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Donor doping supplies conduction electrons to n-type semiconductor; acceptor doping supplies holes to p-type semiconductor. In the nondegenerate, fully ionized limit, majority densities are approximately $n_n=N_D$ and $p_p=N_A$, while minority densities follow $np=n_i^2$. On contact, majority carriers diffuse across the junction, exposing positive donor ions on the n side and negative acceptor ions on the p side. The depletion-region electric field points from n to p and opposes further diffusion. At thermal equilibrium the Fermi level is uniform and drift balances diffusion, with no net current.

For an abrupt junction with uniform doping and permittivity ${\epsilon }_s$, charge neutrality gives $N_D{x}_n=N_A{x}_p$. Poisson's equation then givesPositive applied $V$ means forward bias: it lowers the barrier and narrows the depletion region. Most depletion width lies in the more lightly doped side. The formula assumes the depletion approximation and fails once its operating assumptions break down.

forward bias increases minority densities at the depletion edges by the factor $e^{qV/k_{B}T}$. In the neutral material, the steady diffusion–recombination equation gives exponentially decaying excess minorities, with lengths $L_n=\sqrt{D_n{\tau }_n}$ and $L_p=\sqrt{D_p{\tau }_p}$. Taking the carrier gradients at the junction yieldsThis ideal diode law assumes low injection, nondegenerate statistics, negligible depletion-region recombination and series resistance, and sufficiently long neutral regions. Its exponential comes from the bias-dependent injection barrier; there is no universal abrupt “turn-on” voltage. reverse bias suppresses injection, giving an approximate $-I_S$ current until additional leakage or breakdown matters.^[ref]

Rectification uses the asymmetric current-voltage characteristic; controlled reverse breakdown enables voltage references, with current limiting. Bias-dependent depletion capacitance enables varactors. photodiodes and solar cells collect optically generated carriers through the junction field and diffusion toward it; efficient light-emitting diodes use radiative electron–hole recombination, preferably in a direct-gap material. Series resistance, high injection, trap-assisted recombination, heating and finite carrier storage explain departures from the ideal steady-state law.

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.

For propagation along $z$, a transverse wave has two perpendicular displacement components. With $\varphi =kz-\omega t$, write $u_x=A\cos \varphi$, $u_y=B\cos (\varphi +\delta )$. Their trajectory at a fixed point obeysThe general nondegenerate trajectory is an ellipse: elliptical polarization. For $\delta =0$ or $\pi$, the components remain in phase or antiphase and the trajectory collapses to a line. A single nonzero component is also linearly polarized. Equal amplitudes $A=B$ with $\delta =\pm \pi /2$ give circular polarization. Unequal quadrature amplitudes give an ellipse aligned with the chosen axes, while general phase differences rotate its axes.

For the displayed convention and $\delta =+\pi /2$, at $z=0$ the vector starts at $+x$ and moves towards $+y$ as time increases. Reversing $\delta$ reverses its rotation. Right- and left-handed labels require a stated viewing and phase convention, so rotation is defined explicitly here. A fixed ellipse is a coherent relationship between transverse components; unpolarized light instead has fluctuating orientation and phase relationships on the measurement timescale.

A linear polarizer transmits a linearly polarized beam with intensity proportional to ${\cos }^2\theta$, allowing its direction to be measured by an analyser. circular polarization transmits equal intensity through any linear analyser, but that test alone does not distinguish it from unpolarized light. A quarter-wave plate converts a suitable coherent circular state to linear polarization, allowing the combination to discriminate them. For light the electric field defines the usual polarization; the magnetic field remains perpendicular to it and to propagation. Longitudinal sound in a homogeneous gas lacks the two transverse degrees of freedom needed for these polarization states.

Take the electron's signed charge to be $e=-q$, where $q>0$ is the elementary-charge magnitude. Conventional electric current flows along $+x$, while the mean electron drift velocity is negative. With $\mathbf{B}=B_z\hat{\mathbf{z}}$, the magnetic Lorentz force separates charge across the width until a transverse electric field balances it. The accumulated edge charge then sustains a Hall voltage without a transverse steady current. In a one-carrier model, the Hall coefficient identifies the sign and number density of carriers; real multiband materials require a more careful interpretation.

The acceleration term contains the effective mass describing the local band response. The relaxation term removes drift momentum on the mean momentum-relaxation time $\tau$. The remaining two terms are the electric and magnetic Lorentz force on signed charge $e$; the magnetic force is perpendicular to velocity and does no instantaneous work. The relaxation-time approximation replaces individual scattering events by an averaged drift equation.

Let $n$ be the electron number density, distinct from the negative charge $e$. In transverse open circuit $⟨\mathbf{v}⟩=(v_x,0,0)$ and $j_x=ne{v}_x=I_x/(wt)$. The steady component equations givehenceConsequentlyThere is no sign ambiguity in this field-defined $R_H$ once axes and carrier charge are fixed.

A measured voltage ratio additionally needs probe polarities, which the geometric bar diagram does not label. To reproduce the positive ratio specified for electrons, define the longitudinal voltage drop $V_x=\phi (0)-\phi (l)=E_{x}l$ and the transverse rise $V_y=\phi (w)-\phi (0)=-E_{y}w$. ThenReversing either voltage's probes reverses the measured ratio, without changing the physical carrier density. With both voltages instead defined as drops along their positive axes, the algebraic ratio would have the opposite sign. The figure records the convention used here.

For the density, geometry gives $\rho =Rwt/l=Rt/(l/w)$ and $|R_H|=t|V_y|/(I_x|B_z|)=tR|V_y/V_x|/|B_z|$. ThusandThe mobility is $\mu =|R_H|/\rho =0.0130{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. With $m^{\ast }=m_e$,

A monovalent copper model assigns roughly one delocalized conduction electron per atom to a broad, approximately free-electron-like band. Using $m_e$ is therefore a sensible first estimate for its transport scale. Monovalence alone does not prove that every band curvature or orbit has exactly the free mass: periodic-potential mixing, the nonspherical Fermi surface and direction-dependent scattering limit the model. The calculated $\tau$ is conditional on the stipulated effective-mass approximation.

Experimentally, reverse both current and magnetic field and extract the part of the transverse voltage odd in each. This rejects fixed contact offsets and much longitudinal pickup; use four-terminal longitudinal sensing, measure dimensions independently and check heating before assigning material parameters. These controls are implemented in a reusable Hall/conductivity teaching experiment, although that experiment uses germanium rather than the copper sample here.^[ref]

Doping changes the mobile carrier population by introducing controlled impurity levels. In an intrinsic semiconductor, the filled valence band is separated from the nearly empty conduction band by a band gap $E_g$. Thermal excitation produces an electron-hole pair, so the electron and hole number densities obey $n=p=n_i$. A electron hole represents a missing valence-band electron and behaves as a mobile positive charge carrier. For nondegenerate bands in thermal equilibrium,Here $N_C,N_V$ are the effective density of states, $E_C,E_V$ the band edges, and $E_F$ denotes the chemical potential, conventionally called the Fermi level. The semiconductor mass-action law holds in this nondegenerate equilibrium regime.

In a tetrahedrally bonded elemental semiconductor such as silicon, a substitutional group-V atom has an extra outer electron relative to the host. It is a donor impurity: a shallow level just below $E_C$ can ionise, releasing an electron to the conduction band and leaving a fixed positively charged donor. The resulting n-type semiconductor has electrons as majority carriers and holes as minority carriers. Conversely a group-III substitution supplies one fewer bonding electron. An acceptor impurity can take an electron from the valence band, leaving a mobile electron hole and a fixed negative acceptor; the resulting p-type semiconductor has holes as majority carriers.

The impurity state is shallow because the electron or hole responds to a screened Coulomb potential and an effective mass, rather than the bare atomic potential. A simple hydrogenic donor estimate iswhere ${ϵ}_r$ is the relative permittivity. This is a useful scaling model, with real band anisotropy and central-cell effects omitted. Dopants are fixed in the lattice; their ionisation changes mobile carrier numbers without making a uniformly doped bulk sample carry a large net electric charge.

The charge neutrality condition isWhen shallow dopants are almost fully ionised and $N_D-N_A\gg n_i$, an n-type region has $n\simeq N_D-N_A$ and $p\simeq n_i^2/n$. A p-type region similarly has $p\simeq N_A-N_D$ and $n\simeq n_i^2/p$. Thus compensation by the opposite impurity type reduces the majority-carrier density; the total dopant concentration is not generally the mobile carrier density. At sufficiently high impurity density the degenerate semiconductor regime requires the full Fermi-Dirac distribution and these exponential approximations fail.

There are three useful temperature regimes. At low temperature, dopant freeze-out leaves many shallow levels un-ionised. In the extrinsic regime, the dopants are largely ionised and set the majority-carrier density. At high enough temperature, intrinsic pair creation dominates and $n,p$ approach $n_i$. The Fermi level moves towards the conduction band with donor doping and towards the valence band with acceptor doping. The electrical conductivity isDoping therefore controls conductivity and its carrier type, but ionised impurities also scatter carriers and can reduce charge carrier mobility. Carrier density and mobility must both be considered.

A p-n junction forms when p-type and n-type regions are joined within a semiconductor. Initially the concentration gradients drive diffusion current: electrons diffuse from the n region to the p region and holes in the opposite direction. Recombination removes mobile majority carriers near the join, exposing fixed positive donors on the n side and fixed negative acceptors on the p side. This forms a depletion region and an electric field directed from n to p. Its drift current opposes diffusion. At equilibrium the net current vanishes and the Fermi level is uniform, although the band edges vary spatially through band bending.

For an abrupt nondegenerate junction with fully ionised dopants, its built-in potential isUnder the depletion approximation, the space-charge densities are $-e{N}_A$ on a p-side width $w_p$ and $+e{N}_D$ on an n-side width $w_n$. Integrating Poisson equation and imposing zero field outside gives $N_A{w}_p=N_D{w}_n$ andMost of the depletion width therefore extends into the more lightly doped side. The schematic shows the impurity levels before contact and the resulting space charge and potential after contact; there is no externally extractable equilibrium battery voltage from the built-in field alone.

A forward bias, with p positive relative to n, reduces the barrier from $V_{\mathrm{bi}}$ towards $V_{\mathrm{bi}}-V$ and injects minority carriers across the junction. In the ideal low-injection, diffusion-controlled regime this produces the Shockley diode equation, $I=I_s[e^{eV/(k_{B}T)}-1]$. Reverse bias increases the barrier and the depletion region; only a small thermally generated minority-carrier current flows before breakdown. Reverse-biased depletion regions also give a voltage-dependent junction capacitance. Real devices have recombination, leakage, series resistance and breakdown mechanisms beyond the ideal law. Controlled doping creates majority and minority carriers, and their junction produces a depletion barrier that makes rectification possible.

Take an abrupt equilibrium p-n junction with the p side at $x<0$ and the n side at $x>0$. In the quasi-neutral bulk, electron holes are the majority carriers on the p side and electrons are the majority carriers on the n side. With ionized dopant densities $N_A,N_D$ and nondegenerate statistics,where $n_i$ is the intrinsic carrier concentration. The semiconductor mass-action law $np=n_i^2$ applies at equilibrium. Both mobile populations are small in the depletion region compared with the dopant densities.

The charge and carrier density of an abrupt p-n junction must distinguish mobile populations from net charge. Diffusion leaves negatively charged ionized acceptor impurities on the p side and positively charged ionized donor impurities on the n side. Under the depletion approximation,The neutral bulks have large carrier densities but nearly zero net charge density. The electric field points from the positive donor region toward the negative acceptor region, balancing diffusion at equilibrium.

The figure uses equal dopant densities only to show a simple symmetric sketch. Generally the more lightly doped side has the wider depleted interval. The carrier profiles are smooth, whereas the step-shaped net charge density is the ideal depletion approximation.

Treat sodium as monovalent, with one mobile electron per atom. Its mass density is ${\rho }_m=0.97\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}=970\mathrm{kg}{\mathrm{m}}^{-3}$, and its molar mass is $M=0.023\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Using the Avogadro constant,The electron mobility $\mu =e\tau /m_e$ relates the magnitude of drift velocity to electric field. The Drude model then givesUsing $m_e=9.109\times 10^{-31}\mathrm{kg}$,The inferred time is the momentum relaxation time. Identifying it with the mean time between collisions is the intended simple isotropic Drude model assumption; small-angle scattering need not relax directed momentum in a single event.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.

At an interface, frequency is conserved for stationary media, and tangential wave phase must match. These conditions give Snell's law for refraction and constrain reflected and transmitted directions. Amplitudes follow from continuity of the appropriate fields or mechanical forces. On strings the relevant mechanical impedance is $\sqrt{T\mu }$; for sound it is related to $\rho c_s$, and for electromagnetic waves to $\sqrt{\mu /ϵ}$. A mismatch gives wave reflection, and a lossless boundary conserves the total incident, reflected and transmitted power. Multiple reflections can reinforce or cancel, as in a Fabry–Pérot etalon. Total internal reflection and other boundary conditions confine waves in waveguides. Transverse modes then become discrete, and even an otherwise nondispersive material develops longitudinal cutoffs and wave dispersion. A membrane strip, for example, obeys ${\omega }_n^2=c_m^2[k^2+(n\pi /a{)}^2]$.

Propagation also involves transport and loss. The Poynting vector gives electromagnetic energy flux; mechanical energy transport is force or pressure times velocity, averaged over a cycle. Wave absorption, viscosity, electrical resistance and imperfect boundaries dissipate energy. Scattering redistributes energy into other directions and modes, while geometric spreading reduces local intensity without necessarily dissipating anything. These mechanisms should be distinguished when interpreting a decreasing measured amplitude. In strongly absorbing or rapidly varying dispersive regimes, the derivative called group velocity is not automatically the speed of a sharp signal front; pulse reshaping and the medium's causal response must be considered.

Finally, quantum matter waves illustrate how the same propagation mathematics can describe a different physical quantity. For a free nonrelativistic particle, $\omega =\hslash k^2/(2m)$, so the group velocity $\hslash k/m$ equals its particle velocity and differs from the phase speed $\hslash k/(2m)$. In a periodic crystal, Bloch's theorem replaces the free dispersion by energy bands. Their slope controls wave-packet velocity, $v_g={\hslash }^{-1}dE/dk$, and their curvature gives band effective mass. Thus mechanical confinement, material response and microscopic periodicity all reshape propagation through their allowed modes and dispersion, while the physical interpretation of the wave and its transported quantities remains essential.

The Hall effect is a transverse electric response to a magnetic force on moving charge carriers. Take conventional current density $\mathbf{J}=J_x\hat{\mathbf{x}}$ and magnetic field $\mathbf{B}=B_z\hat{\mathbf{z}}$ in a conducting slab. For one carrier type of number density $n$ and signed electric charge $q$, the drift velocity is ${\mathbf{v}}_d=\mathbf{J}/(nq)$. Carriers initially deflect sideways until a Hall electric field balances the magnetic Lorentz force:The Hall coefficient is thereforeFor electrons it is $-1/(ne)$, whereas a simple hole-dominated conductor gives $+1/(pe)$. The sign reveals the effective carrier sign; the magnitude estimates their density in the one-carrier approximation.

For width $w$ along $y$ and thickness $t$ along $z$, $I=J_{x}wt$. Define the signed Hall voltage by $V_H=E_{y}w$, namely the electric potential at the lower-$y$ edge minus that at the higher-$y$ edge. ThenThis explicit contact convention fixes the voltage sign. A Hall bar with longitudinal and transverse contacts measures both electrical resistivity and the Hall response. Reverse the field and take the part of the transverse voltage odd in $B_z$ to suppress longitudinal-voltage pickup from misaligned contacts; reversing current provides another useful check.

A Hall measurement is useful for carrier identification, semiconductor doping measurements and magnetic-field sensing. Combining a one-carrier $R_H$ with electrical conductivity gives a mobility magnitude $\mu =\sigma |R_H|$. However, mixed electron and hole conduction is not described by a single density: in weak fields,The response weights mobilities differently from ordinary conductivity. Complicated energy bands, anisotropic Fermi surfaces and multiple bands can likewise change the simple interpretation in metals; a positive Hall coefficient does not mean that the material contains positively charged fundamental electrons.

Semiconductor doping introduces controlled impurities to alter equilibrium carrier populations and electrical conductivity. In an intrinsic semiconductor, thermally excited electrons in the conduction band and electron holes in the valence band have equal densities $n=p=n_i$. A shallow donor impurity, such as phosphorus in silicon, supplies an electron from an impurity level just below the conduction-band edge. Once ionized it is positively charged, and the material is n type. An acceptor impurity, such as boron in silicon, accepts a valence electron into a level just above the valence-band edge, leaving a mobile hole and a negatively charged ionized impurity; the material is p type.

With dopant densities $N_D,N_A$, charge neutrality requiresFor fully ionized, nondegenerate dopants at equilibrium, the semiconductor mass-action law gives $np=n_i^2$. In a substantially n-type sample, $N_D-N_A\gg n_i$, soThe corresponding p-type result is $p\simeq N_A-N_D$ and $n\simeq n_i^2/(N_A-N_D)$. The majority population grows while the minority population decreases; adding donors does not simply add equal numbers of free electrons and holes.

The Fermi energy shifts toward the conduction band for n-type material and toward the valence band for p-type material. In the nondegenerate approximation,where $N_C,N_V$ are effective band-edge state densities. At very low temperature, dopants may not be ionized and carrier freeze-out occurs. At intermediate temperatures, ionized dopants dominate the carrier density; at sufficiently high temperature, intrinsic excitation can dominate. Heavy doping can produce degenerate carrier statistics, invalidating these Boltzmann expressions.

The electrical conductivity is $\boxed{\sigma =e(n{\mu }_e+p{\mu }_h)}$. Doping usually raises the majority carrier density, but ionized impurity scattering can lower mobility, so conductivity need not scale indefinitely with dopant concentration. Joining p-type and n-type regions creates a p-n junction, with a depletion region and built-in electric field established by carrier diffusion and compensating drift. The fixed ionized dopants, not an absence of all charge, account for the depleted region's net charge density.

In a metal, a partially filled energy band or overlapping energy bands supplies electrons with nearby unoccupied states. At thermal equilibrium, oppositely directed group velocities are equally populated and the net electric current vanishes. An applied electric field distorts the occupation distribution, producing a small mean drift velocity. A filled energy band alone has no such low-field intraband response because its velocity contributions cancel over the entire Brillouin zone.

In the Drude model, let $n$ be the mobile electron number density, let $e>0$ be the elementary charge, and let $\tau$ be the momentum relaxation time. The mean motion obeysThe steady state is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m$. The current density is opposite to the electron drift velocity:This gives microscopic Ohm's law, with electrical conductivity $\sigma$ in $\mathrm{S}{\mathrm{m}}^{-1}$ and electrical resistivity ${\rho }_{\mathrm{el}}$ in $\Omega \mathrm{m}$. For a uniform wire of length $\ell$ and cross-sectional area $A$, its electrical resistance is $R={\rho }_{\mathrm{el}}\ell /A$. The dissipated power per volume is $\mathbf{J}\cdot \mathbf{E}=\sigma E^2$, the Joule heating supplied by the external source.

The free electron model with Fermi-Dirac statistics improves the microscopic picture. The Pauli exclusion principle fills states up to the Fermi energy at low temperature. For a spherical, parabolic energy band with two spin states,Only states near the Fermi surface change their occupation in a weak-field response; the large equilibrium Fermi velocity is not the small drift velocity. The same $n{e}^2\tau /m$ formula results for this isotropic parabolic model when $\tau$ is evaluated for transport near the Fermi surface. In a crystal, the energy band group velocity, effective mass and possibly direction-dependent scattering replace the simplest bare-particle picture. The mean free path is approximately ${\ell }_{\mathrm{mfp}}=v_F\tau$.

A perfectly periodic static lattice is already included in Bloch states and does not by itself give a finite electrical resistivity. Momentum relaxation comes from defects, impurities, boundaries, and electron-phonon scattering. Independent scattering mechanisms approximately add their rates, giving Matthiessen's rule,The approximately temperature-independent ${\rho }_0$ is the residual resistivity. In ordinary simple metals, the phonon contribution is approximately proportional to temperature at high temperature; a three-dimensional acoustic-phonon transport model gives a $T^5$ contribution well below the relevant Debye temperature. Other mechanisms can dominate in particular materials or regimes, so neither power law is universal.

The classical Drude model incorrectly treats the equilibrium carrier statistics. Fermi-Dirac statistics explains the small electronic heat capacity, while electrons carrying both charge and heat give the approximate Wiedemann-Franz law ${\kappa }_e/(\sigma T)={\pi }^2{k}_B^2/(3e^2)$ for degenerate carriers with appropriate nearly elastic scattering. Both the simple electrical conductivity formula and this thermal relation depend on the assumptions about energy bands and scattering, rather than being exact for all metals.

Take a p-n junction made from one semiconductor, with the p-type semiconductor on the left and the n-type semiconductor on the right. At thermal equilibrium the chemical potential $\mu =E_F$ is horizontal throughout. Before contact the chemical potential lies nearer the valence band on the p side and nearer the conduction band on the n side; after contact the band edges bend to permit one common value.

In this orientation both the conduction band edge $E_C$ and the valence band edge $E_V$ decrease from p to n by $e{V}_{\mathrm{bi}}$. For a homojunction their separation $E_g=E_C-E_V$ is constant. The acceptor impurity level $E_A$ lies just above $E_V$ where acceptors are present, and the donor impurity level $E_D$ lies just below $E_C$ where donors are present. These impurity levels follow the local band bending; they do not form one additional horizontal band across the device.

The shaded depletion region has few mobile electrons or electron holes and contains fixed negative ionized acceptor impurities on the p side and positive ionized donor impurities on the n side. Thus its electric field points from n to p. Its neutral outer regions have negligible electric field, so the band edges become flat there. In the depletion approximation, charge neutrality requires $N_A{x}_p=N_D{x}_n$; the symmetric drawing illustrates $N_A=N_D$ rather than assuming that equality is universal. Diffusion and field-driven drift current cancel at equilibrium: no applied voltage does not mean zero internal electric field or zero built-in electric potential difference.

For a plane wave propagating along $\hat{\mathbf{k}}$ in vacuum, Maxwell equations give $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$, with $\mathbf{E}\perp \hat{\mathbf{k}}$. HenceTaking $E=E_0\cos (\mathbf{k}\cdot \mathbf{r}-\omega t)$, the period average of the squared cosine is $1/2$. ThusThe fields are in phase, so the travelling-wave energy flux never reverses. The peak amplitude convention is essential; using a root-mean-square electric field would remove the factor $1/2$.

Since $\nabla \cdot \mathbf{B}=0$, a magnetic field can locally be represented by a magnetic vector potential,It is not unique: $\mathbf{A}\to \mathbf{A}+\nabla \chi$ leaves its curl unchanged. In magnetostatics, choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Ampère's law then gives ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{J}$ in vacuum. For a localized steady current distribution with the usual decay condition,Its curl gives the Biot–Savart field. For a uniform field, one convenient choice is $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. Stokes's theorem relates ${\oint }_C\mathbf{A}\cdot d\mathbf{l}$ to magnetic flux through a spanning surface. In time-dependent electrodynamics, $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$; changing $\mathbf{A}$ by $\nabla \chi$ requires $\varphi \to \varphi -{\partial }_t\chi$ to preserve the electric field. Gauge freedom simplifies calculation without changing the electromagnetic observables.

Let the unit normal point from layer 1 to layer 2. The electrostatic electric field boundary condition is $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$, while Gauss's law for the electric displacement field gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Here ${ϵ}_i$ denotes absolute permittivity; replace it by ${ϵ}_0{ϵ}_{ri}$ if dielectric constants are relative.

In the steady conducting state, charge conservation makes the normal current density the same in both layers:Writing $\mathcal{D}={\sigma }_2{d}_1+{\sigma }_1{d}_2$, the steady fields in two leaky dielectric layers areThe signed interfacial free surface charge density is the jump of $D_n$:The total free-plus-bound sheet charge instead follows from the vacuum form of Gauss's law,The bound contribution is ${\sigma }_{\mathrm{total}}-{\sigma }_f$. These are signed quantities for the chosen normal and voltage direction; their magnitudes are obtained by absolute values. They describe the long-time DC state, not the initial capacitive transient. Positive conductivities suffice for the derivation; if both vanish, conductivity alone does not determine the field division, which also depends on the initial free interfacial charge.

The fundamental mode of an ideal coaxial cable is a transverse electromagnetic mode: the electric field is radial between the conductors, the magnetic field is azimuthal, and energy flows along the cable through the dielectric. There is no cutoff for this mode; higher modes, conductor losses and dielectric dispersion can matter outside the ideal transmission-line regime.

Ignoring edge fields in the parallel-strip geometry, $E_x=V/a$ and $H_y=I/b$. Per unit length the charge is $ϵbV/a$ and the linked magnetic flux is $\mu aI/b$. Thus the parallel-plate transmission line hasFor a segment $dz$, charge conservation, equivalently the appropriate integral of the Ampère-Maxwell equation, gives $I(z)-I(z+dz)=Cdz{\partial }_{t}V$. Applying Faraday's law around a longitudinal rectangle between the plates gives $V(z)-V(z+dz)=Ldz{\partial }_{t}I$. These are the lossless telegrapher's equations,Differentiate the second with respect to $z$ and use the first, and conversely for $I$:The signal speed is $1/\sqrt{LC}=1/\sqrt{\mu ϵ}$, equal to $c$ for a vacuum-filled line. It is the speed of the coupled electromagnetic disturbance, rather than the electron drift velocity. Electrons locally adjust their motion as the field arrives; one electron need not travel the cable's length to transmit a signal. The Poynting vector in the dielectric integrates to power $VI$.

For an outward radial electric field, Gauss's law gives the enclosed chargeTaking the enclosing sphere to infinity gives $Q_{\mathrm{total}}=0$. A nonzero local field is compatible with zero net charge. Indeed, for $r>0$,Positive inner charge is balanced by negative outer charge. There is no point charge at the origin, since ${\lim }_{r\to 0}4\pi {ϵ}_0{r}^2{E}_r=0$. The field's magnitude does not specify its inward/outward sign, but reversing a globally radial direction leaves the zero total unchanged. This is net charge from the asymptotic radial electric field.

For a fixed circuit, Faraday's law of induction readsSurface orientation and contour direction obey the right-hand rule. The negative sign is Lenz's law: a passive induced current acts against the flux change, consistently with energy conservation. The induced electric field generally has nonzero curl and cannot be represented by one single-valued electrostatic potential.

For a material circuit moving with local velocity $\mathbf{u}$, the measured electromotive force includes the magnetic Lorentz force on its carriers:This distinguishes transformer induction in a stationary coil from motional induction in a changing circuit. A straight sliding rod of length $l$ moving perpendicular to itself and to uniform $B$ has $|\mathcal{E}|=Blv$. With circuit resistance $R$, the electrical power $(Blv{)}^2/R$ equals the external mechanical power needed to oppose magnetic drag. An ideal transformer gives $V_2/V_1=N_2/N_1$ in magnitude from common flux linkage; real leakage, resistance and core losses limit that simple result. pickup coils measure flux change rather than an absolute static magnetic field, so integration and a reference state are needed to recover a field history.

For the dielectric half-space the two formal sources coincide at $(a,0,0)$, outside the region being represented. Their sum is $Q+Q^{\prime }=2{ϵ}_{r}Q/(1+{ϵ}_r)$ in the dielectric Coulomb kernel, henceThere is no charge singularity inside this half-space. The expression satisfies Laplace's equation there, decays at infinity, and matches the vacuum solution's tangential electric field and normal electric displacement. The coincident image is a mathematical representation, not an extra physical charge in the vacuum region.

Insert $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$ into the given identity. For a linear, time-independent, nondispersive medium, $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$, givingThe divergence theorem then yieldsThe left side is outward electromagnetic power flow. The current term is power delivered to matter, becoming Joule heating for an Ohmic conductor, and ${\partial }_{t}u$ is the change in stored electric plus magnetic energy. The minus sign says outward flux reduces stored energy unless supplied locally.

Follow the drawing's coordinates: lower strip at $z=0$, upper strip at $z=d$, current $I$ along $+x$ in the upper strip and back along the lower. The wording “both strips lie in the $xy$-plane” must mean parallel to that plane, since the drawing separates them in $z$. Neglect the side and end fringes using $l\gg a\gg d$. Then $I=V_0/(R+R_L)$ and the top-strip potential is $V(x)=V_0-IRx/l$, while the lower strip is equipotential at zero.

The harmonic function matching those two surfaces isIndeed ${\nabla }^2\varphi =0$, and it has the required potentials at $z=0,d$. ThusThe two current sheets reinforce their magnetic field between the strips; its direction follows the right-hand rule. The Poynting flow in resistive parallel strips follows from their cross product:Power flows along the gap towards the load and upwards into the resistive strip. Across a transverse area $ad$, $P_x(x)=IV(x)$ falls from $I{V}_0$ to $I^2{R}_L$. The upper surface receives ${\int }_0^{l}a{S}_z(z=d)dx=I^{2}R$, exactly the strip's heating; the ideal lower strip receives none. In the charge-free gap, ${\partial }_x{S}_x+{\partial }_z{S}_z=0$, consistent with a static field and no local dissipation. The small longitudinal electric field cannot be discarded if this upward energy transfer is to be represented.