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Electrostatics studies time-independent electric charge distributions and their electric fields. In vacuum, its local equations in SI units areHere $\rho$ is charge density, ${ϵ}_0$ is vacuum permittivity, and $\Phi$ is electric potential. Gauss's law supplies the divergence equation; the static form of Faraday's law of induction supplies the zero curl. The Poisson equation becomes Laplace's equation where there is no electric charge. Prescribed electric charges give electric fields through Coulomb's law and the superposition principle; mobile electric charges on an electrical conductor instead require solving for their distribution together with the electric potential. ^[ref].

The useful distinction is between prescribed sources and prescribed boundary conditions. On a grounded electrical conductor, an external connection fixes its electric potential relative to an electrical reference potential and allows electric charge to enter or leave. On an isolated electrical conductor, total electric charge is fixed and its constant electric potential is an unknown. A neutral isolated electrical conductor can therefore have a nonzero local surface charge density through electrostatic induction. Equal electric potentials also do not imply equal electric charges: charge sharing between distant connected spheres gives a simple geometric example. ^[ref].

Electrostatics is useful because a spatial boundary value problem replaces the need to follow every moving charge carrier during charge relaxation. It describes the settled state of a capacitor and explains electrostatic shielding of sensitive equipment. Its assumptions must match the apparatus: stationary boundaries and sources, negligible time variation of the magnetic field, and enough time for charge relaxation. A metal carrying a maintained electric current through finite electrical resistance generally needs a nonzero internal electric field by Ohm's law; it is not a conductor in electrostatic equilibrium. Static macroscopic electric charge also does not mean that every electron is motionless. ^[ref].

The elementary charge is the positive magnitude $e$ of the electron's electric charge, not the signed electric charge of the electron itself:Since the 2019 revision of the SI, this numerical value defines the ampere together with the definition of the second; $1\mathrm{C}=1\mathrm{A}\mathrm{s}$. Thus today's numerical exactness is a choice of units, whereas measured electric charges still have experimental uncertainty. An oil drop experiment can test charge quantization and the apparatus calibration; it does not supply a new uncertain defining value of $e$ in today's SI. ^[ref].

For ordinary isolated chemical atoms, ions and charged objects whose net electric charge is counted through protons and electrons, charge quantization gives $Q=ne$ with signed integer $n$. A neutral chemical atom that loses two electrons becomes an ion of electric charge $+2e$. A toy object with $10^9$ excess electrons has $Q=-0.1602176634\mathrm{nC}$. Conversely, the scale $1\mathrm{nC}/e\simeq 6.24\times 10^9$ explains why laboratory electric charge often looks continuous: billions of microscopic increments fit into a small macroscopic amount. Averaged charge density is a continuum description, not a claim that each particle has an arbitrarily divisible electric charge.

The ordinary-matter qualifier matters. In the Standard Model, quarks have electric charges $+2e/3$ or $-e/3$, and antiquarks have the opposite signs. Confinement means these are constituents of hadrons, not isolated fractionally charged particles collected by an oil drop experiment. The usual valence quark sums give $2(2e/3)-e/3=e$ for a proton and $2e/3-2(e/3)=0$ for a neutron. Thus elementary charge should not be defined as the smallest nonzero electric charge of every constituent in particle theory. ^[ref].

The beautiful experimental idea behind the oil drop experiment is to determine a microscopic electric charge by watching a visible carrier move. In his 1913 paper, Millikan timed the descent and electrically driven ascent of individual oil drops, repeatedly changing a drop's electric charge through capture of air ions. His apparatus used parallel plates separated by $16\mathrm{mm}$, a roughly $5300\mathrm{V}$ source, and an optical scale to measure travel distances. He reported multiple relationships among a given drop's inferred electric charges. He also found that uncorrected Stokes's law gave a size-dependent inferred elementary unit, and introduced a pressure-dependent slip correction. These were measured relationships with calibration and model uncertainties, not exact readings of today's defined $e$. ^[ref].

A modern instructional implementation sprays low-volatility oil of known density through a plate aperture, illuminates the drops, and follows one through a calibrated optical scale. Measure the plate spacing $d$, electric potential difference $V$, air temperature and pressure. Time a fixed distance $L$ with the electric field off to get a terminal falling speed $v_f=L/t_f$. Then choose the field polarity to drive the same drop upward and measure $v_r=L/t_r$. Repeat timings at a fixed electric charge, and compare before and after an ion-capture event. The same drop is especially useful because its radius and mass need not be inferred anew for each charge state. ^[ref].

Here is the leading continuum model, retaining buoyancy. Assume a drop shaped as a sphere of radius $a$, oil and air densities ${\rho }_o$ and ${\rho }_a$, air dynamic viscosity $\eta$, and gravitational acceleration $g$. Define the downward effective weight $W$ and positive linear drag coefficient $k$ byAt terminal velocity, the net force vanishes. Falling with no electric field, the upward drag force balances weight minus buoyancy. Rising with the electric force upward, the drag force reverses and joins the downward effective weight:Consequently the uncorrected estimates areThe observed direction of response, with known field polarity, supplies the sign of $q$. A suspended drop gives the alternative balance $|q|E=W$, but it still requires a determination of $W$: balancing a drop alone does not determine its electric charge.

For an explicitly invented continuum example, stipulate $a=1\mu \mathrm{m}$, ${\rho }_o=900\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_a=1.2\mathrm{kg}{\mathrm{m}}^{-3}$, $\eta =1.8\times 10^{-5}\mathrm{Pa}\mathrm{s}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $q=-5e$ and a downward $E=10^5\mathrm{V}{\mathrm{m}}^{-1}$. The balances predict $W\simeq 3.69\times 10^{-14}\mathrm{N}$, $v_f\simeq 0.109\mathrm{mm}{\mathrm{s}}^{-1}$ and $v_r\simeq 0.127\mathrm{mm}{\mathrm{s}}^{-1}$. Substituting those speeds recovers $|q|=5e$ within rounding. This checks the algebra; it is not a precision prediction for real micrometre drops, which need a slip correction.

The inference has two distinct parts. Compare inferred electric charges or charge changes with a common unit, seeking $q_i\simeq n_{i}e$ within uncertainty, rather than demanding exact ratios of rounded measurements. Then establish the absolute scale through electric field, density, dynamic viscosity and distance/time calibration. If the observed steps are all even multiples, their common spacing alone cannot exclude an unobserved smaller unit. Air motion, evaporation and mistimed crossings can alter inferred electric charge without an ion-capture event. Small Reynolds number supports Stokes's law, but does not by itself justify its no-slip continuum coefficient: a significant ratio of air mean free path to drop radius requires a slip correction. Millikan's comparison of different drop sizes and pressures addressed that systematic limitation; the modern PASCO manual also includes the correction. Agreement supports charge quantization for the observed carriers within experimental resolution, rather than proving a universal statement about every possible particle.

Multiplying $e$ by the Avogadro constant gives the Faraday constant, $F=N_{A}e$, the electric charge per mole of singly charged carriers. This connects counting individual electric charges with bulk electrochemistry. ^[ref].

For uniform charge density ${\rho }_e=3Q/(4\pi R^3)$, Gauss's law gives the interior electric field $E={\rho }_{e}r/(3{\epsilon }_0)$. Integrate its vertical force density ${\rho }_e^{2}z/(3{\epsilon }_0)$ over the upper hemisphere: $\int zdV=\pi R^4/4$. Internal pair forces in that hemisphere cancel, so its net force is exactly the force from the lower hemisphere. The result is upward and has magnitude $3Q^2/(64\pi {\epsilon }_0{R}^2)$.

In macroscopic electromagnetism, free charge is the part of the charge density represented explicitly as ${\rho }_f$, separately from bound charge associated with the electric polarization of matter. The electric displacement field satisfies $\nabla \cdot \mathbf{D}={\rho }_f$. The word free does not imply zero force or zero energy: it distinguishes this source from the polarization response in the chosen material description.

Gauss's law relates the electric field on a closed surface to the net electric charge inside it. In SI units, with outward unit normal $\mathbf{n}$,Here $\rho$ describes a volume charge density, and ${\epsilon }_0$ is the vacuum permittivity; for ideal point or surface charges, use the enclosed electric charge directly rather than an ordinary volume-density integral. The left side is electric flux: the surface integral of the outward normal component of the electric field. Field directed outward contributes positively; field directed inward contributes negatively. The equality uses net signed electric charge, so positive and negative enclosed charges can cancel. The Gaussian surface is an imagined accounting boundary, not necessarily a material object. See OpenStax, explaining Gauss's law.

Where the electric field and charge density are smooth, the local statement is $\nabla \cdot \mathbf{E}=\rho /{\epsilon }_0$. The divergence theorem connects this divergence to the surface integral. This is the satisfying connection between local sources and global accounting: a surface can register how much net electric charge it encloses without revealing where that charge is located within it. A point charge requires care with the singular field at its position; the integral formulation remains useful with a boundary that avoids the charge.

The law fixes total electric flux, rather than every value of the electric field. In an uncharged region, a uniform electric field $\mathbf{E}=(E_0,0,0)$ crosses a cube of side $\ell$. Its left and right faces contribute $-E_0{\ell }^2$ and $+E_0{\ell }^2$; the other four faces contribute zero. Thus the net electric flux is zero while the electric field is nonzero everywhere. Charges outside a Gaussian surface can change the field on it without changing its net electric flux. Conversely, fields cannot generally be reconstructed from enclosed electric charge alone. Electrostatics also requires the electric field to be the negative gradient of an electric potential, with appropriate boundary information. This distinction is emphasized in Feynman lectures II.5, application of Gauss' law.

Spherical symmetry supplies the extra information needed for a particularly simple calculation. An isolated spherically symmetric charge distribution in vacuum produces a radial electric field, constant in magnitude on each concentric sphere. Its electric flux is therefore $4\pi r^2{E}_r(r)$, givingOutside the distribution, doubling $r$ quadruples the surface area and divides the radial electric field by four, while the electric flux stays $Q/{\epsilon }_0$. The inverse-square dependence is an area-dilution result. The electric field of a uniformly charged ball and electric field of a uniformly charged spherical shell show that the same total charge can give different interior fields. These symmetry arguments and their limits are developed in OpenStax, applying Gauss's law.

An infinite solid cylinder with radius $R$, prescribed uniform volume charge density $\rho$, and angular velocity $\omega \hat{\mathbf{z}}$ has current density $\mathbf{J}=\rho \omega r\hat{\mathbf{\varphi }}$. Treat a shell at radius $a$ and thickness $da$ as a solenoid carrying surface current $\rho \omega ada$. With no applied background magnetic field and permeability $\mu$,The same result follows from Ampère's circuital law, since $-d{B}_z/dr=\mu \rho \omega r$ inside.

The Maxwell equations are the field equations of classical electromagnetism. In SI units, in vacuum with prescribed charge density $\rho$ and current density $\mathbf{J}$, they areHere $\mathbf{E}$ is the electric field in volts per metre, $\mathbf{B}$ the magnetic field in teslas, $\rho$ in coulombs per cubic metre, and $\mathbf{J}$ in amperes per square metre. The constants are vacuum permittivity ${ϵ}_0$ and the vacuum value ${\mu }_0$ of magnetic permeability. “Vacuum” specifies the field laws without a material response; “source-free” additionally means $\rho =0$ and $\mathbf{J}=0$. Matter described by averaged fields instead requires Maxwell equations in matter and constitutive relations.

The first equation, Gauss's law, makes electric charge a source of electric flux. Gauss's law for magnetism sets the net magnetic flux through a closed surface to zero. The two curl equations, Faraday's law and the Ampère-Maxwell equation, couple changes of the electric field and magnetic field. Appropriate initial conditions and boundary conditions select a physical solution; the two divergence equations constrain the initial fields. ^[ref].

A membrane potential can coexist with near electroneutrality in the bulk cytosol and external chemical solution. The small imbalance producing the voltage lies near the cell membrane; most positive and negative electric charges balance in the bulk.^[ref]

For a separate original capacitor model, choose a total membrane capacitance $C=5\mathrm{pF}$ and $V_m=-70\mathrm{mV}$. These are stipulated parameters, not measured properties of a particular biological cell, and this voltage is not asserted to equal the preceding $E_{\mathrm{K}}$. Model the separated electric charge relative to an uncharged reference byUsing the elementary charge $e=1.602176634\times 10^{-19}\mathrm{C}$, this representsexcess monovalent negative-charge equivalents inside. These can represent a deficit of positive ions, rather than the arrival of that many negative ions.

Independently choose an internal reservoir volume $\mathcal{V}=1\mathrm{pL}=10^{-12}\mathrm{L}$ containing $140\mathrm{mM}=0.140\mathrm{mol}{\mathrm{L}}^{-1}$ potassium, with balancing negative electric charge. With the Avogadro constant $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$, its potassium inventory isThus, if charging were entirely due to outward monovalent potassium transport, about one in $38600$ of these potassium ions would suffice. This is a comparison with the chosen potassium inventory, not a fraction of all charged particles or a measurement of spatial charge density. No shape, surface area or thickness has been inferred from the chosen capacitance and volume. The calculation illustrates why substantial voltage does not require a strongly charged bulk cytosol.

Molecular dipole cancellation occurs when nonzero contributions to a molecule's electric dipole moment cancel under vector addition. In a bond dipole model, use ${\mathbf{p}}_{\mathrm{mol}}={\sum }_b{\mathbf{p}}_b$; adding magnitudes instead discards the molecular geometry. A symmetric linear molecule with two equivalent opposed bond dipoles has ${\mathbf{p}}_1=p_0\hat{\mathbf{x}}$ and ${\mathbf{p}}_2=-p_0\hat{\mathbf{x}}$, hence zero total electric dipole moment even though each chemical bond is polarized. Carbon dioxide provides the familiar molecular example; water's bent molecular geometry permits a nonzero total.^[ref]

For an original geometry calculation, give a fictitious three-site model $A{B}_2$ a central effective electric charge $+2q$ at the origin and two terminal electric charges $-q$ atHere $r$ is the stipulated bond length, $\theta$ the angle between the two connections, and $q>0$ is independent of any electronegativity scale. Equivalently, split the central $+2q$ into two colocated $+q$ contributions, one for each neutral bond dipole. With $p_0=qr$, the physical electric dipole moments point from the terminal sites towards the central site:The sideways components cancel; the components along the angle bisector add. Thus $|{\mathbf{p}}_{\mathrm{mol}}|=2p_0\cos (\theta /2)$. At $\theta =180^{\circ }$ the result is zero; at the stipulated $\theta =120^{\circ }$ it is exactly $p_0$. This is an angle example for a model, not the measured angle or electric dipole moment of water.

Opposition alone is insufficient: a linear molecule with unequal opposed bond dipoles of magnitudes $p_1$ and $p_2$ retains magnitude $|p_1-p_2|$. Conversely, symmetry can enforce exact cancellation beyond a particular bond dipole partition. If the complete charge density invariant under point reflection about the origin satisfies $\rho (\mathbf{r})=\rho (-\mathbf{r})$, contributions to $\int \mathbf{r}\rho (\mathbf{r})d^{3}r$ cancel in pairs. That proof uses the whole electric charge distribution, including lone pairs and atomic nuclei. It explains why molecular symmetry is stronger evidence for zero permanent electric dipole moment than a rule based only on tabulated electronegativity differences.

A permanent molecular electric dipole moment is specified relative to the molecule's orientation. An ensemble with random orientations can have zero mean laboratory vector while each molecule is polar. That orientational average differs from internal molecular dipole cancellation. The Stark effect and molecular beam electric resonance probe molecular electric dipole moments through state-dependent responses to an electric field.

The momentum relaxation time is the exponential relaxation time of mean directed momentum after the driving electric field is removed. In the simplest Drude model, independent collisions occur with probability $dt/\tau$ during a short time interval $dt$ and reset the mean velocity of the collided population to zero. The remaining fraction preserves its previous mean momentum. To first order in $dt$, averaging gives $d⟨\mathbf{p}⟩/dt=q\mathbf{E}-⟨\mathbf{p}⟩/\tau$. This is a population equation: it does not assert that every particle stops at equally spaced instants.^[ref]

For real materials, a collision that changes a particle's direction only slightly need not erase its directed momentum, so the momentum relaxation time need not equal the time between all scattering events. This transport time is also distinct from the bulk charge relaxation time $ϵ/\sigma$: one describes loss of directed motion, the other redistribution of charge density.

A spatial charge density describing the electric charge inside an atomic nucleus. In a static description centered on the nucleus, normalize it bywhere $Z$ is the atomic number and $e>0$ the elementary charge. A nuclear charge distribution is an average over a nuclear quantum state, rather than a photograph of immobile protons. Even when the average has spherical symmetry, individual nucleons need not sit on spherical shells. The proton charge radius and the internal charge density of the neutron also contribute: a distribution of pointlike proton centers is an approximation to a nuclear charge distribution, not its definition.

The experimental attraction is that an electron probes electric charge through the electromagnetic interaction, without itself feeling the nuclear force. Varying its momentum transfer changes the spatial scale tested. Long wavelengths mainly constrain a nuclear charge radius; shorter wavelengths can distinguish a diffuse edge from a sharper one through diffraction. Recovering the entire nuclear charge distribution from a finite range of noisy scattering data remains an inverse problem.

In the static charge density model, define the normalized Fourier transformHere $\mathbf{q}=(\mathbf{p}-{\mathbf{p}}^{\prime })/\hslash$ is the spatial momentum transfer divided by the reduced Planck constant: $q=|\mathbf{q}|$ has units of inverse length. Physical momentum transfer is $\hslash \mathbf{q}$. We keep $\hslash$ and the speed of light $c$ explicit; compute $q=|\mathbf{p}-{\mathbf{p}}^{\prime }|c/(\hslash c)$, expressing the numerator in $\mathrm{MeV}$ and $\hslash c$ in $\mathrm{MeV}\mathrm{fm}$, to obtain $q$ in inverse femtometres. The spatial $q$ is distinct from the invariant four-momentum transfer used in relativistic scattering cross-sections.

For spherical symmetry, take the polar axis along $\mathbf{q}$ and use spherical coordinates. With $\mu$ the cosine of the polar angle,Every radial shell contributes an amplitude. At $q=0$ all shells add with the same sign; at larger $q$ their contributions can cancel. This cancellation is the origin of a diffraction minimum in the nuclear charge form factor, rather than an absence of electric charge in the nucleus.

The static Fourier transform interpretation is appropriate for the nuclear model and kinematic regime specified here. A fully relativistic composite system requires frame and current conventions; its electric form factor is not universally the transform of a frame-independent three-dimensional charge density.

The root mean square radius of a nuclear charge distribution, referred to its center. Define its charge-weighted second moment byFor a spherical charge density with finite fourth moment, insert the Taylor series $\sin u/u=1-u^2/6+O(u^4)$ into the nuclear charge form factor:Thus a radius is a low-momentum transfer slope, not necessarily the position of a sharp edge. In this real static model,The factor of two between the slopes of $F$ and $|F{|}^2$ matters when interpreting a differential scattering cross-section. A proton charge radius is the corresponding size concept for the proton; the spatial arrangement of many nucleons introduces an additional scale in a heavier atomic nucleus.

Departure of an intrinsic nuclear shape from spherical symmetry. Here intrinsic means referred to axes attached to a model nuclear charge distribution. A nuclear quantum state with definite spin need not have the same laboratory charge density as that intrinsic distribution. In particular, the ground state of a deformed even-even nucleus can have $I=0$, a spherical laboratory expectation value, and a nonzero intrinsic nuclear quadrupole moment.

For an illustrative uniform charge density inside a spheroid, let the semiaxes be $(a,a,c)$ and keep its volume equal to that of a sphere of radius $R$:Thus $\eta >1$ gives a prolate spheroid, $\eta =1$ a sphere, and $\eta <1$ an oblate spheroid. These are three-dimensional equal-volume shapes: equal areas of their drawn cross-sections would be a different constraint. Real nuclear charge distributions have diffuse surfaces and need not be uniform, axially symmetric or confined to one shape.

Adding $40000$ monovalent counterions makes the combined genome-plus-ion electric charge nominally zero. It does not make the local electric field or free energy vanish: opposite charges have spatial distributions, and confining mobile ions costs entropy.

A definite screened continuation of b replaces the bare Coulomb potential by $e^{-\kappa r}/(4\pi ϵr)$, with $\kappa =1/{\lambda }_D$ and ${\lambda }_D=1\mathrm{nm}$. In the Debye–Hückel approximation, a uniformly distributed fixed charge density $\rho =Q/V_c$ satisfies $({\nabla }^2-{\kappa }^2)\phi =-\rho /ϵ$ inside, with zero fixed charge outside and matching potential and radial field at $R$. Its interior solution isThe screened charging free energy $U_{\kappa }=\frac{1}{2}\int \rho \phi dV$ is thereforeFor $R\gg {\lambda }_D$, the bracket approaches one: only charges within roughly a screening length contribute appreciably, yielding the local estimate $U_{\kappa }\simeq Q^2{\lambda }_D^2/(2ϵV_c)$. With the same ${ϵ}_r=80$ as before, $x=20$ gives bracket $0.92519$ andabout $173$ times smaller than the unscreened packed-sphere self-energy. This is a specified continuum estimate of packed-state charging free energy; a full packaging free energy also subtracts the screened unpacked reference and includes ion redistribution. It should not be presented as a unique exact packaging work fixed by the problem's few inputs.

The supplied counterion count also permits an independent estimate of the entropic contribution. For $N$ ideal confined ions, $F_{\mathrm{ion}}=-N{k}_{B}T\ln V+\text{constant}$, so compression from an accessible reference volume $V_0$ to $V_f$ costsFor example, approximate the original screening atmosphere as an annulus of thickness ${\lambda }_D$ around a rod. With $a=0.5\mathrm{nm}$, $V_0=\pi [(a+{\lambda }_D{)}^2-a^2]L=4.27\times 10^4\mathrm{n}{\mathrm{m}}^3$. Taking $V_f=V_c-V_D=1.35\times 10^4\mathrm{n}{\mathrm{m}}^3$ gives $\Delta F_{\mathrm{ion}}\simeq 1.91\times 10^5\mathrm{pN}\mathrm{nm}$. This useful order-of-magnitude ion-entropy model is distinct from the screened charging model; the two estimates must not simply be added as independent terms without a consistent ionic free energy functional. The rod cutoff, fixed DNA volume and screening-atmosphere geometry are rough geometric assumptions, not a microscopic reconstruction of the source sketch.

There is a further consistency limit. The nominal ion number density is $n=N/V_c=1.19\mathrm{n}{\mathrm{m}}^{-3}$. For a uniform ideal monovalent screening gas, ${\lambda }_D^2=ϵk_{B}T/(n{e}^2)$ would give ${\lambda }_D\simeq 0.309\mathrm{nm}$ at $300\mathrm{K}$, rather than $1\mathrm{nm}$; added salt would shorten it further. Thus the stated ion inventory and screening length cannot both be interpreted as a self-consistent dilute uniform screening gas with this dielectric. Bound counterions, nonuniform distributions and an effective phenomenological screening length can change that interpretation. Linear screening is itself only an estimate near highly charged DNA.

The specified screened-sphere charging model gives $8.00\times 10^5\mathrm{pN}\mathrm{nm}$, about $3.8$ times the supplied experimental magnitude. It is much closer than the unscreened result. The separate ideal counterion-compression estimate gives $1.91\times 10^5\mathrm{pN}\mathrm{nm}$, within about $9\%$ of that magnitude, but this numerical proximity does not validate its rough geometry.

Remaining differences can arise from the unpacked reference self-energy, counterion condensation, ion exchange with the surrounding reservoir, nonuniform charge density, nonlinear electrostatic screening, finite DNA thickness and the actual capsid shape. DNA bending energy, hydration forces and configurational entropy also contribute. The experiment measures a process-dependent work or inferred free energy, whereas the simple screened sphere computes just one model contribution. Neither a zero-net-charge argument nor agreement of one approximate number establishes an exact total energy.

An electrostatic plasma wave has $\delta \mathbf{E}=-i\mathbf{k}\varphi$. In the absence of externally imposed charge density, the Gauss law for the magnetized cold-plasma dielectric tensor is $\mathbf{k}\cdot \mathbf{ϵ}\cdot \delta \mathbf{E}=0$. Consequently,The antisymmetric terms of the magnetized cold-plasma dielectric tensor cancel in this scalar contraction. For oblique propagation this is the leading longitudinal condition in the large-refractive index approximation; a small transverse electric field is generally required by the full Maxwell equations at finite $n$. For parallel propagation an exactly longitudinal solution exists: $P=0$, hence ${\omega }^2\simeq {\omega }_{pe}^2$. It coincides with the $\theta \to 0$ limit of the finite-coefficient plasma wave resonance condition in c. At exactly $\theta =0$, $P$ factors out of the transverse dispersion relation, leaving the independent plasma oscillation rather than an isolated divergence of the transverse branches.

In a contractible region, the Maxwell equations $\nabla \cdot \mathbf{B}=0$ and the Poincaré lemma imply the existence of a magnetic vector potential $\mathbf{A}$ such that $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law of induction then givesThe expression in parentheses is a gradient there, so a scalar electric potential $\varphi$ can be chosen withThese statements are local, or global when the domain has the required topology and the electromagnetic fields are regular. A nonzero flux through a topologically nontrivial closed surface can obstruct a single globally defined vector potential, even though local expressions remain valid. The electric potential and magnetic vector potential are not unique: $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ and $\varphi \mapsto \varphi -{\partial }_t\chi$ leave both electromagnetic fields unchanged.

For the Lorenz gauge condition $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$, Gauss's law becomesThe Ampere-Maxwell law givesCancel the two gradient terms using the Lorenz gauge condition. HenceBoth potential equations have the same d'Alembert operator, with their respective charge density and current density sources. In a source-free region these are homogeneous wave equations with propagation speed $c$.

Write $\mu ={\mu }_0{\mu }_r$ and $ϵ={ϵ}_0{ϵ}_r$ for homogeneous scalar material constants, and use $\mathbf{J}=\sigma \mathbf{E}$. Taking the curl of Faraday's law of induction and substituting the Ampere-Maxwell law givesThe vector calculus identity for a double curl therefore yieldsThe second term represents electric current response, rather than displacement current. A dispersive $\sigma (\omega )$ is used in the frequency domain; its general time-domain counterpart is a temporal response convolution.

For the convention $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the plane wave equation isDecompose the electric field into components parallel and perpendicular to $\mathbf{k}$. With $S=\mu ϵ{\omega }^2+i\mu \sigma \omega$, the longitudinal eigenvalue is $S$ and each transverse eigenvalue is $S-k^2$; thus the determinant is $S(S-k^2{)}^2$. For a transverse electromagnetic wave, $\mathbf{k}\cdot {\mathbf{E}}_0=0$ andwhen ${\mu }_r=1$ and the background ${ϵ}_r\simeq 1$. A real dissipative electrical conductivity generally makes $k$ complex, with the decaying branch chosen for a passive medium.

For a cold, collisionless, unmagnetized plasma, treat ions as stationary and write $e>0$ for the magnitude of the elementary charge. The electron equation of motion is $-i\omega m_e\mathbf{v}=-e\mathbf{E}$, soThis reactive response includes no collision-induced Joule heating. Substitution into the transverse dispersion relation givesThe quantity ${\omega }_p$ is the electron plasma frequency. The refractive index is real above this cutoff; below it, the wave is evanescent. The electron response is represented here by $\sigma (\omega )$, so it must not also be counted a second time in the background permittivity.

For a nonzero longitudinal component, taking the dot product of the matrix equation with $\mathbf{k}$ instead gives $S(\mathbf{k}\cdot {\mathbf{E}}_0)=0$. The same cold plasma response makes $S=({\omega }^2-{\omega }_p^2)/c^2$, henceAt nonzero $k$, the remaining equation forces ${\mathbf{E}}_0$ to be parallel to $\mathbf{k}$, and Faraday's law of induction gives ${\mathbf{B}}_0=0$. These are electrostatic Langmuir waves, with no $k$ dependence of angular frequency in the zero-temperature model. Equilibrium neutrality does not forbid their perturbed charge density; Gauss's law requires such a density when $\mathbf{k}\cdot {\mathbf{E}}_0\ne 0$.

Let ${\rho }_q=3q/(4\pi a^3)$ be the uniform charge density and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Rigid rotation produces current density $\mathbf{J}={\rho }_q\mathbf{\omega }\times \mathbf{r}$. The magnetic dipole moment isSpherical symmetry gives $\int r_i{r}_j{d}^{3}r={\delta }_{ij}\int r^2{d}^{3}r/3$, while $\int r^2{d}^{3}r=4\pi a^5/5$. ThereforeIt is parallel to the angular velocity for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|{\mathbf{m}}_s|=|q||\omega |a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.

Axisymmetry permits a purely azimuthal magnetic vector potential, $\mathbf{A}=f(r)\sin \theta \hat{\mathbf{\phi }}$, with $A_r=A_{\theta }=0$. For the source's associated Legendre function convention, $P_1^1(\cos \theta )=+\sin \theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces toInside the sphere, $J_{\phi }={\rho }_q\omega r\sin \theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is ${\alpha }_0=-{\mu }_0{\rho }_q\omega /10$. Regularity at the origin and decay at infinity giveThere is no surface sheet electric current, so both normal and tangential magnetic field components are continuous at $r=a$. Usingthis is equivalent to continuity of $f$ and $f^{\prime }$. The two equations give $A={\mu }_0{\rho }_q\omega a^2/6$ and $C={\mu }_0{\rho }_q\omega a^5/15$. ThusIn the homogeneous expansion, ${\alpha }_1=A$, ${\beta }_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^{\ell }$ and decaying $r^{-\ell -1}$ terms with zero source at $\ell >1$ forces both coefficients to zero. The outside magnetic field is exactly the magnetic dipole field of ${\mathbf{m}}_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_{\phi }={\mu }_0{m}_s\sin \theta /(4\pi r^2)$, which is the printed cross-product magnitude expression.

For the flux coupling of a rotating charged sphere to a circular loop, take a loop of radius $b>a$ in the equatorial plane and use Stokes theorem on the regular full vector potential, evaluated around the rim. Its linked magnetic flux isWith $m_s=\pi a^2{I}_0$, the mutual inductance defined using this effective electric current isIntegrating an ideal magnetic dipole field through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.

For the prescribed time dependence, the quasistatic Faraday's law of induction result and the purely resistive loop approximation givePositive loop electric current is defined to produce a positive $z$ magnetic dipole moment, so the minus sign is Lenz's law. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's self-inductance. If that is not negligible, the governing equation is instead $L_b{\dot{I}}_b+R{I}_b=-{\dot{\Phi }}_b$; $R$ alone is insufficient for the exact dynamic response.