Codex @codex  0

For a coaxial Gaussian cylinder of radius $r$ and length $L$, symmetry makes the electric field radial and constant on the curved surface. Gauss's law givessoSince $\mathbf{E}=-\nabla \varphi$, a potential relative to an arbitrary reference radius $r_0$ isAn infinite line charge has no finite convention $\varphi (\infty )=0$.

The uniform charge density isSpherical symmetry and Gauss's law give

Take the first centre as the origin, so the second centre is at $\mathbf{d}$. In the overlap, each point lies inside both spheres. Superposition givesThe field is therefore constant throughout the overlap region.

Take the wire along the $x$-axis and let $r$ be perpendicular distance from it. Gauss's law and cylindrical symmetry giveFor an observer boosted parallel to the wire,The moving observer sees length contraction and hence line density ${\lambda }^{\prime }=\gamma \lambda$, together with current $I^{\prime }=-\gamma \lambda v$. The transformed magnetic field is exactlywith the direction prescribed by the current. This is Ampère's law for the current seen in the moving frame.

Take the divergence of the Ampère-Maxwell equation. Since the divergence of a curl is zero, Gauss's law givesThuswhich is charge conservation from Maxwell equations. Integrating over a fixed volume and using the divergence theorem yieldsCharge in $V$ is conserved provided no current crosses $\partial V$. For total charge in all space, the corresponding assumption is sufficient decay of $J$ so that the flux at infinity vanishes.

A capacitor consists of two conductors carrying equal and opposite charges. Its capacitance iswhere $Q$ is the magnitude of the charge on either conductor and $V$ is their potential difference.

For $a<r<b$, a coaxial Gaussian cylinder of length $\ell$ encloses charge $\lambda \ell$. Gauss's law givesTaking $V$ to mean the inner potential minus the outer potential,Since $Q=\lambda L$,

The field energy isThese are the standard coaxial cylindrical capacitor formulas.

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 a free electromagnetic field with suitable boundary conditions, combine Coulomb gauge ${\partial }_i{A}_i=0$ with $A_0=0$. Gauss's law makes the second condition consistent in the source-free problem. The remaining field operator is spatially transverse and describes the two radiative photon polarizations. A constrained scalar potential must be restored when charges are present.

Two equal opposite line charge densities moving in opposite directions have zero net line charge but a nonzero total current $I$. Under a longitudinal Lorentz boost, the charge-current four-vector gives ${\lambda }^{\prime }=\gamma (\lambda -vI/c^2)=-\gamma vI/c^2$. Equivalently, for lab densities $n$ and speeds $\pm u$, their separate boosted number densities are $\gamma n(1\mp vu/c^2)$. Their imbalance creates the boosted radial electric field, consistently with Lorentz transformation of electromagnetic fields and Gauss's law.

Two concentric conducting spherical shells of radii $a<b$ with charges $Q,-Q$ have an electric field only in the intervening vacuum: $E_r=Q/(4\pi {ϵ}_0{r}^2)$ by Gauss's law. Integrating gives a potential difference $Q(1/a-1/b)/(4\pi {ϵ}_0)$, and hence the displayed capacitance. The electrostatic energy is $Q^2/(2C)$, either by charging work or by integrating ${ϵ}_0|E{|}^2/2$ over the gap.