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Use the constitutive relations $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{H}=\mathbf{B}/\mu$, where $\mathbf{D}$ is the electric displacement field, $\mathbf{H}$ the magnetic field strength, $ϵ$ the permittivity, and $\mu$ the magnetic permeability. Let ${\rho }_f$ be the charge density of free charge and ${\mathbf{J}}_f$ its current density. In this homogeneous medium, Gauss's law and the magnetostatic Ampère law are, respectively,Here $W$ is a volume with closed boundary $\partial W$, and $S$ is an oriented spanning surface whose boundary orientation follows the right-hand rule. Thus the electric field form is $\nabla \cdot \mathbf{E}={\rho }_f/ϵ$ and ${\oint }_{\partial W}\mathbf{E}\cdot \mathbf{n}dS=Q_f/ϵ$; the magnetic field form is $\nabla \times \mathbf{B}=\mu {\mathbf{J}}_f$ and ${\oint }_{\partial S}\mathbf{B}\cdot d\mathbf{\ell }=\mu I_f$.

The remaining static Maxwell equations areThese express Gauss's law for magnetism and the static limit of Faraday's law. The volume and circulation forms follow from the divergence theorem and Stokes theorem. The electric charge and electric current on the right of the macroscopic laws are free sources; material electric polarization and magnetization are already included in $\mathbf{D}$ and $\mathbf{H}$.

With charge density $\rho$, current density $\mathbf{J}$, and speed of light $c=({ϵ}_0{\mu }_0{)}^{-1/2}$, the four Maxwell equations in vacuum areThe electric potential and magnetic vector potential determine the electric field and magnetic field byThese definitions satisfy Gauss's law for magnetism and Faraday's law identically. The Lorenz gauge condition isSubstituting into Gauss's law and using the Lorenz gauge condition givesFor the Ampère-Maxwell equation, the double-curl identity givesThe two gradient terms cancel by the Lorenz gauge condition. Defining the d'Alembert operator with the sign convention $\square ={\nabla }^2-c^{-2}{\partial }_t^2$, the remaining equations areConversely, if the electric potential and magnetic vector potential satisfy these wave equations and the Lorenz gauge condition, the same substitutions in reverse give Gauss's law and the Ampère-Maxwell equation. The other two Maxwell equations follow identically from the definitions, proving equivalence within the Lorenz gauge.

The moving point charge has distributional charge density and current densityLetThe given retarded fundamental solution identity immediately implies that the retarded potentialssolve $\square \varphi =-\rho /{ϵ}_0$ and $\square \mathbf{A}=-{\mu }_0\mathbf{J}$ in the sense of distributions. Integrating the spatial Dirac delta function first givesand the analogous integral for $\mathbf{A}$ has numerator ${\mu }_{0}q{\mathbf{v}}_0(t^{\prime })/(4\pi )$.

Assume the physical trajectory has subluminal speed and the retarded time exists. At its root $t_r$, with ${\mathbf{n}}_r=\mathbf{R}(t_r)/R(t_r)$,The positive factor $\kappa =1-{\mathbf{n}}_r\cdot {\mathbf{v}}_0/c$ also proves uniqueness of retarded time for a subluminal source. The Dirac delta function substitution formula gives $\delta (g(t^{\prime }))=\delta (t^{\prime }-t_r)/\kappa$, soThese are the Lienard-Wiechert potentials. The denominator includes the time-substitution factor, rather than just the retarded distance. Away from the trajectory the electric potential and magnetic vector potential are ordinary smooth functions; on the trajectory their wave equations have the indicated distributional sources.

For completeness, the charge continuity equation ${\partial }_{t^{\prime }}\rho +{\nabla }^{\prime }\cdot \mathbf{J}=0$ also verifies the Lorenz gauge. Using $\nabla K=-{\nabla }^{\prime }K$ and ${\partial }_{t}K=-{\partial }_{t^{\prime }}K$, integration by parts in the retarded integrals giveswith the retarded boundary prescription. Thus the construction also has the required Lorenz gauge condition.

Fix the electric potential to vanish at infinity and assume a localized, nonsingular charge density of finite electrostatic energy. Assemble the charge by scaling ${\rho }_s=s\rho$ from $s=0$ to $1$. Linearity gives ${\varphi }_s=s\varphi$, so the work isThe factor $1/2$ prevents double counting of pair interactions. Using Gauss law, $\rho =-{ϵ}_0\Delta \varphi$, and integration by parts,The boundary integral at infinity vanishes for a localized distribution since $\varphi =O(r^{-1})$ and ${\partial }_r\varphi =O(r^{-2})$. Singular point-charge self-energies require a separate regularization and are not finite instances of this identity.

For the uniform solid sphere, spherical symmetry and Gauss law giveIntegrating inward from infinity and matching the continuous electric potential at $R$ givesThe value of the electric field at the centre is the zero vector.

Writing $Q=4\pi \rho R^3/3$, the electrostatic energy of a uniformly charged solid sphere follows either from the charge-potential integral or from the field integral including the exterior:For the nuclear scaling, $Q=Ze$ while constant volume per proton gives $R\propto Z^{1/3}$. Therefore the electric contribution scales as $Q^2/R\propto Z^{5/3}$ in this uniform-density model.

The uniform charge density is ${\rho }_e=3Q/(4\pi R^3)$. Spherical symmetry makes the electric field radial and constant in magnitude on each concentric sphere. A sphere of radius $r<R$ encloses charge $Q(r/R{)}^3$. Gauss's law therefore gives $4\pi r^2{E}_r=Q{r}^3/({\epsilon }_0{R}^3)$, soThis also defines the zero electric field at the centre by continuity.

For the aligned magnetic dipole field,Since ${\mathbf{e}}_z=\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta }$,The corotation velocity is $\mathbf{u}=\Omega r\sin \theta {\mathbf{e}}_{\varphi }$. The current density is therefore the advected charge density,

For a point charge, the charge density and current density areInsert these into the retarded electromagnetic potential. The argument of the Dirac delta function depends on $x^{\prime }$ throughAt its unique zero $x^{\prime }=y(t_{\mathrm{ret}})$, the Jacobian determinant contributes the factorSince the source is subluminal, this quantity is positive. Writing $\mathbf{R}=x-y(t_{\mathrm{ret}})$, $R=|\mathbf{R}|$, and evaluating $\mathbf{v}$ at $t_{\mathrm{ret}}$ gives the Lienard-Wiechert potentialsThe identity ${\mu }_0{ϵ}_0{c}^2=1$ equivalently gives $\mathbf{A}=\mathbf{v}\varphi /c^2$.

Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is

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.

For the thin spherical charge layer limit, smoothness of $h$ givesThus the shell's total electric charge tends to $4\pi {\rho }_0{a}^3{h}^{\prime }(1)$ while its thickness tends to zero. At each fixed $r>a$, the observation point eventually lies outside the shell, whereas at each fixed $r<a$ the interior electric field is unchanged. The limiting electric field away from the sphere is thereforeIts outward-minus-inward jump isPhysically, the limit is a uniformly charged ball of charge density ${\rho }_0$, together with an infinitesimally thin spherical shell of uniform surface charge densityIndeed its electric charge is $4\pi a^2\sigma$, and the standard normal electric field jump is $\sigma /{ϵ}_0$. The surface charge density can have either sign, or vanish, according to $h^{\prime }(1)$.

The order of limits matters. For every finite $\delta >0$, taking the two one-sided limits at $r=a$ first gives zero jump. The displayed nonzero jump is obtained by first taking $\delta \to 0$ at fixed points on either side, then approaching $a$. The value at the surface itself need not coincide with either trace of the limiting electric field.

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.

In the first octant, spherical polar coordinates have $0\le \theta \le \pi /2$ and $0\le \varphi \le \pi /2$. The charge density is $\lambda xy=\lambda r^2{\sin }^2\theta \cos \varphi \sin \varphi$. Hence the total electric charge isThe three factors are $a^5/5$, $2/3$, and $1/2$, respectively. For the middle integral, set $u=\cos \theta$ to obtain ${\int }_0^1(1-u^2)du=2/3$. Thus

For a vacuum electric potential, put $\Phi =R(r)Y(\theta )$ in Laplace's equation in spherical coordinates. Multiplication by $r^2/(RY)$ separates the radial and angular terms. With separation constant $\ell (\ell +1)$ and $x=\cos \theta$, the angular Legendre differential equation isRegularity at both polar axes selects nonnegative integers $\ell$ and the Legendre polynomials. The radial equation is $(r^2{R}^{\prime }{)}^{\prime }-\ell (\ell +1)R=0$, whose independent solutions are $r^{\ell }$ and $r^{-\ell -1}$. Hence the angularly regular axisymmetric vacuum solution isThe radial domain and its boundary conditions determine which terms are admissible.

For the given charge density, only $P_1(\cos \theta )=\cos \theta$ is forced. Write $\Phi =f(r)\cos \theta$. The interior radial Poisson equation is $f^{\prime \prime }+2f^{\prime }/r-2f/r^2=-4$. The identity for the Poisson equation with a spherical harmonic source gives the particular solution $f_p=-r^2$, since the radial operator takes $r^2$ to four. Finiteness at zero and decay at infinity therefore giveContinuity of the electric potential and its radial derivative at $a$ requires $Aa-a^2=B/a^2$ and $A-2a=-2B/a^3$. Solving gives $A=4a/3$, $B=a^4/3$, soAll other angular modes have zero coefficients after imposing regularity, matching and decay. At the origin the interior expression is $(4a/3)z-rz$, which extends with a continuous gradient and value zero. The angular charge density has no unique value exactly at that point; Poisson's equation holds classically for $r>0$ and as a weak solution through the origin. There is no additional point charge.

A material segment of the line with rest length $d{x}^{\prime }$ has laboratory length $dx=d{x}^{\prime }/\gamma$, while its charge $dQ$ is unchanged. ThusEquivalently the line-integrated four-current transforms from $(c{\lambda }^{\prime },0)$ to $(\gamma c{\lambda }^{\prime },\gamma u{\lambda }^{\prime })$. Gauss's and Ampère's laws in $S$ now give $E_r=\lambda /(2\pi {ϵ}_{0}r)$ and $B_{\phi }={\mu }_{0}I/(2\pi r)$, exactly the fields found in part (a). Charge density is frame-dependent even though the charge of the same set of particles is invariant.

In $S$, adding the stationary opposite line cancels charge density and electric field but contributes no current. HenceThe neutral current filament in a boosted frame is not neutral. Transform its line-integrated four-current:The same result follows by adding the original resting line $({\lambda }^{\prime },0)$ to the new negative line, whose primed density and current are $(-\gamma \lambda ,+\gamma \lambda u)$. In particular ${\lambda }^{\prime }-\gamma \lambda =-\gamma \lambda u^2/c^2$ because $\lambda =\gamma {\lambda }^{\prime }$.

Transforming the total fields givesFor positive $\lambda$, $\mathbf{u}\times \mathbf{B}$ points inward, explaining the electric-field sign. The primed Gauss and Ampère fields calculated from ${\lambda }_{\mathrm{tot}}^{\prime }$ and $I_{\mathrm{tot}}^{\prime }$ reproduce these expressions. Both a moving charge distribution and a moving current must be transformed; boosting just one constituent would give an inconsistent total.

Let an object be at rest in $S_0$, with proper charge density ${\rho }_0$ and no rest-frame current. In a frame moving at $\mathbf{v}$ relative to it, the object moves with velocity $-\mathbf{v}$. A simultaneous volume element is Lorentz-contracted along the motion, $dV=d{V}_0/\gamma$, while its charge is invariant. ThereforeThe object's four-velocity in this frame is $U^{\mu }=\gamma (c,-\mathbf{v})$, soA scalar proper density times a Lorentz four-vector transforms as a Lorentz four-vector, demonstrating the claim. With the alternative convention that $\mathbf{v}$ denotes the object's velocity in the observer frame, the spatial component is $+\gamma {\rho }_0\mathbf{v}$ instead. Summing the four-currents of different carrier species preserves this transformation law even when the combined current has no rest frame, as for a lab-neutral current-carrying wire.

The prescribed surface charge density is that of a uniformly polarized sphere and has zero total charge because $\int \cos \theta d\Omega =0$. Its electric dipole moment at $t=0$ isThe transverse components cancel by axisymmetry, while the $z$ component isThusThe angular surface distribution is a pure degree-one spherical harmonic. Hence its static external electric potential is exactly that of this dipole, with no higher electric multipole moments; it is not an assertion that the potential inside the finite sphere equals a singular point-dipole potential.

For right-handed uniform rotation about $+\hat{\mathbf{x}}$, the rigidly attached dipole isConsequently $|\ddot{\mathbf{p}}{|}^2=p_0^2{\omega }^4$ is constant. The electric dipole radiation formula givesThere is no extra factor of one-half: the squared sine and cosine contributions of the two perpendicular oscillating components sum to one. This is the leading electric-dipole approximation for radiation, requiring $\omega R/c\ll 1$ and observation in the radiation zone. The stated permanent surface distribution is interpreted as the actual charge producing the given external dipole; adding another dielectric screening factor would contradict the dipole moment the question asks to establish. The relative permittivity enters the induced-polarization problem in part (d), not this prescribed total dipole moment.

The information in the original question is insufficient to determine a unique spin-induced change of charge density. It specifies ideal dielectric filaments with a prescribed excess linear charge density, but not the charge carrier species, their spin state, charge carrier mobility, binding potential or relaxation mechanism. A classical line charge has no spin variable at all. The following conditional conclusions are supported by the fields already calculated.

For a carrier moving with its filament, the local magnetic field in its instantaneous rest frame is, for the transverse fields here,where $\mathbf{u}$ is that carrier's laboratory velocity. Exclude its filament's singular self-field and first retain only the prescribed straight convective motion.

In Case 1 there is a common rest frame in which both filaments and the excess charges are stationary and the interfilament magnetic field vanishes. Equivalently, in the laboratory ${\mathbf{B}}_A=\mathbf{v}\times {\mathbf{E}}_A/c^2$, so B's co-moving charges have ${\mathbf{B}}_{A,\mathrm{rest}}=0$. Thus uniform common translation alone creates no extra mutual rest-field spin splitting. This does not exclude intrinsic spin-orbit coupling of electrons moving within their atomic binding potentials; that material physics was never specified.

In Case 2 there is relative motion and no frame in which both source distributions are stationary. Put $E_{\ast }=\eta /(2\pi {ϵ}_{0}d)$ as in part (c). At the respective filament axes, the external fields experienced in each carrier distribution's own rest frame areFor the first expression B is at rest in $O$; for the second A is at rest in $O^{\prime }$ and the field comes from B moving at $-v\hat{\mathbf{x}}$. For a carrier with magnetic dipole moment ${\mathbf{\mu }}_s$, the interaction $H_s=-{\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}}$ distinguishes opposite transverse spin orientations. A field gradient could exert a spin-dependent force $\nabla ({\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}})$ on mobile carriers; spin relaxation could instead produce a spin polarization. These are possible responses, not consequences of charge density alone. In B's frame the displayed interaction is directly a magnetic-moment coupling to the moving source's field; treating the moving carrier's electrostatic orbital term in another frame also requires consistent binding and acceleration dynamics, including any applicable Thomas precession.

For example, if $\eta >0$ and mobile electron spins can relax, the lower magnetic-energy orientation has the electron's magnetic moment toward $+\hat{\mathbf{y}}$ and its spin toward $-\hat{\mathbf{y}}$. Oppositely oriented moments would experience opposite gradient forces. The sign changes if the source charge changes sign. None of these statements fixes a net charge redistribution without a confinement and response model, and the stipulated zero transverse size supplies no resolved “sides” on which charge could accumulate.

Therefore Case 1 has no new mutual spin effect from common bulk motion alone; Case 2 permits motion-induced spin-dependent interactions, but neither case determines an additional charge-density profile from the supplied data. In particular, the factor $\gamma$ relating proper and laboratory linear charge densities in part (b) follows from length contraction and is not an effect of spin-orbit coupling. If the charges remain pinned in an ideal dielectric, their imposed density remains the prescribed density in either case.

Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:In electrostatics, the electric field is the negative gradient of the electrostatic potential, $\mathbf{E}=-\nabla V$. Substitution into the differential equation form of Gauss's law gives Poisson's equation ${\nabla }^{2}V=-\rho /{ϵ}_0$. Consequently, in a region containing no charge density,which is Laplace's equation. Charges on the bounding conductors enter through the boundary conditions; they need not vanish for Laplace's equation to hold inside the empty gap.

Take the divergence of the proposed Ampère's law. Since the divergence of a curl vanishes identically, it implies $\nabla \cdot \mathbf{J}=0$. However, local charge conservation requires the continuity equationFor example, electric current entering a charging capacitor accumulates electric charge on its plates, so the charge density is time dependent. The proposed equation cannot describe that process.

From Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Thus $\nabla \cdot ({ϵ}_0{\partial }_t\mathbf{E})={\partial }_t\rho$, which supplies precisely the missing divergence. Adding the displacement current density gives the Ampère-Maxwell equation:Its divergence now yields the correct continuity equation. Through a fixed surface, the associated displacement current is $I_D={ϵ}_{0}d{\Phi }_E/dt$, where ${\Phi }_E$ is the electric flux. In a vacuum capacitor gap, this equals the conduction electric current charging the plates, even though no charges cross the gap. Charge conservation fixes the required divergence of the correction; by itself it does not exclude adding a further divergence-free vector field.