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In the inertial frame, rigid corotation gives $\mathbf{u}=\mathbf{\Omega }\times \mathbf{r}$. The ideal magnetohydrodynamics condition is the motional electric fieldApply Gauss's law in Gaussian units and use the divergence and curl of a cross product:The exterior field is produced by currents inside the star, so $\nabla \times {\mathbf{B}}_0=0$ there, while rotation with constant angular velocity has $\nabla \times \mathbf{u}=2\mathbf{\Omega }$. Hence the required Goldreich-Julian charge density isHere $n_q$ denotes electric charge per unit volume, as in the question. If each carrier has charge $q$, its signed number density is $n_q/q$.

Split the microscopic sources into free and bound parts and defineSubstitution of $\rho ={\rho }_{\mathrm{f}}-\nabla \cdot P$ into Gauss's law gives $\nabla \cdot D={\rho }_{\mathrm{f}}$. Similarly, using $J=J_{\mathrm{f}}+{\partial }_{t}P+\nabla \times M$ in the Ampère-Maxwell equation gives $\nabla \times H=J_{\mathrm{f}}+{\partial }_{t}D$. The homogeneous Maxwell equations are unchanged. Thus

Using $\rho ={\epsilon }_0\nabla \cdot E$, $E=-\nabla \varphi$, integration by parts, and $\varphi =0$ on the boundary givesPut $C=q/(4\pi {\epsilon }_0)$. Gauss's law gives the radial fieldTaking zero potential outside,The charge formula givesThe field formula gives the same result after integrating $4\pi r^2{E}_r^2$ over the two nonzero annuli.

Maxwell equations areTaking the divergence of the Ampère-Maxwell equation and using Gauss's law givesThereforebecause $J$ vanishes on the boundary. Componentwise integration by parts also giveswith the boundary term again zero.

Gauss's law gives $E=Q\hat{r}/(4\pi {\epsilon }_0{r}^2)$ outside any spherically symmetric charge distribution. The shell in (ii) has $Q=2\pi {\rho }_0(b^2-a^2)$, and differentiating its exterior potential gives exactly this field.

On a simply connected region, $\nabla \times E=0$ gives $E=-\nabla \varphi$. Then Gauss's law gives ${\nabla }^2\varphi =-\rho /{\epsilon }_0$.

For any bounded volume $V$ with piecewise smooth closed boundary and outward unit normal vector $\mathbf{n}$, Gauss's law in integral form isThus the outward electric flux equals the enclosed electric charge divided by the vacuum permittivity. To prove it, apply the divergence theorem to the electric field and use its given divergence:The classical proof assumes the requisite differentiability of the electric field. For piecewise smooth charge densities, one may apply the divergence theorem piece by piece and cancel the internal fluxes.

Set $r=R$ in the enclosed electric charge formula:Outside the entire charge distribution, Gauss's law gives $E_r(r)=Q/(4\pi {ϵ}_0{r}^2)$. Integrating $E_r=-{\phi }^{\prime }$ with $\phi (\infty )=0$ givesThus the exterior electric potential and electric field are exactly those of a point charge $Q$ placed at the center. This is the electrostatic shell theorem: outside a spherically symmetric charge distribution, only its total electric charge affects the electric field, not how that electric charge is distributed radially. If positive and negative electric charges cancel to give $Q=0$, both exterior quantities vanish.

By the assumed spherical symmetry of the electric potential, the electric field is radial: $\mathbf{E}=E_r(r){\mathbf{e}}_r$ for $r>0$. On a sphere of radius $r$, Gauss's law therefore readsPut $R=a(1+\delta )$. For $r<a$, the enclosed electric charge is $4\pi {\rho }_0{r}^3/3$. In the spherical shell, the factor $s^2$ in the volume element cancels $s^{-2}$ in the charge density, soIt follows thatAt the origin the first formula gives $\mathbf{E}=0$ without needing a radial unit vector. At both shell interfaces the adjacent formulas agree: for every fixed $\delta >0$, there is no surface charge concentrated on either sphere, and the electric field is continuous there. No sign restriction on $h^{\prime }$ has been assumed, so the shell charge density and the radial electric field can be negative.

Since the electric field is irrotational, introduce an electrostatic potential with $\mathbf{E}=-\nabla \phi$. Gauss's law becomes the Poisson equationThe curl equation is automatic because the curl of a gradient vanishes. Independence of $z$ makes $E_z$ constant, and its required decay forces it to be zero, so a $z$-independent potential suffices.

Differentiate the Green-function representation, valid away from the charged sheet:For $0<x<L$, the two inverse tangent limits at $y\downarrow 0$ are $\pi /2$ and $-\pi /2$; approaching from below reverses both signs. ThusThis agrees with the normal electric field jump obtained by applying Gauss's law to a thin pillbox enclosing unit surface charge density.

Let $\hat{\mathbf{r}}$ point transversely away from the line and define $\hat{\mathbf{\phi }}=\hat{\mathbf{x}}\times \hat{\mathbf{r}}$. Cylindrical Gauss's law in the line's rest frame givesThe transverse distance is unchanged by the axial boost: $r=r^{\prime }$. Applying the inverse of the given field transformation to a boosted uniform line charge,Here $\gamma =(1-u^2/c^2{)}^{-1/2}$ and all longitudinal field components vanish. Substituting back verifies ${\mathbf{B}}^{\prime }=\gamma (\mathbf{B}-\mathbf{u}\times \mathbf{E}/c^2)=0$ and ${\mathbf{E}}^{\prime }=\gamma (1-u^2/c^2)\mathbf{E}=\mathbf{E}/\gamma$. The sign of ${\lambda }^{\prime }$ determines the radial and circulation directions.

Use ideal large plates and neglect edge effects. A Lorentz boost normal to the plates does not contract either transverse plate dimension. Charge is invariant, so charge per plate area remains ${\sigma }^{\prime }=\sigma$. Gauss's law then givesThe plate separation contracts, but the infinite-plate field is independent of separation. It is therefore incorrect to infer a stronger field merely from the shorter gap. The ideal field is parallel to the boost, so there is no induced transverse magnetic field from this boost.

Gauge freedom. The magnetic vector potential satisfies $\mathbf{B}=\nabla \times \mathbf{A}$. The transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient is zero. For time-dependent fields, transform the electric scalar potential as $\varphi \mapsto \varphi -{\partial }_t\chi$ as well, so that $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ is unchanged. The scalar $\chi$ must have the regularity and single-valuedness appropriate to the physical region and its boundary conditions. Local values of $\mathbf{A}$ therefore depend on the gauge transformation.

Gauge and nonzero component. In the standard axisymmetric Coulomb gauge representative for the long-solenoid approximation,The divergence vanishes since $A_{\varphi }$ is independent of the azimuth. Inside, this agrees with the usual symmetric gauge representative $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. The Coulomb gauge condition alone does not force the other two components to vanish; the symmetry, regularity and customary boundary choice select this representative from the remaining gauge freedom.

Radial dependence. Write the interior magnetic field as $B_0\hat{\mathbf{z}}$, where $B_0={\mu }_{0}nI$ for a vacuum solenoid of solenoid turn density $n$. Its exterior magnetic field vanishes in the infinitely long idealization. Applying Stokes theorem to a coaxial circle gives $2\pi r{A}_{\varphi }(r)=\Phi (r)$. Therefore the Coulomb-gauge vector potential of a long solenoid isIt grows linearly inside, is continuous at $R$, and decays as $r^{-1}$ outside. Although its exterior curl is zero, its circulation is the enclosed magnetic flux $\Phi =B_0\pi R^2$. No regular single-valued gauge function removes it throughout the exterior while preserving this circulation.

Electric field of the rotating charged surface. For clarity retain absolute permittivity $ϵ$ and permeability $\mu$ in the cylinder formulae; setting them to unity gives the rationalized-unit convention in the question. If the stated unity means relative material constants in SI, use ${ϵ}_0,{\mu }_0$ instead. Assume prescribed charge densities in the laboratory frame and the nonrelativistic, steady-rotation model.

The axisymmetric surface charge is stationary as a distribution even while the material rotates. A Gaussian cylinder of length $L$ has enclosed charge zero for $r<R$ and $2\pi RL\sigma$ for $r>R$. Gauss's law therefore givesThe radial electric field jumps from zero to $\sigma /ϵ$ at the surface and then decays as $r^{-1}$. Its direction reverses for negative $\sigma$. The angular velocity does not enter this electric-field result for a specified surface charge.

Magnetic field of the volume-charged cylinder. A shell of radius $a$ and thickness $da$ has surface charge $\rho da$, azimuthal surface current $d{K}_{\varphi }=\rho \omega ada$, and the field of an infinite solenoid: $d{B}_z=\mu \rho \omega ada$ inside that shell and zero outside it. At radius $r<R$, only shells with $a>r$ contribute. Thus the magnetic field of a rotating charged solid cylinder isWith no applied background magnetic field, $\mathbf{B}=0$ outside. The profile is parabolic, largest on the axis and zero at the boundary. The sign follows $\rho \omega$. As a check, the current density is $J_{\varphi }=\rho \omega r$, and Ampère's circuital law gives $-d{B}_z/dr=\mu J_{\varphi }$, as required.

Electron impulse when the current is switched off. Choose positive $I$ so the initial magnetic flux is ${\Phi }_i={\mu }_{0}nI\pi R^2$ along $+z$. Faraday's law of induction gives, outside the solenoid,For an electron of electric charge $q=-e$, treat its position as fixed while the short pulse acts. The inductive impulse outside a switched solenoid isConsequently the nonrelativistic final velocity isThe induced electric field circulates in the positive azimuthal direction while a positive flux decreases, so the electron accelerates in the negative direction. Once the pulse has passed, there is no exterior field and it travels along the tangent at the impulse point, rather than continuing in a circular orbit. If the resulting momentum is relativistic, use ${\mathbf{v}}_f=c{\mathbf{p}}_f/\sqrt{m_e^2{c}^2+|{\mathbf{p}}_f{|}^2}$ instead. The sudden-switch result assumes negligible displacement during the pulse and the usual quasistatic electromagnetic induction approximation; “instantaneous” switching does not mean instantaneous propagation of fields to arbitrary distance.

One can check the sign using conserved canonical angular momentum: axial symmetry conserves $p_{\varphi }^{\mathrm{can}}=L_z+qr{A}_{\varphi }$. Initially $L_z=0$, and finally $A_{\varphi }=0$, so $L_z=q{\Phi }_i/(2\pi )=-e{\Phi }_i/(2\pi )$.

Interferometric measurement at constant current. Use a coherent electron beam split into two paths passing on opposite sides of a shielded long solenoid, then recombine them to form interference fringes. Keep the current constant during the passage and suppress leakage fields on the paths. In the Aharonov-Bohm effect, the phase difference due to the magnetic vector potential is the closed-loop integralThe overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta \phi /(2\pi )$, so the measured fringe shift is linear in $I$ and repeats whenA gauge transformation adds a gradient whose closed-loop integral vanishes for a single-valued $\chi$. Endpoint changes in electron wavefunction phase cancel in the interference measurement. The measured Aharonov-Bohm phase is therefore gauge invariant: it determines the flux, or equivalently the circulation of $\mathbf{A}$. In the chosen Coulomb gauge this implies $A_{\varphi }(d)={\mu }_{0}nI{R}^2/(2d)$, but the experiment does not select or measure a unique local vector potential independently of gauge.

Far field of the rotating sphere. Interpret uniform charge as uniform volume charge of total $Q$, and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. The magnetic dipole moment of a rigidly rotating charge distribution isIts axial component is $m_z=(\rho \omega /2)\int r_{\perp }^2{d}^{3}r$. For a solid sphere, $\int r_{\perp }^2{d}^{3}r=8\pi R^5/15$ and $\rho =3Q/(4\pi R^3)$, soThe magnetic field of a rotating charged spherical body is a static magnetic dipole field at large distances:In particular, $|\mathbf{B}|={\mu }_0|\mathbf{m}|\sqrt{1+3{\cos }^2\vartheta }/(4\pi r^3)$. Both the angular mean of the magnitude and its root mean square therefore scale as $\boxed{r^{-3}}$; the latter is $B_{\mathrm{rms}}=\sqrt{2}{\mu }_0|\mathbf{m}|/(4\pi r^3)$. The angular average of the vector is zero, which is different from average strength. Steady rotation leaves the charge and current distributions stationary, so this is magnetostatics, not a radiative $r^{-1}$ field. If uniform charge instead means a thin spherical surface, replace the dipole moment by $Q{R}^2\mathbf{\omega }/3$; the angular pattern and $r^{-3}$ scaling are unchanged.

In SI units, Gauss's law for a closed surface with outward unit normal vector isSpherical symmetry and the absence of an imposed external field make the electric field radial, $\mathbf{E}=E_r(r){\mathbf{e}}_r$. For $0\le r\le a$ (with the physical radius $a>0$),Outside the charge distribution this becomes $Q(a)=8\pi K{a}^5/15$. Applying Gauss's law on spheres givesBoth formulas agree at $r=a$, and the first tends to zero at the origin. Thus the total charge is $8\pi K{a}^5/15$. Negative $K$ reverses the field direction. If the radius is zero the distribution vanishes; a negative value of $a$ gives no nonnegative-radius support under the printed piecewise inequalities and likewise gives the zero distribution.

In SI units the Maxwell's equations in vacuum, with sources, areTaking the divergence of the last equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot j=0$, the local conservation of electric charge. Thus sources that violate charge conservation cannot satisfy all the Maxwell's equations. Equivalently, with this conservation law the curl evolution preserves the electric divergence constraint if it holds initially; Faraday's law similarly preserves $\nabla \cdot B=0$.

With $\rho =j=0$, use $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-\Delta E$ and substitute the two curl equations. This gives $-\Delta E=-{\mu }_0{ϵ}_0{\partial }_t^{2}E$. Applying the same calculation to $B$ yieldsA monochromatic plane wave has one angular frequency $\omega$ and constant-phase planes $\mathbf{K}\cdot r=\omega t-\text{constant}$, normal to the wavevector $\mathbf{K}$. Its wavelength is $2\pi /|\mathbf{K}|$.

For a nontrivial traveling wave with $\omega \ne 0$, substitution of the displayed electric field into the divergence equation gives $\mathbf{K}\cdot e_1=\mathbf{K}\cdot e_2=0$. Faraday's law is satisfied byIndeed ${\partial }_{t}B=\mathbf{K}\times E_{\varphi }$ and $\nabla \times E=-\mathbf{K}\times E_{\varphi }$. The magnetic divergence is zero, while Ampère's law reduces to $|\mathbf{K}{|}^2{E}_{\varphi }/\omega =(\omega /c^2)E_{\varphi }$, giving the dispersion relation. The amplitudes may be arbitrary transverse vectors; they need not be mutually orthogonal or have equal lengths. Their combination describes the polarization.

Ignoring edge fields, Gauss's law gives $E=\sigma /{ϵ}_0$ between the parallel plates. Their voltage is $V=Ea=\sigma a/{ϵ}_0$, and their charge magnitude is $Q=\sigma s^2$. Therefore

For a point charge $q_j$, Coulomb's law gives ${\mathbf{E}}_j=q_j\hat{\mathbf{r}}/(4\pi {ϵ}_0{r}^2)$. The flux through an oriented surface element isHere $d\Omega$ is its signed solid angle seen from the charge. A closed surface subtends total oriented solid angle $4\pi$ for a point inside and zero for a point outside. Thus the flux of each inside charge is $q_j/{ϵ}_0$ and each outside charge has zero net flux. By superposition,This is Gauss's law; a surface passing through a point-charge singularity is excluded.

Place charges $+Q$ and $-Q$ on the inner and outer conductors. In the vacuum gap, Gauss's law gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$. Their electric potential difference isConsequently the spherical capacitor hasFor a narrow gap $b-a\ll a$, this approaches ${ϵ}_0(4\pi a^2)/(b-a)$, the local parallel-plate approximation. For $b\to \infty$, it becomes the capacitance $4\pi {ϵ}_{0}a$ of an isolated sphere.

For a charging parallel-plate capacitor of area $A$, neglect fringing. The field in the gap is $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isAn Amperian loop around the feed wire can bound a surface cutting the wire, which has conduction current $I$, or a bulging surface passing through the dielectric gap, which has no conduction current. Including displacement current makes both surfaces give the same magnetic circulation. It is also required by charge conservation: the divergence of the Ampère-Maxwell equation and Gauss's law yields $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the continuity equation.