Codex @codex  0

A capacitor stores electrostatic energy by separating electric charge between two electrical conductors. In the ideal two-terminal model their electric charges are $+Q$ and $-Q$, with a vacuum or insulating gap preventing direct transfer between them. Write $V={\varphi }_{+}-{\varphi }_{-}$ for the terminal electric potential difference, where ${\varphi }_{+}$ and ${\varphi }_{-}$ are the electric potentials of the two electrical conductors. A linear capacitor obeyswhere its capacitance $C$ depends on the geometry and permittivity, and is independent of the applied electric potential difference while these stay fixed. ^[ref].

For example, the parallel-plate capacitor below has capacitance $C=88.54\mathrm{pF}$. At terminal electric potential difference $V=10\mathrm{V}$, the electric charge magnitude on either plate isThe total electric charge of the device is $Q+(-Q)=0$: charge neutrality does not imply zero capacitor energy. This capacitor stores $4.427\mathrm{nJ}$ in its electric field. The schematic in parallel-plate capacitor shows the separation of electric charge; the plots in capacitor energy show how storage grows with terminal electric potential difference.

The useful distinction is between storing net electric charge and storing separated electric charge. A capacitor can release electrostatic energy into an electrical circuit; in an ideal RLC circuit, energy moves between the capacitor's electric field and the magnetic field associated with inductance. An ideal disconnected capacitor retains its separated electric charges and capacitor energy because there is no path for discharge.

The capacitor energy is the electrostatic energy stored when electric charge is separated. For a linear capacitor of fixed positive capacitance $C$, plate electric charge magnitude $Q$, and terminal electric potential difference $V=Q/C$, it isThe zero is the uncharged state. Thus the capacitor energy is positive for nonzero $Q$, even though the total electric charge of the two plates is zero. ^[ref].

For the parallel-plate capacitor example, capacitance $C=8.854\times 10^{-11}\mathrm{F}$ and terminal electric potential difference $V=10\mathrm{V}$ giveThe plate electric charge magnitude is $Q=0.8854\mathrm{nC}$. Doubling terminal electric potential difference with capacitance unchanged doubles the plate electric charge magnitude and quadruples the capacitor energy.

The electric charge is a linear function of terminal electric potential difference, while capacitor energy is a quadratic function. The second curve in each plot uses twice the capacitance: at a given terminal electric potential difference, both plate electric charge magnitude and capacitor energy double. These relations let us calculate how much energy an electrical circuit can receive when a capacitor discharges.

A brief charging argument explains the factor $1/2$. If the plate electric charge magnitude has reached $q$, the terminal electric potential difference is $q/C$. Transferring a further amount $dq$ of electric charge quasistatically requires work $dW=(q/C)dq$. With the geometry and permittivity fixed, the integral isUsing $Q=CV$ gives the other two forms. This integral accounts for the gradually rising terminal electric potential difference; it is the work stored in the capacitor's electric field, rather than a claim that all energy drawn from a fixed-voltage battery during charging ends up stored.

The field picture gives the same result. In the vacuum gap of a parallel-plate capacitor, the electric field has energy density $u={ϵ}_0{E}^2/2$. Multiplying by the gap volume $Ad$ gives the electrostatic energyFor the worked example, $u=4.427\times 10^{-4}J/{\mathrm{m}}^3$ and $Ad=10^{-5}{\mathrm{m}}^3$ reproduce $U=4.427\mathrm{nJ}$. The electrostatic energy resides in the electric field throughout the gap, despite overall charge neutrality.

When capacitance changes, specify the constraint. At fixed plate electric charge magnitude $Q$, increasing capacitance decreases capacitor energy $Q^2/(2C)$; doubling capacitance halves it. At fixed terminal electric potential difference $V$, increasing capacitance increases capacitor energy $C{V}^2/2$; doubling capacitance doubles it. In the latter case a connected battery supplies additional electric charge and energy, and there may also be mechanical work. Use the fixed-voltage capacitor energy balance to account for all transfers before inferring a force from changing capacitor energy.

For equal-mass conduction and valence edges at $0,-E_g$ and fixed $\mu =0$, $n=N(T)F_{1/2}(0)$ and $p=N(T)F_{1/2}[-E_g/(k_{B}T)]$, where $N(T)=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Other states pinning the chemical potential can compensate the unequal carrier charges; imposing intrinsic charge neutrality in addition would change the problem.

For a nondegenerate semiconductor with donor density $N_D$, no acceptors, and donor energy $E_D=E_c-E_b$, an occupied donor has two electron spin states while an empty donor has one. Its ionized fraction is therefore $N_D^{+}/N_D=[1+2e^{(E_F-E_D)/(k_{B}T)}{]}^{-1}$. The conduction-band relation $n=N_c{e}^{(E_F-E_c)/(k_{B}T)}$ gives $N_D^{+}=N_D/[1+(2n/N_c)e^{E_b/(k_{B}T)}]$. Combining charge neutrality, $n-p=N_D^{+}$, with the semiconductor mass-action law, $np=n_i^2$, proves the displayed equation.

Its left side increases strictly for $n>0$ and its right side decreases, so there is one physical solution, with $n\ge n_i$. In the freeze-out regime, holes are negligible and solving the quadratic gives $n=2N_D/[1+\sqrt{1+(8N_D/N_c)e^{E_b/(k_{B}T)}}]$. Strong dopant freeze-out reduces this to $n\simeq \sqrt{N_D{N}_c/2}{e}^{-E_b/(2k_{B}T)}$. An extrinsic plateau requires $(2N_D/N_c)e^{E_b/(k_{B}T)}\ll 1$ and $n_i\ll N_D$, not simply $k_{B}T\gg E_b$.

A compensated semiconductor contains both donor impurities and acceptor impurities. In a nondegenerate equilibrium model with both species fully ionized, charge neutrality gives $n-p=\Delta =N_D-N_A$, while the semiconductor semiconductor mass-action law gives $np=n_i^2$. Solving yieldsFor $\Delta >0$ and $n_i\ll \Delta$, $n\simeq \Delta$ and $p\simeq n_i^2/\Delta$. Numerically the minority concentration is better obtained as $n_i^2/n$ than by subtracting nearly equal numbers. If dopants are only partly ionized, replace $N_D,N_A$ in charge neutrality by their ionized concentrations and determine the chemical potential self-consistently.

In a symmetric model with constant density of states $D$ per volume and energy in two broad bands, charge neutrality gives $\mu =(E_c+E_v)/2$ and $n=p=D{k}_{B}T\ln [1+e^{-E_g/(2k_{B}T)}]$. In the nondegenerate limit, $n\simeq D{k}_{B}T{e}^{-E_g/(2k_{B}T)}$. Thus the leading exponential activation law can coexist with an algebraic temperature prefactor.

The fraction of a specified population of chemical atoms that has become ionized. For singly ionized atoms with no external charge source, charge neutrality gives equal free-electron and ion number densities. Its equilibrium value depends on both temperature and total number density through the Saha ionization equation.

The depletion approximation replaces the junction's transition region by a region without mobile carriers, leaving only fixed ionized dopants. For a uniformly doped abrupt p-n junction, $\rho =-e{N}_A$ on the p side, $\rho =+e{N}_D$ on the n side and $\rho \simeq 0$ outside. Charge neutrality gives $N_A{x}_p=N_D{x}_n$. Integrating Poisson equation gives the electric field and the electrostatic band bending.

Let an isolated donor impurity admit an empty ionized state with statistical weight $g_0$ and a singly occupied neutral state with statistical weight $g_1$, with double occupation excluded. Its grand canonical ensemble weights are $g_0$ and $g_1{e}^{-\beta ({ϵ}_D-\mu )}$, so $r=g_1/g_0$ multiplies the occupied-to-empty ratio. A nondegenerate empty state and two allowed electron spin orientations give $r=2$; a simplified single-state occupation model uses $r=1$. With negligible electron holes and uncompensated donor impurities, charge neutrality and nondegenerate conduction occupation imply $n^2/(N_D-n)=(N_c/r)e^{-E_i/(k_{B}T)}$, where $E_i$ is the donor binding energy and $N_c$ the effective density of states.

Electric charge is an additive, signed property of matter that determines its coupling to the electromagnetic field. Write a particle's electric charge as $q$, or a collection's total electric charge as $Q$; the SI unit is the coulomb, $\mathrm{C}$. A stationary test particle in an electric field experiences electric force $\mathbf{F}=q\mathbf{E}$, while motion adds the magnetic part of the Lorentz force. Electric charge both produces an electric field and determines the response to one. For two stationary point particles in vacuum, Coulomb's law makes like signs repel and opposite signs attract. The choice of which sign to call positive is a convention, used consistently: protons carry $+e$ and electrons carry $-e$, where $e$ is the elementary charge. ^[ref].

Additivity means adding signed electric charges, rather than their magnitudes:for ordinary matter containing $N_p$ protons, $N_e$ electrons, and electrically neutral neutrons. A toy collection with seven protons and ten electrons has $Q=-3e$, however much neutral material accompanies it. Charge neutrality means $Q=0$; it does not mean an absence of charged constituents. Nor does it guarantee an absence of electric force: the spatial arrangement of positive and negative electric charges matters, as illustrated by an electric dipole. This makes a tiny imbalance in an otherwise nearly neutral object physically important. ^[ref].

Charge conservation states that a collection closed to transport of electric charge keeps its total signed $Q$. If $n$ electrons move from object A to object B without other exchange, then $\Delta Q_A=+ne$ and $\Delta Q_B=-ne$; the sum is unchanged. The amount of positive electric charge alone need not be conserved: creation of an electron and a Positron adds opposite electric charges with zero sum. These bookkeeping rules let us distinguish separation or transport of electric charge from creation of net electric charge. ^[ref].

A simple motivating experiment uses two charge-producing paddles, a conducting pail on an insulating support, a grounded outer shield and a high-input-impedance electrometer. Zero the pail, then isolate it. Rub the initially neutral paddles together, keeping their charged surfaces away from other objects. Lower each deeply into the pail without touching: the electrometer gives opposite signs. Inserting both together tests cancellation. For a separate contact test, insert one charged paddle without touching and withdraw it: the reading returns to zero. Repeat after zeroing, this time touching the inner wall before withdrawal: a residual reading shows transfer of electric charge. Zero the pail and reinsert the paddle to test how much electric charge remains on it. PASCO describes these procedures in Basic Electrostatics System manual, demonstrations 1A and 1B, printed pp. 13–16.

The crucial control is withdrawal without contact: a response during insertion can come from electrostatic induction, which redistributes the pail's existing electric charge. In the fixed-geometry approximation, the electrometer measures electric potential difference proportional to enclosed electric charge, with conversion set by effective capacitance; a voltage reading is not already a measurement in coulombs. Opposite readings and their cancellation test additivity and charge conservation within resolution. Leakage or accidental contact spoils that interpretation, and contact need not transfer all the paddle's electric charge. This macroscopic experiment cannot resolve elementary charge steps: that requires the much smaller particles of the oil drop experiment.

A finite electric dipole consists of equal and opposite electric charges separated by a nonzero displacement. It has charge neutrality but a nonzero electric dipole moment $\mathbf{p}=q\mathbf{d}$. An ideal point electric dipole is the mathematical limit $d=|\mathbf{d}|\to 0$ with $\mathbf{p}$ fixed, so $q$ must increase as $1/d$. Shrinking the separation while holding $q$ fixed instead makes the electric dipole moment vanish. The ideal electric dipole describes an exterior electric field; it does not reproduce the two near-charge singularities or the internal structure of a molecule. ^[ref].

Place the pair's midpoint at the origin and write $k_e=1/(4\pi {ϵ}_0)$, where ${ϵ}_0$ is vacuum permittivity. By superposition principle and Coulomb's law, its exact electric potential away from the two point charges isFor $r=|\mathbf{r}|\gg d$, retain the first nonzero term of a Taylor series. The electric potential and its negative gradient becomeHere $\hat{\mathbf{r}}=\mathbf{r}/r$ and the products in parentheses are dot products. These are exact exterior formulas for the ideal electric dipole and distant approximations for a finite pair. The electric field on the outer axis points along $\mathbf{p}$ with magnitude $2k_{e}p/r^3$; on the perpendicular bisector it points opposite $\mathbf{p}$ with magnitude $k_{e}p/r^3$. This directional structure is what a scalar total electric charge misses. ^[ref].

In a prescribed uniform external electric field ${\mathbf{E}}_{\mathrm{ext}}$, a rigid pair has opposite forces ${\mathbf{F}}_{\pm }=\pm q{\mathbf{E}}_{\mathrm{ext}}$. Its net force is zero, but summing their moments about the midpoint gives the torqueThe cross product makes the torque perpendicular to both the electric dipole moment and the external electric field. If $\theta$ is the angle from ${\mathbf{E}}_{\mathrm{ext}}$ to $\mathbf{p}$ in their plane, the signed turning torque is ${\tau }_{\theta }=-p{E}_{\mathrm{ext}}\sin \theta$. It tends to reduce a small positive $\theta$. ^[ref].

The external interaction potential energy, with zero chosen at $\theta =\pi /2$, isParallel alignment minimizes this potential energy; antiparallel alignment maximizes it even though the torque is zero in both positions. This omits the pair's constant internal binding potential energy and assumes a fixed permanent electric dipole moment. ^[ref].

A nonuniform external electric field can exert a net force on a neutral electric dipole. Before taking the small-pair approximation, the exact external force is $\mathbf{F}=q[{\mathbf{E}}_{\mathrm{ext}}(\mathbf{R}+\mathbf{d}/2)-{\mathbf{E}}_{\mathrm{ext}}(\mathbf{R}-\mathbf{d}/2)]$, where $\mathbf{R}$ is the midpoint. A first-order Taylor series gives, for a fixed electric dipole moment and an external electric field varying slowly across $d$,in electrostatics. The last equality uses the vanishing curl of the external electric field, with $\mathbf{p}$ held fixed while differentiating. Thus uniform external electric fields turn a misaligned electric dipole, whereas gradients can also move its midpoint. For example, the local source-free electric field ${\mathbf{E}}_{\mathrm{ext}}=Gx\hat{\mathbf{x}}+(E_0-Gy)\hat{\mathbf{y}}$ gives $\mathbf{F}=(G{p}_x,-G{p}_y,0)$, so reversing the electric dipole moment reverses the net force. This stipulated local model has zero divergence and zero curl.

A motivating experiment is Filsinger and colleagues' study of electrostatic deflection of polar molecules. Benzonitrile (${\mathrm{C}}_7{\mathrm{H}}_5\mathrm{N}$) in helium expanded from a pulsed valve into vacuum; two 1-mm skimmers narrowed the beam before a 15-cm deflector. A grounded trough and a high-voltage rod produced a vertical electric field gradient. Scanning a focused laser vertically downstream and recording benzonitrile ions from photoionization measured the beam's relative density. With voltage off, the profile spanned about 2 mm; applying voltage shifted and broadened it upwards. At 10 kV, many molecules lay outside the original profile. ^[ref].

Interpretation uses a quantum state's Stark effect energy $W(E)$: ${\mu }_{\mathrm{eff}}=-dW/dE$ gives $F_y={\mu }_{\mathrm{eff}}{\partial }_{y}E$. States whose energy falls with increasing electric field seek stronger fields. The voltage-off profile controls the mechanical aperture. Inferring state populations needs calculated trajectories, speed distributions and a rotational temperature model; profile displacement alone does not measure a permanent electric dipole moment uniquely. This is an electric Stark effect experiment, with photoionization occurring after deflection. ^[ref].

For the symmetric finite pair, the first omitted electric multipole expansion term after the electric dipole is third order, so typical relative errors scale as $(d/r{)}^2$. This stronger error estimate need not hold for an arbitrary neutral molecule with a nonzero electric quadrupole moment. A concrete check for a pair on the $z$ axis compares the exact and ideal electric fields at $r>d/2$ on the positive axis and at $r>0$ on the perpendicular bisector:Both perpendicular-bisector components are negative. At $r=5d$, the ratios are approximately $1.0203$ and $0.9852$: errors of $+2.03\%$ and $-1.48\%$ relative to the ideal electric field. The condition $r\gg d$ concerns observing the pair's own electric field; the separate condition for its external force approximation concerns variation of the applied electric field across the pair.

The electric dipole moment measures the first spatial moment of signed electric charge. For localized point charges $q_i$ at positions ${\mathbf{r}}_i$, or an integrable charge density $\rho (\mathbf{r})$ with a finite first moment, define^[ref].

The electric dipole moment is a vector with SI units $\mathrm{C}\mathrm{m}$: coulombs times length in metres. Opposite electric charges can cancel in $Q$ while contributing in the same direction to $\mathbf{p}$. For a neutral pair with positive electric charge $q>0$ at ${\mathbf{r}}_{+}$ and negative electric charge $-q$ at ${\mathbf{r}}_{-}$,Thus the electric dipole moment points from negative to positive electric charge, not along the attraction between the two point charges. ^[ref].

Changing the coordinate origin changes the position vectors, not the physical electric charge distribution. If the new origin is at $\mathbf{a}$ relative to the old origin, ${\mathbf{r}}_i^{\prime }={\mathbf{r}}_i-\mathbf{a}$, so direct substitution givesConsequently charge neutrality makes the electric dipole moment independent of the origin. For a system with nonzero total electric charge, always specify the origin: a single point charge at the old origin has $\mathbf{p}=0$ but ${\mathbf{p}}^{\prime }=-Q\mathbf{a}$. This is coordinate bookkeeping, not a newly created electric dipole. The full electric potential stays the same at the same physical point; the terms of its electric multipole expansion and the observation coordinates change together.

As a stipulated numerical example, put $+2\mathrm{nC}$ at $(1,0,0)\mathrm{mm}$ and $-2\mathrm{nC}$ at $(-1,0,0)\mathrm{mm}$. Their total electric charge is zero and $\mathbf{p}=(4\times 10^{-12},0,0)\mathrm{C}\mathrm{m}$. Translating the origin by $(3,-2,0)\mathrm{mm}$ leaves this electric dipole moment unchanged. Adding $+1\mathrm{nC}$ at the old origin instead gives $Q=1\mathrm{nC}$ and ${\mathbf{p}}^{\prime }=(1,2,0)\times 10^{-12}\mathrm{C}\mathrm{m}$. Distinguishing these two cases prevents the mistaken claim that every electric dipole moment is independent of the origin.

The usefulness of the electric dipole moment is compression: many internal electric charges determine one vector governing the leading distant electric potential of a neutral localized object when $\mathbf{p}\ne 0$. ^[ref]. A polar molecule need not literally contain two point charges; its atomic nuclei and electron density collectively give an electric dipole moment. For example, the chemical polarity of water reflects its bent molecular geometry, whereas the symmetric linear molecular geometry of carbon dioxide makes its permanent electric dipole moment vanish. ^[ref]. A collection of such electric dipole moments per unit volume gives electric polarization. ^[ref].

Symmetry of the excitation spectrum under exchange of conduction band electrons and valence band holes, after choosing a central zero of energy. In a two-band model with equal effective masses and equal degeneracy factors, charge neutrality in an intrinsic semiconductor fixes the chemical potential at that central energy. Indeed the Fermi-Dirac distribution obeys $1-f(-E;\mu =0)=f(E;\mu =0)$, so excitation populations are equal at every positive energy.

For $T\ll m_e$ and negligible chemical potentials, the combined equilibrium density of electrons and positrons divided by the photon number density isThe exponential suppression eventually makes pair equilibrium incompatible with the residual electrons required by charge neutrality.

Thermal electron-positron pairs in chemical equilibrium with photons have opposite chemical potentials when the one-particle energy includes rest energy. For negligible degeneracy and $k_{B}T\ll m_e{c}^2$, the zero-chemical-potential density per charge species is $n_0=2[m_e{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}{e}^{-m_e{c}^2/(k_{B}T)}$. In that nondegenerate limit, charge neutrality and the classical equilibrium product $n_{-}{n}_{+}=n_0^2$ give $n_{+}=(\sqrt{n_b^2+4n_0^2}-n_b)/2$, with net electron density $n_b$. Pairs are important when $n_0\gtrsim n_b$. At arbitrary temperature the zero-chemical-potential marker uses the integral $n_0=({\pi }^2{\hslash }^3{)}^{-1}{\int }_0^{\infty }{p}^2[\exp (\sqrt{m_e^2{c}^4+p^2{c}^2}/k_{B}T)+1{]}^{-1}dp$. Its ultrarelativistic limit is $3\zeta (3)(k_{B}T/\hslash c{)}^3/(2{\pi }^2)$. A nonzero net chemical potential shifts the exact abundance boundary; the zero-potential curve is an approximate crossover marker.

For a chosen homogeneous composition ${\mathrm{La}}_{2-x}{\mathrm{Ba}}_x{\mathrm{CuO}}_{4-\delta }$, assign formal oxidation states $+3$ to lanthanum, $+2$ to barium and $-2$ to oxygen. If $v$ is the mean formal copper oxidation state, charge neutrality requiresThe excess formal positive charge per copper relative to $+2$ is therefore $p_{\mathrm{formal}}=x-2\delta$. For stipulated $x=0.12$ and $\delta =0.01$, it is $0.10$; at $\delta =0.06$, it is zero. These are bookkeeping examples, not measurements of mobile carriers. The symbol $x$ here belongs to this declared formula, not the original mixed-phase recipe. A whole-pellet elemental ratio cannot simply be assigned to each constituent phase.

Charge neutrality alone does not locate holes. The lecture reviews oxygen-related carrier evidence and charge-$2e$ pairing from the Josephson effect and fluxoid and flux quantization in a superconductor. Pair charge does not establish phonon-mediated BCS theory for cuprate superconductors.^[ref]

The free electron model approximates conduction electrons by noninteracting particles with the displayed parabolic dispersion relation, filling one-particle states according to Fermi-Dirac distribution. A uniform positive background supplies charge neutrality. Periodic boundary conditions give one spatial state per $k$-space volume $(2\pi {)}^d/V_d$ in $d$ dimensions; the physical spin degeneracy is two. The model captures basic state counting and degeneracy but omits the periodic lattice potential, interaction corrections and detailed electronic band structure.

A sensitive charge neutrality test lets nominally neutral gas escape from an electrically isolated vessel connected to an electrometer. A net charge per escaping molecule would leave a measurable opposite charge on the vessel. Removing stray ions and charged particles is essential. Hydrogen and helium measurements constrain charge imbalance despite different particle binding and internal motions, providing evidence for velocity-independent additive charge rather than a charge proportional to relativistic particle energy. The 1960 experiment used this efflux principle. ^[ref]

For nondegenerate conduction bands and valence bands, charge neutrality gives $N_c{e}^{-(E_c-{\mu }_i)/(k_{B}T)}=N_v{e}^{-({\mu }_i-E_v)/(k_{B}T)}$. Taking logarithms yieldsFor single spin-degenerate parabolic extrema this becomes $(E_c+E_v)/2+(3k_{B}T/4)\ln (m_h^{\ast }/m_e^{\ast })$. Valley multiplicities or additional hole bands belong in $N_c,N_v$ and change the logarithm. The intrinsic chemical potential lies exactly at midgap only when the two effective density of states values agree.

The fraction of nuclei in a specified ionized state. For pure singly ionized hydrogen, $x=n_e/(n_H+n_e)$ by charge neutrality; a different species or normalization requires a different definition.