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Put the charged sheet at $z=0$ and the grounded plates at $z=\pm d/2$. By symmetry the electric field points away from the positive sheet with equal magnitude in the two gaps. A Gaussian pillbox crossing the sheet givesThe plate boundary conditions fix their electric potentials to zero. For the charged sheet between grounded parallel plates, the central potential isWith vacuum permittivity ${\epsilon }_0=8.85419\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $V_s\simeq 28.24\mathrm{V}$. An electron has electric charge $-e$, so moving from a grounded plate to this positive potential decreases its potential energy by $e{V}_s$. Conservation of energy then givesThe unrounded value is $3.15\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, about $0.0105c$. Either plate gives the same speed. The ideal calculation neglects the electron's disturbance of the prescribed field and image-force effects at its release point, consistently with treating it as a test particle.

Reflection and central spot. At a boundary from lower to higher refractive index, the near-normal electric-field reflection coefficient is negative, $r=(n_1-n_2)/(n_1+n_2)$. This gives a $\pi$ wave phase reversal on reflection from the glass plate through the air film. Reflection at the lens's glass-to-air underside has no such reversal. These are consequences of the Fresnel equations for the reflected electric field.

At ideal contact the film thickness is zero, so the two reflected contributions have zero geometrical optical path difference but a relative $\pi$ phase reversal. They produce destructive interference, and the central spot is dark. This conclusion concerns the film reflection; other lens surfaces can add a weak background.

Rings and their radii. At radial distance $r$, let the air thickness be $t$. The two relevant rays reflect at the upper and lower film interfaces. The lower reflection accumulates an extra round-trip optical path length $2t$, plus the relative $\pi$ reflection shift. As $t$ increases away from contact, alternating constructive and destructive thin-film interference forms Newton's rings. At near-normal incidence, a bright ring satisfiesThe extended source can illuminate these localized film fringes: each incident contribution interferes with its own two reflected paths; unrelated source points need not be mutually coherent.

For the spherical lens surface, geometry gives $r^2+(R-t{)}^2=R^2$, so $r^2=2Rt-t^2$. Since $t\ll R$, $t\simeq r^2/(2R)$. Combining the equations yieldsThe $m=0$ bright ring is the first bright ring around the dark contact point.

Curvature from ring diameters. Writing $D_m=2r_m$, subtract the squared diameters to eliminate the unknown initial order:Using SI units,

Liquid-filled film. With film refractive index $n=1.4$, still smaller than both glass indices, there remains exactly one relative reflection phase reversal. The round-trip optical path length is now $2nt$, soHere $\lambda =500\mathrm{nm}$ is the illuminating vacuum wavelength, approximately its air value, rather than its shorter value inside the liquid. The ring labelled $m=2$ therefore has diameterFilling the gap contracts each ring radius by $1/\sqrt{n}$ at fixed order and curvature.

An electric dipole is a localized pair of opposite electric charges, or more generally a neutral electric charge distribution with nonzero first electric moment. For charges $-q$ and $+q$ separated by displacement $\mathbf{d}$ directed from negative to positive, the electric dipole moment isA neutral molecule can have a permanent electric dipole moment when its positive and negative charge centres do not coincide. An applied electric field can also produce an induced electric dipole moment by distorting an initially nonpolar electric charge distribution.

At distance $r\gg |\mathbf{d}|$ in vacuum, expand the difference of the two point-charge potentials to first order in the separation. The leading electric potential isThe monopole term cancels because the net electric charge is zero. Taking $\mathbf{E}=-\nabla \Phi$ gives the electric dipole fieldOn the dipole axis its magnitude is $2p/(4\pi {\epsilon }_0{r}^3)$ and its direction is along $\mathbf{p}$; in the equatorial plane the magnitude is $p/(4\pi {\epsilon }_0{r}^3)$ and the field points opposite to $\mathbf{p}$. The $r^{-3}$ decay is faster than the $r^{-2}$ field of an isolated point charge.

In a uniform external electric field, the opposite forces on the two charges cancel in their resultant but form a couple. Its torque isFor a fixed permanent electric dipole moment, the interaction potential energy isDifferentiating with respect to orientation recovers the torque. Parallel alignment minimizes $U$ and is stable against a small rotation; antiparallel alignment maximizes $U$ and is unstable. Rotating a fixed dipole slowly from parallel to antiparallel requires external work $2pE$.

A nonuniform electrostatic electric field can also exert a net force. Expanding the two charge forces gives $\mathbf{F}\simeq (\mathbf{p}\cdot \nabla )\mathbf{E}$, which equals $\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed dipole in a curl-free electrostatic field. Thus a suitably aligned dipole is attracted toward stronger field. An induced electric dipole moment requires care in computing energy: when $\mathbf{p}=\alpha \mathbf{E}$ for constant isotropic electric polarizability, building the dipole costs energy and the effective interaction is $U=-\alpha E^2/2$, rather than substituting directly into the fixed-dipole formula. These field, torque and energy effects explain molecular orientation and electric polarization.

Faraday's law of induction relates an induced electromotive force to the rate of change of magnetic flux. Choose a positive orientation around a loop and its associated surface normal by the right-hand rule. For a stationary conducting loop,Unlike an electrostatic field, the induced electric field generally has nonzero curl; the differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a moving circuit, magnetic flux transport for a moving circuit includes the magnetic Lorentz force on the charges:where $\mathbf{u}$ is the local wire velocity. Thus changing the field or moving a conductor can produce induction.

The minus sign is Lenz's law: an induced electric current, when a closed conducting path permits it, produces a magnetic field opposing the change of linked flux. It need not oppose the original field itself. For example, increasing flux through a loop produces a current tending to reduce that increase; decreasing the same flux reverses the current. The opposition is consistent with conservation of energy: external mechanical or electrical work is needed to maintain a change against its inductive reaction.

For a coil of $N$ turns, define the magnetic flux linkage $\Lambda ={\sum }_{j=1}^N{\Phi }_j$, reducing to $N{\Phi }_B$ when every turn links the same flux. For fixed geometry and a linear magnetic response, self-inductance $L$ is defined by $\Lambda =LI$, and the coil's induced electromotive force isThe corresponding terminal voltage drop, with the passive sign convention, is $V_L=LdI/dt$. It opposes changes in electric current; it does not require the current itself to vanish.

Building the current from zero requires power $P=I{V}_L=LIdI/dt$. Integrating gives the energy stored by an ideal inductor:This energy is stored in the magnetic field and can be returned when the current falls. In a series RL circuit driven by a constant voltage $V_0$, Kirchhoff's voltage law gives $L\dot{I}+RI=V_0$. Starting with zero current, $I=(V_0/R)(1-e^{-tR/L})$, with time constant $L/R$. This makes the resistance to rapid current changes directly observable. Two coils also have mutual inductance: changing primary current induces secondary electromotive force $-Md{I}_1/dt$ with a polarity determined by the chosen winding orientations, the principle behind a transformer.

Point-charge field and superposition. Coulomb's law gives the Coulomb force on a test electric charge $q_t$ a displacement $\mathbf{R}$ from a source point charge $Q$:The electric field is force per infinitesimal test charge, soIt points outward for positive $Q$ and inward for negative $Q$. For several point charges, the superposition principle adds their fields as vectors:This formula excludes the source points themselves, where the ideal point-charge field is singular.

Exact field components of the pair. PutThe displacement vectors from the positive and negative charges to the observation point are $(x-a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$ and $(x+a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$. Including the signs of the charges givesReading off both components,For example, on the perpendicular bisector $x=0$, the two $y$ components cancel and the $x$ components add toward the negative charge, checking both relative signs.

Far-field polar components. The pair's electric dipole moment points from negative to positive charge:Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$, with $\theta$ measured from that positive dipole axis. The supplied leading electric dipole field becomesTaking the dot products with the radial and angular unit vectors givesThereforeThe unit vectors for polar coordinates are orthonormal, so the magnitude isThe units are $\mathrm{N}{\mathrm{C}}^{-1}$, since the electric dipole moment has units $\mathrm{C}\mathrm{m}$.

Angular sketch and comparison. At fixed large $r$, the magnitude has period $\pi$. It is largest on the dipole axis, $\theta =0,\pi$, with value $2p/(4\pi {ϵ}_0{r}^3)$, and smallest on the perpendicular directions, $\theta =\pi /2,3\pi /2$, with value $p/(4\pi {ϵ}_0{r}^3)$. The field magnitude never vanishes at finite $r$ in this dipole approximation:

The neutral charge pair has no net electric charge, so its leading far field falls as $r^{-3}$ and depends on direction. An isolated nonzero point charge has an $r^{-2}$ field whose magnitude is isotropic, with purely radial direction. The dipole also generally has a nonzero $E_{\theta }$ component.

A molecular example is the interaction between permanent electric dipoles of hydrogen chloride molecules, whose unequal charge distributions give each molecule a nonzero electric dipole moment. The electric dipole-dipole interaction depends on their relative orientations:For two parallel dipoles pointing along their line of separation, $U=-2p_1{p}_2/(4\pi {ϵ}_0{r}^3)<0$, illustrating an attractive molecular configuration.

Keep $+x$ along the upper-sheet current and $+z$ upward, putting the lower sheet at $z=0$ and upper sheet at $z=d$. Between them, the upper sheet's lower-side magnetic field is $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The lower sheet carries current in $-x$, so its upper-side magnetic field is also $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The superposition principle therefore givesEach perfect conductor is an equipotential. The circuit's resistor carries current from upper to lower, so Ohm's law fixes their electric potential difference:The broad parallel faces and their constant potentials make the electric field perpendicular to them and independent of the coordinates along them, away from edges. There is no volume electric charge in the gap. Apply Gauss's law to a small pillbox wholly within the gap, with equal face areas $A$ at heights $z_1,z_2$. Its side flux is zero, soThus $E_z$ is independent of height as well. Since $V_{\mathrm{upper}}-V_{\mathrm{lower}}=-{\int }_0^d{E}_{z}dz$,It points from the higher-potential upper sheet to the lower sheet. Gauss's law requires the planar symmetry argument: zero enclosed charge alone would not prove a uniform electric field in a general geometry. These combined fields characterize parallel current sheets with a resistive termination.

The electric displacement field is $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$. A homogeneous linear isotropic dielectric has $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, so $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. At a free-charge-free interface the normal component of D is continuous, while electrostatics requires continuity of tangential E.

At half immersion the conductor centre lies in the flat air-liquid interface. Set $\varphi =C/r$ in both media, with potential zero at infinity. The sphere is an equipotential, and the potential is continuous across the liquid plane. On that plane the radial E field is tangential, so tangential E agrees while both normal D components vanish. Each half-space also satisfies the source-free potential equation. The fields outside the sphere areGauss's law for D on a sphere gives $Q=2\pi {ϵ}_0(1+{ϵ}_r)C$. ThereforeThe half-immersed conducting sphere in a dielectric has radial E lines with equal field strength at equal radius on either side, whereas the D flux density is ${ϵ}_r$ times larger in the liquid. The corresponding free surface charge is larger on the lower hemisphere. The unchanged field direction at this particular interface follows from its tangential geometry, not from a claim that electric fields never refract at dielectric interfaces.

The outward electrostatic pressure is $p_e=\mathbf{E}\cdot \mathbf{D}/2$. Integrating its vertical component over the two hemispheres uses ${\int }_{\mathrm{upper}}\cos \theta dA=\pi a^2$ and the negative of that integral below. ThusThe force is downward for ${ϵ}_r>1$. Initially, buoyancy on the uncharged sphere gives ${\rho }_{\ell }gf(4\pi a^3/3)=Mg$. At half immersion the upward buoyancy is therefore $Mg/(2f)$, leaving a required downward electric force $Mg(1/(2f)-1)$. Equating the two givesThis uses the ideal flat-interface, linear-dielectric model with surface tension and interface deformation neglected. If ${ϵ}_r=1$, there is no electric force of this kind, consistently preventing this mechanism from pulling an initially less-than-half-immersed sphere down to the specified level.

A modulation-doped GaAs-AlGaAs electron gas is confined on the GaAs side of the interface. The larger-gap AlGaAs supplies a conduction-band barrier; charge transfer from remote donors and the resulting electric field bend the bands into an approximately triangular well. transverse electronic mode wavefunctions are confined normal to the interface but free in its plane. electrons settle into the well until their electrochemical potential agrees with the donor/contact environment. An undoped spacer separates mobile electrons from ionized donors, reducing impurity scattering. Equilibrium has one Fermi level across the structure; the conduction-band edge is not flat. Surface depletion and the cap also affect the profile. The original drawing illustrates one occupied confinement transverse electronic mode; a single spatial electron gas need not in general imply only one occupied vertical transverse electronic mode.

molecular beam epitaxy can grow a clean GaAs buffer/channel, an undoped AlGaAs spacer, a Si-doped AlGaAs supply region and a protective GaAs cap on a suitable substrate. Control composition, interface quality and background impurities during growth; placing dopants directly in the channel defeats the principal mobility advantage. Define a mesa or isolation region, fabricate low-resistance ohmic contacts to the buried gas, and pattern fine metal split gates by electron beam lithography and the lift-off process. Negative gate bias depletes the underlying gas, leaving a narrow channel between the gates. Check contact resistance, leakage, carrier density and mobility; operation with ballistic electron transport requires channel length shorter than the relevant scattering lengths and smooth confinement on the wavelength scale. The lithographic gap does not equal the electrical width because of lateral depletion.

For a long uniform channel supporting ballistic electron transport, transverse confinement creates thresholds $E_n$ with continuous longitudinal propagation above each threshold. One directed spin branch of a transverse electronic mode has states per length $dk/(2\pi )$ and group velocity $v={\hslash }^{-1}dE/dk$. Hence $vdk/(2\pi )=dE/h$: velocity cancels the one-dimensional density of states. Including two unresolved spin states,For low temperature and small bias, a fully open transverse electronic mode has $T_n(E_F)=1$, givingThe assumptions are ideal reservoirs/contacts, negligible reflection and interactions that would invalidate the independent-subband description, and unresolved spin splitting. A resolved spin branch contributes $e^2/h$. The finite two-terminal electrical resistance is a contact/mode-injection effect, despite the absence of dissipative scattering inside the ideal channel.

For voltage partition in a ballistic constriction, a short saddle has strong local electric-field and electrostatic screening variation near its entrances and barrier, rather than the uniform Ohmic voltage gradient of a diffusive wire. Right-moving incident states inherit the source distribution and left-moving states the drain distribution; there is no unique local equilibrium chemical potential inside an ideal ballistic saddle. The electrostatic potential must be determined self-consistently. In a symmetric rigid-barrier approximation, the barrier follows the mean lead potential and half the electrochemical bias lies on each side. This is a stated approximation, not a universal spatial voltage profile.

For a saddle $U=U_0-m^{\ast }{\omega }_x^2{x}^2/2+m^{\ast }{\omega }_y^2{y}^2/2$, transverse electronic mode thresholds are $E_n=U_0+\hslash {\omega }_y(n+1/2)$. The smooth-barrier transmission is $T_n(E)=[1+\exp (-2\pi (E-E_n)/(\hslash {\omega }_x)){]}^{-1}$, approaching a step for a sharp threshold on the bias/thermal scale.^[ref]

Let ${\mu }_{S,D}=E_F\pm eV/2$ and assume no additional bias dependence of $T_n$. At zero temperature,When the source and drain sample $N+1$ and $N$ thresholds, the differential electrical conductance is $(N+1/2)2e^2/h$. These half plateaus arise from the two ends of the bias window, not fractional carriers. The dc ratio $I/V$ need not share their plateau values. Heating, spin splitting and interaction-induced features must be distinguished experimentally.

In bias spectroscopy of one-dimensional subbands, plot differential electrical conductance or its gate derivative against gate bias and source-drain bias. Threshold branches occur where $E_n(V_g)={\mu }_S$ or ${\mu }_D$. When one bias-window edge meets $E_n$ and the other $E_{n+1}$,Thus the intersections of adjacent threshold branches calibrate the one-dimensional spacing directly in energy. A gate lever arm $\alpha =-e^{-1}\partial E_n/\partial V_g$ can then convert gate intervals to energy; it should be measured rather than guessed. Asymmetric bias partition, a moving barrier or a gate-dependent spacing modifies slopes and plateau weights, requiring the corresponding self-consistent bias model.

In vacuum, the source-free Maxwell equations arePut $\psi =\omega (t-z/c)$ and denote the constant electric amplitude by $E_0$. A corresponding traveling-wave magnetic field isBoth divergences vanish because both the electric field and the magnetic field depend only on $z,t$ and have zero $z$ component. Direct differentiation gives $\nabla \times \mathbf{E}=(\omega E_0/c)(\cos \psi ,\sin \psi ,0)=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{B}=(\omega E_0/c^2)(-\sin \psi ,\cos \psi ,0)=c^{-2}{\partial }_t\mathbf{E}$. A time-independent background magnetic field satisfying the homogeneous equations could be added; the formula is the wave component.

Introduce the magnetic vector potential and electric potential by $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. These identities automatically give the divergence equation for $\mathbf{B}$ and Faraday's law. For this wave choose $\varphi =0$ and $\mathbf{A}=(c/\omega )\mathbf{B}$; the derivatives just calculated verify both potential equations.

The electromagnetic force is the Lorentz force, $Q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. On the smooth constraint plane $z=z_0$, the constraint force is normal to the plane. Both $\mathbf{v}$ and $\mathbf{B}$ are tangential, so $\mathbf{v}\times \mathbf{B}$ is normal and is balanced by that constraint force; it does not affect the tangential motion. ThusA particular motion is ${\mathbf{r}}_{\parallel }={\mathbf{r}}_0-[Q{E}_0/(m{\omega }^2)](\cos \psi ,\sin \psi )$, a circle of radiusThe general planar motion also includes a constant drift velocity; the particle moves on a circle when that drift is zero. The constraint force can do no work because it is perpendicular to the velocity.

In a p-n junction, diffusion and electron-hole recombination near the contact leave a depletion region of ionized dopants. Its built-in electric field opposes further majority-carrier diffusion. At thermodynamic equilibrium drift and diffusion currents cancel. Forward bias lowers the electrostatic barrier and injects electrons and electron holes across the junction; reverse bias raises the barrier and suppresses that majority-carrier injection. For an ideal diode below breakdown,Thus a modest forward voltage gives a rapidly increasing current, whereas reverse current remains approximately $-I_s$. This asymmetric current-voltage relation makes the junction a rectifier. Real leakage, electron-hole recombination and breakdown limit the ideal expression.

Charge carrier mobility measures weak-field drift speed per electric field. For electrons and electron holes,Both mobilities are positive, with SI units ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$; the opposite drift signs reflect the opposite carrier charges. In a simple isotropic relaxation-time model, electron mobility is ${\mu }_e=e{\tau }_e/m_e^{\ast }$ and hole mobility is ${\mu }_h=e{\tau }_h/m_h^{\ast }$. These are momentum relaxation times, not the minority carrier lifetime for electron-hole recombination.

Scattering by phonons, ionized impurities and defects limits mobility. Increasing phonon populations commonly lowers mobility at high temperature; ionized-impurity scattering can dominate at low temperature and high semiconductor doping. Approximate addition of independent inverse relaxation times is the Matthiessen's rule. At sufficiently strong field, velocity saturation invalidates the constant-mobility drift law.

The Hall effect provides a measurement: a current-carrying bar in a perpendicular magnetic field develops a transverse voltage. In a one-carrier model $|R_H|=1/(en)$ and $\mu =\sigma |R_H|$, combining the transverse voltage with longitudinal electrical conductivity. Multiple carrier types and scattering-dependent Hall factors require a more detailed interpretation.

In an isolated energy band, semiclassical motion satisfies $\hslash \dot{k}=F$ and $v={\hslash }^{-1}dE/dk$. The band is periodic in crystal momentum, so a constant force sweeps the state through equivalent points of the Brillouin zone. Its velocity repeats rather than increasing without bound. For an electron in a constant electric field, $F=-e\mathcal{E}$, and the Bloch oscillation period isFor $E(k)=E_0-2J\cos ka$, $v=(2Ja/\hslash )\sin ka$ and integration gives a bounded position oscillation of amplitude $2J/|F|$ about its mean. A filled band has no corresponding net transport oscillation after summing uniformly over all $k$.

Observation requires scattering times longer than $T_B$ and negligible interband tunnelling. Ordinary crystals often scatter too rapidly before a full zone traversal; larger-period structures and clean atom-optical lattices help. In the cold-atom Bloch oscillation experiment, accelerating a standing-wave lattice supplied a controlled force, and releasing the atoms at different times revealed an oscillating momentum distribution.^[ref]

A liquid crystal flows like a liquid while retaining some molecular order. In a nematic liquid crystal, elongated molecules have long-range orientational order but no ordinary positional lattice. The nematic director $\mathbf{n}$ specifies the preferred axis, with $\mathbf{n}$ and $-\mathbf{n}$ equivalent for head-tail-symmetric molecules. The nematic order parameter $S=⟨(3{\cos }^2\theta -1)/2⟩$ distinguishes an isotropic distribution, $S=0$, from perfect parallel alignment, $S=1$. A smectic phase additionally has a periodic layered density modulation, while molecules remain mobile within layers.

The broken rotational symmetry allows birefringence and distinct elastic costs for nematic splay, nematic twist and nematic bend. Under crossed polarizers, rotating a thin aligned sample changes the transmitted intensity because its two optical polarization components acquire different phases. Heating through a nematic-to-isotropic transition removes that orientational birefringence. An electric field can rotate the director if the dielectric anisotropy of a nematic favors alignment, explaining the optical switching principle of a liquid-crystal cell. Molecular shape, temperature, concentration and surface anchoring determine which ordered phase appears.

In a nematic liquid crystal, elongated molecules have orientational order around a nematic director $\mathbf{n}$, but lack the long-range positional order of a crystal. Head-tail symmetry identifies $\mathbf{n}$ and $-\mathbf{n}$. Molecular anisotropy and sufficient density favor collective alignment: for long hard rods this can increase translational entropy by reducing excluded-volume constraints, while attractive interactions can also favor alignment. Thermal orientational disorder competes with these effects.

The scalar nematic order parameter convention here isThus $S=0.44$ means $⟨{\cos }^2\theta ⟩=0.6267$. The effective orientation angle from nematic order isThis is not enough information to determine the literal mean angle $⟨\theta ⟩$: different orientation distributions can have the same second moment. Isotropic orientations have $S=0$, and perfect axial alignment has $S=1$.

In the supplied threshold expression, $d$ is the cell thickness; $K_1,K_2,K_3$ are the splay, twist and bend elastic constants of the Frank elastic energy. The dimensionless relative dielectric anisotropy of a nematic is $\Delta ϵ={ϵ}_{\parallel }-{ϵ}_{\perp }$, and ${ϵ}_0$ is the vacuum permittivity. Positive anisotropy favors a director along the applied electric field. Strong surface anchoring fixes orthogonal in-plane directions on the two plates. The low-field director therefore twists by a quarter-turn through the cell. Between crossed polarizers, with the entrance polarization matched to the first surface director, adiabatic optical following in a twisted nematic rotates the polarization and allows transmission. A sufficiently large normal field tilts the interior director towards the field, suppressing that rotation and making the crossed-polarizer cell dark; the anchored boundary layers remain distorted. Reversing the polarizer arrangement can reverse the optical contrast.

The threshold is a Fréedericksz transition: the electric energy gain overcomes the restoring elastic energy of an allowed distortion. Using the expression explicitly supplied by the paper givesThe thickness cancels for this strong-anchoring model. This is the onset of director reorientation, not an assertion that every molecule instantaneously becomes perfectly aligned. The quoted elastic combination is taken as given for this numerical task; a threshold derived from a specified full director-energy model must use that model's own elastic combination consistently. The companion illustration in A2(b) shows the plate anchoring, polarizers and bulk field-aligned state.

In a Hall effect measurement, a current flows along $x$ and a magnetic field is applied along $z$. The transverse Lorentz force initially deflects the mobile charges, building a transverse electric field until the net transverse current vanishes. Defining $R_H=E_y/(j_x{B}_z)$, the one-carrier Drude model gives $R_H=1/(nq)$. Thus electron-like carriers give $R_H=-1/(ne)$ and hole-like carriers give $R_H=+1/(pe)$.

A nearly filled band can conduct through holes: its missing electrons behave as positive mobile charges. Metals can have both electron-like and hole-like Fermi surface sheets. In the weak-field two-carrier approximation the Hall coefficient isThe sign depends on the carrier densities and squared mobilities; it need not identify the most numerous carrier. Experimentally the Hall voltage changes sign when the field or current reverses, allowing subtraction of unwanted longitudinal-voltage pickup. Anisotropic bands and different scattering rates complicate a literal conversion of $R_H$ into a carrier count.

A positive electric potential difference between gate and substrate attracts electrons towards the insulating interface and repels the mobile electron holes of the p-type semiconductor substrate. The exposed, negatively charged acceptor impurities form a depletion region; the thin mobile electron layer at its surface is the semiconductor inversion layer. The conduction band and valence band bend downwards towards the positively biased gate, because the electron potential energy is $-e\psi$.

To specify the conventional MOS inversion threshold, let $N_A$ be the ionized acceptor density, $n_i$ the intrinsic carrier concentration, andIn the neutral bulk the intrinsic level lies $e{\varphi }_F$ above the semiconductor chemical potential. A positive surface potential ${\psi }_s$ lowers both band edges by $e{\psi }_s$, giving, in the nondegenerate approximation,At ${\psi }_s={\varphi }_F$, the surface is intrinsic: $n_s=p_s=n_i$. The usual onset of strong inversion instead meansThus the surface electron density has grown to the bulk majority-carrier density. The chemical potential need not actually enter the conduction band. The phrase “just forms” can also refer to the earlier weak-inversion crossover; the diagram uses the conventional strong-inversion threshold and labels it explicitly.

Within the depletion approximation the negative space charge is approximately $-e{N}_A$ over a widthwhere ${ϵ}_s$ is the absolute semiconductor permittivity. Its outer edge is where the bands become flat and mobile electron holes recover their bulk density. The electrical insulator blocks steady carrier transport while sustaining an electric field; the metal gate supplies positive surface charge balancing depletion and inversion charge. Under gate bias, the gate and substrate electrochemical potentials are different, so a single equilibrium Fermi level should not be drawn through the whole biased stack.

A solar cell uses the photovoltaic effect to turn photon energy into electrical work. In a semiconductor, above-gap photon absorption generates an electron in the conduction band and an electron hole in the valence band. A p-n junction, together with selective electrical contacts, collects the two carrier species at opposite terminals. Carrier diffusion and the built-in electric field help separate them before electron-hole recombination. Illumination maintains different quasi-Fermi levels for electrons and electron holes; their splitting, divided by $e$, sets the available terminal voltage in an ideal device. The equilibrium built-in potential alone cannot continuously deliver work.

For an ideal junction, the illuminated output current density isHere $J_{\mathrm{sc}}$ is the photogenerated short-circuit current density and $J_0$ the diode saturation current. The useful operating point maximizes $VJ$, rather than either current or voltage separately. With solar-cell fill factor $\mathrm{FF}=P_{\max }/(V_{\mathrm{oc}}{I}_{\mathrm{sc}})$,Series resistance, shunts, incomplete photon absorption and nonradiative recombination reduce the attainable efficiency.

The Shockley-Queisser limit removes these avoidable losses but retains the following physical constraints. There is one semiconductor junction with one band gap; each absorbed above-gap photon yields at most one collected electron-hole pair; carriers thermalize to the cell temperature; and the inverse optical process produces unavoidable radiative recombination. If ${\Phi }_{\odot }(E)$ is the incident photon flux per unit energy, ideal photon absorption givesThe same ability to absorb light requires the junction to emit it. At a forward voltage $V$, the emitted photon population has a chemical potential $eV$; in the ordinary nondegenerate limit the resulting recombination current is the exponential diode term above. Maximizing $VJ$ with this radiative $J_0$, and then choosing $E_g$, gives the limiting efficiency.

A small band gap collects many photons but gives a low voltage and large thermalization losses. A large band gap permits a larger voltage but transmits too much of the incident spectrum. Below-gap photons are lost to transmission, excess photon energy above the gap becomes heat, and radiative emission and its associated entropy loss keep $e{V}_{\mathrm{oc}}$ below $E_g$. The optimum single-junction conversion is roughly thirty percent under ordinary unconcentrated sunlight; the precise ceiling depends on spectrum, cell temperature and optical geometry. The original detailed-balance calculation gave about $30\%$ at $E_g\simeq 1.1\mathrm{eV}$ under its specified assumptions.^[ref] Multiple junctions and optical concentration change the assumptions and consequently change the bound.

In the Lorentz oscillator model, each participating bound electron is displaced by an applied electric field against a restoring force. The model replaces complicated interband transitions by a driven damped harmonic oscillator. If $u$ is the electron displacement along the field and the electron charge is $-e$, its equation of motion isHere ${\omega }_0$ describes binding and $\gamma >0$ describes loss of coherent motion. For $E(t)=\mathrm{Re}(E_0{e}^{-i\omega t})$, the complex amplitude isThe induced electric dipole moment is $-eu$, so the electric polarization amplitude for number density $n$ isUsing $D={ϵ}_{0}E+P={ϵ}_0{ϵ}_{r}E$ givesThe paper's $ϵ$ is this dimensionless relative permittivity, rather than the absolute permittivity ${ϵ}_0{ϵ}_r$. Independent classes of bound electrons contribute additive susceptibilities, usually with different resonance frequencies and optical oscillator strengths.

Writing $D_{\omega }=({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2$ separates the response intoAt zero frequency the real part is $1+{\omega }_p^2/{\omega }_0^2$ and the imaginary part vanishes. Near resonance the imaginary part has a positive photon absorption peak and the real part changes rapidly, passing through $1$ at ${\omega }_0$. Above resonance the real part can become negative when damping is sufficiently weak; at very high frequency it approaches $1$ from below while the imaginary part tends to zero. Finite damping removes the undamped divergence.

For a nonmagnetic material take the passive square-root branch $N=\eta +i\kappa =\sqrt{{ϵ}_r}$, with $\kappa \ge 0$. A plane wave then has an intensity attenuation coefficient $\alpha =2\omega \kappa /c$. Thus the imaginary permittivity expresses dissipation, whereas the actual photon absorption coefficient is determined by the imaginary refractive index. The normal-incidence power reflectance isA large refractive-index mismatch produces wave reflection even without loss. In the undamped model ${ϵ}_r<0$ between ${\omega }_0$ and $\sqrt{{\omega }_0^2+{\omega }_p^2}$, so the field is evanescent and wave reflection is total in a semi-infinite sample. With damping the response includes both wave reflection and photon absorption; a peak in the imaginary permittivity should not simply be identified with a reflectance peak.

To infer the short-wavelength background from $R=0.35$, assume weak photon absorption and take the physical semiconductor branch $\eta >1$. ThensoThis is an estimate of the valence-electron background in a frequency window above the doped-carrier response and below the effective bound-electron resonance. One reflectance value alone does not determine both parts of a complex dielectric constant. Even for real ${ϵ}_r$, a reciprocal-index branch gives the same $R$; the ordinary background index greater than unity selects the branch used here.

Using the supplied conventional cubic lattice parameter, the participating valence-electron number density isUsing $m=m_e$ for the stipulated free-electron comparison givesAn angular frequency “in eV” means its quantum energy $\hslash \omega$, not $h\omega$. No doped-carrier effective mass should be inserted into this valence-electron count.

The low-frequency limit of the bound-electron contribution obeysConsequentlyThis effective resonance is much larger than the quoted $0.2\mathrm{eV}$ band gap. There is no contradiction: the single oscillator has been assigned the strength of all sixteen participating valence electrons per conventional cell, whereas the band gap is only the lowest interband excitation threshold. A real spectrum distributes optical oscillator strength over many transitions. Equating the oscillator resonance to the lowest threshold would misrepresent the assumed background response. Here “low frequency” means low relative to ${\omega }_0$; the static response of the doped conducting sample also contains its free carriers.

The additional mobile carriers in doped indium antimonide contribute a Drude frequency response, obtained by setting their restoring frequency to zero. At infrared frequencies well below the effective valence resonance, the sum is approximatelyAt the shorter wavelengths the carrier term is small and $R$ approaches the background value. As wavelength increases, the negative free-carrier contribution lowers the real permittivity towards unity. For weak damping, refractive-index matching to vacuum gives the Drude-Lorentz reflectivity minimum atThis minimum is near ${ϵ}_r^{\prime }=1$, not its zero. At the slightly lower screened plasma frequency ${\omega }_{pc}/\sqrt{{ϵ}_b}$, the real permittivity crosses zero; at longer wavelengths it becomes negative and the material becomes strongly reflective. Finite damping prevents a perfectly vanishing minimum and rounds the edge. At still lower frequency the large conducting response also drives $R$ towards unity.

For fixed carrier effective mass and background permittivity, ${\omega }_{pc}\propto \sqrt{n_c}$ and ${\lambda }_{\min }\propto n_c^{-1/2}$. The observed shift of the minimum towards shorter wavelength as doping increases is therefore the expected free-carrier trend. The figure is an original model illustration, not a digitization or quantitative fit of the supplied curves; the carrier effective mass, damping and their density dependence have not been specified.

For one isotropic carrier species in the Drude model, the longitudinal electrical resistivity at zero transverse electric current is ${\rho }_{xx}=1/(ne\mu )$, and the Hall electrical resistivity has magnitude $|{\rho }_{yx}|=B/(ne)$. Equivalently, the electric field balances the magnetic deflection through $E_y=v_{x}B$, while $|v_x|=\mu |E_x|$. Therefore the magnitude of the Hall angle obeysThe required ratio givesThe electron density cancels. With electric current along $+x$ and field along $+z$, electrons have negative Hall coefficient and hence $E_y<0$; the requested one-percent condition concerns the magnitude. This calculation assumes electrons dominate transport and the transverse contacts draw no electric current, so a Hall field can develop.

Dopant atoms and carrier type. Substitutional semiconductor doping places a small fraction of foreign atoms on host lattice sites. In a tetrahedrally bonded group-IV semiconductor, a group-V substitution such as phosphorus in silicon has an extra valence electron. A shallow donor level just below the conduction edge can release that electron, leaving a positively charged donor. In an n-type semiconductor, mobile electrons are the majority carriers and the chemical potential shifts toward the conduction band.

A group-III substitution such as boron in silicon has one fewer valence electron. An acceptor level just above the valence edge captures an electron from the valence states, leaving a mobile electron hole and a negatively charged acceptor. In a p-type semiconductor, electron holes are the majority carriers and the chemical potential shifts toward the valence band. The material can remain electrically neutral in both cases: mobile carrier electric charge is compensated by fixed ionized impurity electric charge. Dopant freeze-out, compensation and degeneracy qualify the simplest one-carrier-per-dopant picture.

Making the junction. Choose p material at $x<0$ and n material at $x>0$. Electrons initially diffuse from n to p, and electron holes from p to n, recombining near the interface. Negative ionized acceptors are exposed on the p side and positive ionized donors on the n side. They produce an electric field directed from n to p. The field opposes further majority-carrier diffusion until drift and diffusion balance and the equilibrium electrochemical potential is uniform.

The depletion regions are depleted of mobile majority carriers, not of the fixed dopant charges that create the space charge. At equilibrium the conduction and valence edges bend together, while the chemical potential is flat. The specified $\Delta {\varphi }_0$ is a electric potential difference: the corresponding initial chemical-potential energy separation is $e\Delta {\varphi }_0$. This convention is required by the quoted units and width formulas. Write ${\epsilon }_s=ϵ{ϵ}_0$ for the absolute permittivity; the paper's $ϵ$ is relative permittivity.