vacuum reference. For the following synthetic example, stipulate $A=0.010{\mathrm{m}}^2$, $d=1.00\mathrm{mm}$ and the rounded vacuum permittivity ${ϵ}_0\approx 8.854\times 10^{-12}F/m$. Then $C_0=88.54\mathrm{pF}$ in this model, and relative permittivity four gives $C=354.16\mathrm{pF}$. A residual parallel capacitance of $20\mathrm{pF}$ would make an uncorrected sample/reference ratio $374.16/108.54\approx 3.447$, rather than four. This additive error does not cancel in a ratio.

At nonzero frequency $f$, measure both magnitude and phase of the electric current relative to the applied electric potential difference. Define $Y=I/V=1/Z$, where $Z$ is electrical impedance. For the stated complex permittivity convention, the ideal uniform specimen givesThus a parallel capacitor and electrical conductance describe the readout at that frequency; a single scalar capacitance cannot describe both components. The diagram's assumed dielectric relaxation gives $C_p=265.62\mathrm{pF}$ and $C_0{ϵ}_r^{\prime \prime }=88.54\mathrm{pF}$ at $1\mathrm{kHz}$. These are model predictions, not observations.

Repeat the reference measurement after removing the specimen and vary the small applied electric potential difference to test drift and linearity. Record specimen thickness, electrode area, contact conditions, frequency and temperature. Air gaps act as additional layers in series and can lower inferred relative permittivity; roughness, uncertain thickness and the fringing field of a capacitor limit the geometric formula. At sufficiently high frequency, the fixture itself changes the electromagnetic field, so low-frequency guarding and a lumped capacitor model require reassessment. ^[ref].

Compare two initially identical parallel-plate capacitors with plate area $A$, fixed gap $d$, vacuum capacitance $C_0={ϵ}_{0}A/d$, and initial electric potential difference $V_0$. Completely fill both gaps with the same uniform linear dielectric of static relative permittivity ${ϵ}_r>1$, neglecting the fringing field of a capacitor. Keep one connected to an ideal voltage source and isolate the other. The geometric capacitance becomes $C={ϵ}_r{C}_0$ in both cases; the electrical constraint determines what else changes.^[ref]

Write $Q_0=C_0{V}_0$, $E_0=V_0/d$ and $U_0=C_0{V}_0^2/2$. Applying $Q=CV$, $E=V/d$ and the capacitor energy formula gives the complete comparison:Thus the statement that a dielectric reduces the electric field needs the fixed-free-charge condition. With fixed electric potential difference, the voltage source supplies additional free plate electric charge, maintaining $E=V_0/d$ despite the new bound electric charges.

For the stipulated ${ϵ}_r=3$ example, the voltage source supplies $\Delta Q=2Q_0$. At constant $V_0$, its delivered electrical work is $V_0\Delta Q=4U_0$, whereas the increase in capacitor energy is $2U_0$. In an ideal reversible quasistatic insertion, the remaining $2U_0$ is mechanical work delivered by the assembly, as accounted for in fixed-voltage capacitor energy balance. For the isolated assembly, capacitor energy falls from $U_0$ to $U_0/3$, allowing $2U_0/3$ of mechanical work in the same ideal limit. Actual losses reduce recoverable work; energy increases in the connected case do not arise from the dielectric supplying free energy.

The generally complex, frequency- and wavevector-dependent relative permittivity describing linear polarization response. Its imaginary part represents dissipative response with the sign fixed by the time convention; optical prefactors connect it to an absorption coefficient.

Modification of the interaction between electric charges by a medium's electrical response. In the simplest Wannier-Mott exciton model, the Coulomb potential is reduced by a constant relative permittivity; actual screening can depend on spatial and temporal scales.

An electrical insulator has sufficiently small transport of electric charge for its operating conditions. Insulation is a practical requirement involving electric potential difference, temperature, geometry and duration, rather than a universal claim of zero electrical conductivity. Electric polarization can be substantial in an electrical insulator; dielectric breakdown can end the insulating regime.

For an idealized uniform slab with small constant electrical conductivity $\sigma$, the Ohmic conductor model predicts leakage electrical resistance $R=d/(\sigma A)$ in parallel with capacitance $C=ϵA/d$. An isolated capacitor then hasThis is a derived model of slow discharge, not an assertion that every leakage mechanism obeys Ohm's law. Its time scale matches charge relaxation. The fixed-charge comparison in dielectric capacitor response needs an observation time short compared with $RC$, but long enough for electric polarization to settle. During a maintained voltage, leakage consumes power $V^2/R$ as Joule heating; delayed electric polarization can add dielectric loss under changing electric fields.^[ref]

Large relative permittivity alone does not rule out either mechanism.^[ref]

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

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

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

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

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

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

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

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

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

The static Lindhard dielectric function describes the real, zero-frequency linear response of a uniform free electron gas. Here ${ϵ}_1=\mathrm{Re}ϵ$, $\mathbf{r}$ is the position operator, and $\mathbf{q}$ is the perturbation wavevector, $q=|\mathbf{q}|$, $V$ the normalization volume, $e>0$ the elementary charge, and ${ϵ}_0$ the vacuum permittivity. The normalized one-electron states are ${\psi }_{\mathbf{k}}=V^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}$ with $E_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$; $f(E)$ is their Fermi-Dirac distribution occupation. Both momentum sums exclude spin; the prefactor two counts the two equal spin populations. The density-operator matrix element is ${\delta }_{{\mathbf{k}}^{\prime },\mathbf{k}+\mathbf{q}}$. Coincident energy denominators are interpreted by the static limit or Cauchy principal value.

A weak total electric potential ${\varphi }_{\mathrm{tot}}$ perturbs an electron by $-e{\varphi }_{\mathrm{tot}}$. First-order state mixing produces $\delta n=e{\Pi }_0(q){\varphi }_{\mathrm{tot}}$, whereThe induced charge is $-e\delta n$. Self-consistent Poisson equation screening then gives ${\varphi }_{\mathrm{ext}}=ϵ{\varphi }_{\mathrm{tot}}$, with $ϵ=1+e^2{\Pi }_0/({ϵ}_0{q}^2)$. This is the random phase approximation: use the independent-electron response to the self-consistent total field.^[ref]

At zero temperature $f(E_{\mathbf{k}})=\Theta (k_F-k)$. Shifting the integration variable in the second occupation term givesThe two angular integrals are equal. Using the supplied integral, with $x=q/(2k_F)$, givesThe limits are $F(0)=1$, $F(1)=1/2$ and $F(x)\sim 1/(3x^2)$ for large $x$. The first recovers Thomas-Fermi screening; the logarithm produces the Kohn singularity of static Lindhard screening at $q=2k_F$.

Use the Fourier transform convention $\rho (\mathbf{q})=\int d^{3}r{e}^{-i\mathbf{q}\cdot \mathbf{r}}\rho (\mathbf{r})$. A point charge $Ze$ has total screened charge $Ze/ϵ(q)$ and induced charge $-Ze[1-ϵ(q{)}^{-1}]$. Angular inversion therefore gives the radial induced-charge screening kernelIn particular, ${\rho }_{\mathrm{ind}}(\mathbf{q}=0)=-Ze$: the total screening charge is negative although its local density can oscillate in sign.

Integrate the sine transform by parts twice; $g(0)=g(\infty )=0$ removes the boundary terms. For the Friedel density tail from a derivative singularity, write $Q=2k_F$ and $g^{\prime \prime }(q)\sim C/(q-Q)$. ThenIndeed, near $Q$ the principal value transform is ${\int }_{-\infty }^{\infty }\sin [(Q+s)r]ds/s=\pi \cos (Qr)$ for $r>0$. Smooth terms do not remove this leading oscillatory long-range contribution. Thus ${\rho }_{\mathrm{ind}}\propto \cos (2k_{F}r)/r^3$.

For a three-dimensional, single-valley gas with two spin states, the Friedel oscillation wavelength from carrier density isLow-temperature scanning tunneling spectroscopy near exposed dopants can map the oscillatory local density of states, with a Fourier feature near $2k_F$ at the Fermi energy. An energy-resolved tunnelling image measures local spectral weight, not directly the bulk energy-integrated charge: comparing occupied-bias maps and accounting for the surface geometry is necessary. In actual GaAs, the background relative permittivity and band effective mass change screening strengths; they do not change this wavelength at the stated density.

A Wannier-Mott exciton is a bound conduction-band electron and valence-band hole whose envelope extends over many lattice cells. That large radius permits an effective mass description and dielectric screening of the electron–hole attraction. With electron and hole masses $m_e^{\ast },m_h^{\ast }$, define reduced mass $\mu =m_e^{\ast }{m}_h^{\ast }/(m_e^{\ast }+m_h^{\ast })$ and center-of-mass mass $M=m_e^{\ast }+m_h^{\ast }$. The bulk relative-motion Hamiltonian is hydrogenic with an appropriate relative permittivity ${ϵ}_r$:A semiconductor quantum well confines the electron and hole perpendicular to the plane while leaving approximately free in-plane center-of-mass motion. If the well width is small compared with the bulk exciton Bohr radius, their relative motion becomes quasi-two-dimensional. The ideal zero-width, same-dielectric limit isIts ground-state binding is $4R_X$, compared with $R_X$ in three dimensions, and the ground envelope decays on the scale $a_X/2$. Stronger binding follows because confinement prevents the pair from separating in the third direction.

For a finite-width quantum-well exciton, average the Coulomb attraction over the normalized electron and hole confinement envelopes:This smooths the strict planar singularity and changes the binding. The ideal factor four is a useful reference, not a universal value for every real well. Finite barriers, dielectric mismatch, several confinement subbands and nonparabolic valence bands can require a coupled treatment.

Optical absorption creates a nearly zero in-plane center-of-mass wavevector exciton. For the lowest confinement subbands, its line lies approximately atThe discrete bound-state peaks occur below the free electron–hole continuum. Dipole-allowed envelope states have nonzero relative wavefunction at zero separation; the absorption strength also contains the confinement overlap $|\int {\chi }_e^{\ast }(z){\chi }_h(z)dz{|}^2$. Confinement can increase the overlap and the binding, making excitonic optical structure more prominent. A perpendicular electric field shifts and separates the confinement envelopes, changing the transition energy and usually reducing their optical overlap: this is the quantum-confined Stark effect.

Quantum wells tune both the continuum threshold and the bound-state energy; an optical line shift is not automatically a change in binding alone. The hydrogenic envelope model requires a Wannier-sized state, rather than an exciton localized to one lattice cell.

The sharp solid curve is the $4.2$ K spectrum; the broadened dashed curve is the $77$ K spectrum. At low temperature, a strong discrete exciton transition appears just below the continuum edge, with weaker higher members converging toward the band gap. Heating broadens those resonances through scattering, reduces their spectral resolution, and shifts the band edge to lower photon energy. The lower-energy broad dashed maximum is therefore not evidence for a second unrelated low-temperature band gap.

Read approximate energies from the native figure: the main low-temperature peak is near $1.610$ eV and a weaker feature is near $1.617$–$1.618$ eV, with the continuum edge around $1.620$ eV. A three-dimensional hydrogenic series $E_n=E_g-R^{\ast }/n^2$ then givesFor example, $E_2-E_1=3R^{\ast }/4$ predicts about $7.5$ meV, consistent with those feature spacings. These are graphical estimates at roughly meV precision, not extra digits extracted from the line widths. The strong direct interband onset and subgap series support a direct-gap Wannier-Mott interpretation. The continuum absorption scale is of order $10^6$ m${}^{-1}$, corresponding to an optical absorption length of order a micrometre. At $77$ K, $k_{B}T\simeq 6.6$ meV is already comparable with the inferred binding scale, explaining why a sharp Rydberg series is poorly resolved; at $4.2$ K, $k_{B}T\simeq 0.36$ meV.

With $m_e^{\ast }=m_h^{\ast }=0.2m_e$, the reduced mass is ${\mu }^{\ast }=0.1m_e$. In the screened hydrogen model,It follows thatA radius of several nanometres is consistent with an extended exciton in an ordinary semiconductor lattice. The relative permittivity inferred here is the effective dielectric screening parameter of that model; identifying the material chemically or determining a full dielectric spectrum is not possible from these data alone.

Seek the steady forced displacement as $x=x_0{e}^{-i\omega t}$. The oscillator equation givesFor $n$ ion pairs per volume, the ionic polarization is $P_{\mathrm{ion}}=nqx$. Add the electronic polarization ${ϵ}_0({ϵ}_{\infty }-1)E$, assumed frequency-independent over this infrared range. The complex relative permittivity is thereforeFor a longitudinal free oscillation without external free charge, $\mathbf{k}\cdot \mathbf{D}=0$ and the longitudinal electric field requires ${ϵ}_r({\omega }_{\mathrm{LO}})=0$ in the lossless limit. HenceEliminating the microscopic oscillator strength givesHere ${\omega }_{\mathrm{LO}}$ denotes the undamped longitudinal frequency; with finite damping the zero is at a complex frequency. The sign of the imaginary term follows the stated $e^{-i\omega t}$ convention and gives positive absorptive $\mathrm{Im}{ϵ}_r$ for $\omega >0$, $\gamma >0$. At zero frequency the same model yields the Lyddane-Sachs-Teller relation ${ϵ}_r(0)/{ϵ}_{\infty }={\omega }_{\mathrm{LO}}^2/{\omega }_{\mathrm{TO}}^2$. The high-frequency limit here is above the phonon resonance but below electronic resonances, not arbitrarily high photon energy.

Take the longitudinal scalar piezoelectric coefficient appropriate to the load and electrode direction. For a uniformly stressed slab of face area $A$ and thickness $t$, the source charge is $\Delta Q=d\Delta F=dA\Delta \sigma$. Its capacitance is $C={ϵ}_0\kappa A/t$, so the ideal initially uncharged, electrically isolated sensor givesThe sign depends on the poling direction and electrode convention. Equivalently, the incremental constitutive law is $\Delta D=d\Delta \sigma +{ϵ}_0\kappa \Delta E$; an ideal open circuit keeps free electrode charge, hence $D$, fixed and gives $\Delta E=-d\Delta \sigma /({ϵ}_0\kappa )$. area cancels because both generated charge and capacitance scale with area. Thickness does not cancel: the material comparison is through $d/\kappa$ only when geometry is fixed. Here $\kappa$ must be the relative permittivity under the relevant mechanical and frequency conditions.

The prescribed surface charge density is that of a uniformly polarized sphere and has zero total charge because $\int \cos \theta d\Omega =0$. Its electric dipole moment at $t=0$ isThe transverse components cancel by axisymmetry, while the $z$ component isThusThe angular surface distribution is a pure degree-one spherical harmonic. Hence its static external electric potential is exactly that of this dipole, with no higher electric multipole moments; it is not an assertion that the potential inside the finite sphere equals a singular point-dipole potential.

For right-handed uniform rotation about $+\hat{\mathbf{x}}$, the rigidly attached dipole isConsequently $|\ddot{\mathbf{p}}{|}^2=p_0^2{\omega }^4$ is constant. The electric dipole radiation formula givesThere is no extra factor of one-half: the squared sine and cosine contributions of the two perpendicular oscillating components sum to one. This is the leading electric-dipole approximation for radiation, requiring $\omega R/c\ll 1$ and observation in the radiation zone. The stated permanent surface distribution is interpreted as the actual charge producing the given external dipole; adding another dielectric screening factor would contradict the dipole moment the question asks to establish. The relative permittivity enters the induced-polarization problem in part (d), not this prescribed total dipole moment.

Let $\mathcal{I}$ be the sphere's mechanical moment of inertia about its rotation axis, and let ${\omega }_i=\omega (0)$. For an undriven rigid sphere the rotational energy is $E_{\mathrm{rot}}=\mathcal{I}{\omega }^2/2$. In the weak-radiation-loss approximation, conservation of energy givesIntegrating $d({\omega }^{-2})/dt=2K/\mathcal{I}$ givesEquivalently, the characteristic spin-down time isThe radiative spin-down of a rotating electric dipole therefore has angular frequency that decays as $t^{-1/2}$ at late times within this model, rather than exponentially. The original paper gives no mass or mass-density profile, so $\mathcal{I}$ cannot be determined from $R$ and the relative permittivity. If a homogeneous solid sphere of mass $M$ is additionally assumed, insert $\mathcal{I}=2M{R}^2/5$; uniform dielectric response alone does not imply this mechanical mass distribution.

The approximation is important. If the dipole angle is $\varphi (t)$ with $\dot{\varphi }=\omega (t)$, direct differentiation gives $|\ddot{\mathbf{p}}{|}^2=p_0^2({\omega }^4+{\dot{\omega }}^2)$. Using the constant-frequency radiation law instantaneously neglects the second term and assumes $|\dot{\omega }|\ll {\omega }^2$. This gives the standard leading slow spin-down solution above. An exact nonadiabatic radiation reaction evolution would require the mechanical and near-field dynamics beyond the information supplied, rather than treating this leading law as exact at arbitrarily rapid damping.

The given static induced surface charge and the dipole integral from part (b) giveHere $ϵ$ is the dimensionless relative permittivity, as required by the stated $ϵ\pm 1,2$ expression; ${\alpha }_E$ is the SI electric polarizability. Take ${\mathbf{E}}_{\mathrm{inc}}(t)=E_0\cos \omega t\hat{\mathbf{e}}$ and $k=\omega /c=2\pi /\lambda$. In the Rayleigh scattering regime the field is approximately uniform across the sphere. Its induced dipole radiates mean powerThe incident time-averaged light intensity in vacuum is $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$. Dividing power by intensity gives the total cross section,The units are ${\mathrm{m}}^2$; the comparison is with the geometric projected area $\pi R^2$, not with a length:The inequality holds for an ordinary nonabsorbing dielectric, with finite contrast factor and $kR\ll 1$. It explains why a sufficiently small transparent sphere scatters weakly and why its scattered power has the characteristic ${\lambda }^{-4}$ dependence. Near a material resonance where $ϵ+2$ is small, this simple inequality and static-polarizability approximation need reconsideration, including loss and radiative corrections. For the stated low-frequency ordinary dielectric the supplied polarizability is the intended approximation.

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.

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.