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The electric potential of a point charge $Q$ at the origin isFor the two charges forming the dipole,The first-order Taylor expansion at large $|\mathbf{x}|$ isand therefore, with the electric dipole moment $\mathbf{p}=Q\mathbf{d}$,to leading order. Taking the interaction of a second dipole with the corresponding electric field gives the stated electric dipole-dipole interaction

Let the lattice spacing be $d$ and write the central dipole asIts two horizontal neighbours have moment $p(\cos \theta ,-\sin \theta )$. For either one, the expression in parentheses, after extracting $d^{-3}$, isIts two vertical neighbours have moment $p(-\cos \theta ,\sin \theta )$, and each contributes insteadAdding all four nearest-neighbour interactions and using the Pythagorean trigonometric identity givesThe angle has cancelled, so the energy is independent of $\theta$.

Let $r$ be distance from the common centre. By Gauss's law, the electric field is radial and equalsTaking the electric potential to vanish at infinity and requiring continuity across each shell gives

The electrostatic energy is the integral of the electric-field energy density:Only the two nonzero-field regions contribute, soTherefore

These potentials give the uniform magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$ and zero electric field. With the signed frequency ${\omega }_c=qB/m$, the Lorentz force law becomesThusand explicitlyThe transverse trajectory is a circle about the fixed guiding centre $(x+\dot{y}/{\omega }_c,y-\dot{x}/{\omega }_c)$. The positive cyclotron frequency is

With Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$, the relativistic momentum and four-momentum areThe Newtonian formula $a=F/m$ fails because $p$ is not $mv$: both the magnitude and direction of $v$ affect $dp/dt$. Sincethe relativistic force isTaking the dot product with $v$ gives $F\cdot v=m{\gamma }^{3}v\cdot a$, and substitution yields the inverse relationThis is the requested sum of a vector parallel to $F$ and one parallel to $v$.

In the constant electric field, $dp/dt=qE$ and the particle starts with $p(0)=0$, so $p=qEt$. The relativistic energy-momentum relation impliesThus the speed increases monotonically but remains subluminal, and $v$ tends to $c$ in the direction of $qE$ as $t\to \infty$.

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 $\varphi (\infty )=0$.

A dielectric is an electrically insulating material whose bound positive and negative charges can be displaced slightly by an electric field, producing electric polarization without sustained conduction current. Its macroscopic electric response is described by its permittivity.

The temporal component found in part (a) gives directlyFor a constant electric field, this is the work done by the field along the particle trajectory:

In electrostatic equilibrium, a nonzero electric field inside a conductor would move its free charges. Hence $E=0$ in the conducting material and the potential is constant throughout each connected conductor. Immediately outside its surface the tangential electric field is zero. A Gaussian pillbox across the surface givesSince $E_{\mathrm{in}}=0$, the electrostatic boundary conditions at a conductor give

For the widely separated shells, let their charges be $Q_1,Q_2$. Their common potential requiresThus the charge sharing between distant connected spheres is

For the charge on connected concentric spherical shells, the potentials at the two radii areEquality forces $Q_1=0$. Henceso all charge lies on the exterior of the outer shell.

For the neutral sphere in the uniform field, the far-field condition gives $\alpha =-E$. Constancy of the potential on $r=R$ gives $\beta =E{R}^3$. ThereforeThe outward normal field at the surface isThe induced charge on a conducting sphere in a uniform electric field is consequentlyFinally,confirming neutrality.

A dielectric is a medium whose constituent charges become polarized by an applied electric field; it may also acquire a magnetization. Spatial variation of $P$ leaves uncompensated dipole charge, giving ${\rho }_{\mathrm{bd}}=-\nabla \cdot P$. Time-dependent polarization transports charge and aligned microscopic current loops contribute magnetization current, giving the bound charge and bound current

With $H=p_j{p}_j/(2m)$, the stated brackets giveandThus $m\ddot{\mathbf{q}}=e\dot{\mathbf{q}}\times \mathbf{B}$, the Lorentz-force equation with zero electric field. The magnetic interaction has been moved from the Hamiltonian into the noncanonical symplectic structure.

The dispersion relation is $\omega =ck$. At a perfect conductor the tangential electric field and normal magnetic field vanish. Thus $E_{ref}=-E_0\hat{x}{e}^{i(-kz+\omega t)}$ and $B_{ref}=-E_0\hat{y}{e}^{i(-kz+\omega t)}/c$. At the surface the magnetic field is $-2E_0\hat{y}/c$, so with outward normal $\hat{z}$ the surface current is $K=2E_0\hat{x}/({\mu }_{0}c)$ (real parts understood).

At the surface of a perfect conductor, the tangential electric field and normal magnetic field vanish. An incident plane electromagnetic wave is therefore accompanied by a reflected wave whose tangential electric field cancels the incident field at the surface.

Permittivity describes a medium's response to an electric field. In a linear isotropic medium, the electric displacement is $\mathbf{D}=ϵ\mathbf{E}$; vacuum permittivity is its vacuum value ${ϵ}_0$.

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.

In a constant magnetic field $B\hat{z}$ with no electric field, a charged relativistic particle moves in a circle in the perpendicular plane. Its angular frequency with respect to proper time iswhereas its angular frequency in an inertial coordinate time is ${\omega }_t=qB/(\gamma m)$. The orbit radius at speed $v$ is

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.

The even, spherically symmetric hydrogen ground state has no permanent electric dipole, so a uniform electric field gives no first-order energy shift. Odd-parity excited states mix in at second order, yielding a negative shift quadratic in the field.

The vector Dirac current has an even temporal component and odd spatial components under time-reversal symmetry. Invariance of its real minimal coupling therefore gives the displayed transformation of the gauge field, up to a gauge transformation. Differentiation at $x_T=(-x^0,\mathbf{x})$ yields $F_{\mu \nu }(x)\mapsto -t_{\mu }{t}_{\nu }{F}_{\mu \nu }(x_T)$, with $t=(1,-1,-1,-1)$. Thus the electric field is even and the magnetic field is odd.