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The image charge for a grounded conducting sphere is obtained by the Kelvin transform:It lies inside the electrical conductor, outside the region where it represents the physical electric field. The electric potential isAt the physical point charge, the electric potential has its usual Coulomb's law singularity. There is no physical singularity at the image charge: the interior expression is identically zero.

For $|\mathbf{r}|=a$, direct expansion of squared distances givesConsequently the two terms cancel exactly, proving the zero Dirichlet boundary condition on the sphere. At infinity each term is $O(r^{-1})$, and more explicitlyThus the Dirichlet boundary condition at infinity also holds. The candidate solves the Poisson equation with the one physical point charge; uniqueness of Poisson equation, with these Dirichlet boundary conditions, identifies it as the required electric potential.

Put $k=b/a>1$ and ${\mathbf{e}}_z=(0,0,1)$. The complete construction of image charges between concentric grounded spheres consists of two families, indexed by every integer:The term $(q_0,{\mathbf{r}}_0)$ is the physical point charge; every other term is an image charge. In the first family $n\le -1$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. In the second family $n\le 0$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. Thus no image charge lies in the physical gap.

To derive and verify these families, reflection of an axial point charge $(Q,z)$ in a grounded sphere of radius $R$ replaces it by the image charge $(-RQ/z,R^2/z)$. Reflection in $a$ sends $(q_n,{\mathbf{r}}_n)$ to $({\tilde{q}}_{-n},{\tilde{\mathbf{r}}}_{-n})$; reflection in $b$ sends it to $({\tilde{q}}_{1-n},{\tilde{\mathbf{r}}}_{1-n})$. Successive reflections therefore generate exactly the listed families. The method of images givesThe physical point charge itself is excluded when evaluating this formula. On $r=a$, pair the first-family term $n$ with the second-family term $-n$; their electric potentials cancel by the single-sphere distance identity. On $r=b$, pair $n$ with $1-n$ instead. These reindexings are justified by absolute convergence, as follows.

For $n\to -\infty$, both source positions tend to the origin while their electric charges have size $O(k^n)$. On the closed gap their distances are eventually at least $a/2$, so both electric potential terms are $O(k^n)$. For $n\to +\infty$, both distances are at least a positive constant times $k^{2n}$ and the electric charges are $O(k^n)$, giving $O(k^{-n})$ terms. Both tails are bounded by convergent geometric series. This proves absolute convergence and uniform convergence of the tails on the closed gap. On any compact set avoiding the physical point charge, the differentiated tails obey the same type of bounds: for example, the large positive tail of the electric field is $O(k^{-3n})$. Thus termwise differentiation is valid and the sum is a harmonic function except at the physical point charge, where it has the correct Coulomb's law singularity. The boundary cancellations and uniqueness of Poisson equation complete the construction.

Let $r_0=|{\mathbf{r}}_0|$ and let ${\Omega }_0=({\theta }_0,{\phi }_0)$ describe its direction in spherical coordinates; write $\Omega =(\theta ,\phi )$. The given Dirichlet Green function between concentric spheres has normalization ${\nabla }^{2}G=-4\pi {\delta }^{(3)}$, so the electric potential is $\varphi =qG/(4\pi {ϵ}_0)$. On the inner sphere the unit normal from metal into the gap is $+{\mathbf{e}}_r$, hence its induced surface charge density isNear this boundary $r_{<}=r$ and $r_{>}=r_0$. The needed radial derivative isIt cancels the factor $2\ell +1$ in the Dirichlet Green function. Therefore the answer in spherical harmonics isThis spherical harmonic expansion converges on the boundary: after the angular sum, the two radial factors decay geometrically with ratios $a/r_0<1$ and $a{r}_0/b^2<1$, while the angular factors grow at most polynomially. This justifies the boundary derivative and the subsequent surface integral.

Translate the concentric spheres so that their centre is $(0,0,-a)$, take $b=a+d$, and keep the point charge fixed at $(0,0,h)$. Its distance from the centre is $r_0=a+h$. As $a\to \infty$, the upper portions of the inner and outer spheres approach the planes $z=0$ and $z=d$. The preceding total induced electric charge becomesThe outer sphere has total induced electric charge $Q_b=-q-Q_a$: a Gaussian surface within its conducting material has zero electric flux, and the grounded exterior is field-free by uniqueness of Poisson equation. Therefore $Q_b\to -qh/d$.

One can justify the total-charge limit without assuming that local convergence of electric fields preserves their surface integrals. Use induced charge as harmonic measure: if a harmonic function $w$ is one on the selected grounded boundary and zero on the other, Green's second identity with the electric potential givesIndeed, ${\nabla }^2\varphi =-q{\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)/{ϵ}_0$, $\varphi =0$ on the boundaries, and their unit normal outward from the gap satisfies $\sigma ={ϵ}_0{\partial }_n\varphi$. For the spheres the weighting harmonic function is $w_a(r)=a(b-r)/(r(b-a))$, exactly the factor already obtained. For the slab the two weighting harmonic functions are $w_0=1-z/d$ and $w_d=z/d$. Applying the same identity to the decaying grounded-slab electric potential, whose boundary term at lateral infinity vanishes, establishes directlyTheir sum is $-q$, and their relative sizes measure how close the point charge lies to each grounded plane.

For this radially layered coaxial cable, defineUse the solution regular at the axis with bounded exterior electric potential, so that the empty interior and exterior have the constant electric potentials fixed by their adjacent conducting cylinders.

There is no free charge in either electrical insulator or on their interface. Gauss's law and continuity of normal electric displacement field give $D_{\rho }=C/\rho$ in both layers. The electric potential difference fixes the constant becauseThe electric potential, including the empty regions bounded by the conducting cylinders, is thereforeBoth inner and outer Dirichlet boundary conditions hold, and the electric potential is continuous at the material interface, whereTaking the negative gradient of the electric potential yieldsThe radial electric field generally jumps at $\rho =c$, but $ϵE_{\rho }(c^{-})={ϵ}^{\prime }{E}_{\rho }(c^{+})$, as required by the absence of free surface charge density there. The conducting cylinders carry opposite linear charge densities $\lambda =2\pi C$ and $-\lambda$.

For an Amperian loop of radius $\rho$, the enclosed electric current is zero, $I$, or zero according as $\rho <a$, $a<\rho <b$, or $\rho >b$. The magnetostatic Ampère law therefore gives $H_{\phi }=I/(2\pi \rho )$ in the gap, and zero outside. Using the two magnetic permeabilities,The tangential magnetic field strength is continuous at the material interface, while the tangential magnetic field need not be. At each ideal conducting surface and at the material interface, the displayed electric field and magnetic field formulas specify one-sided limits; a single interface value is unnecessary when the limiting values differ.

Orient the unit normal $\mathbf{n}$ from $R_1$ to $R_2$ and write $\mathbf{H}$ for the magnetic field strength. For static electric fields, $\mathbf{E}=-\nabla \varphi$; the static Ampère's law gives $\nabla \times \mathbf{H}={\mathbf{J}}_f$. The curl product rule therefore gives the local Poynting flux and terminal power identityIntegrate over an oriented piece ${\Sigma }_R$ of the separating surface and apply Stokes theorem:Once all the wires are included, the first surface integral is ${\sum }_k{\varphi }_k{I}_k$: the electric potential is constant on each perfect conductor and the integral of its normal current density is its signed electric current. Equivalently, remove the wire cross-sections and apply Stokes theorem in the surrounding insulator. Each resulting hole has negative boundary orientation, and the magnetostatic Ampère law around it gives the positive contribution ${\varphi }_k{I}_k$.

For a closed separating surface the boundary term is absent. For a noncompact separating surface, take $R\to \infty$ with the usual isolated-circuit condition ${\oint }_{\partial {\Sigma }_R}\varphi \mathbf{H}\cdot d\mathbf{\ell }\to 0$. In either case,For a steady closed circuit the signed electric currents sum to zero, so shifting the reference of electric potential does not change the terminal power.

For a fully general noncompact surface, the hypotheses as printed need an additional condition at infinity. The exact result before imposing it isTime independence alone does not remove this boundary term or ensure a finite Poynting flux; for example, spatially constant crossed electric fields and magnetic fields carry a constant Poynting vector through an infinite plane. The stated terminal-power formula holds with the boundary condition specified above.

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.

Use the sign $s=\pm 1$ to label a point charge $sq$ at $s\mathbf{p}/(2q)$, with velocity $s{\mathbf{p}}^{\prime }/(2q)$. All quantities on the right below are evaluated at $t_0=t-r/c$. To first order,The denominator in each Liénard–Wiechert potential therefore expands assoIn the sum of the two electric potentials, the $q/r$ terms cancel and the finite terms add. This gives the retarded potentials of a point electric dipole:For the magnetic vector potential, the signed electric charge and signed velocity multiply to give the same sign for both contributions:Since ${\mu }_0{ϵ}_0{c}^2=1$,The ${\mathbf{p}}^{\prime }$ term in the electric potential arises from the moving-source denominator and remains finite in the point-electric dipole limit; discarding it would fail to give the full retarded electric field.

Set $K=1/(4\pi {ϵ}_0)$, $t_0=t-r/c$ and $\mathbf{n}=\mathbf{r}/r$, and evaluate $\mathbf{p}$, ${\mathbf{p}}^{\prime }$, ${\mathbf{p}}^{\prime \prime }$ at $t_0$. The chain rule givesIn particular, for $\mathbf{u}(t_0)$ and any positive integer $m$,Apply this gradient identity to the electric potential $\varphi =K[(\mathbf{n}\cdot \mathbf{p})/r^2+(\mathbf{n}\cdot {\mathbf{p}}^{\prime })/(cr)]$:The magnetic vector potential is $\mathbf{A}=K{\mathbf{p}}^{\prime }/(c^{2}r)$, with time derivative ${\partial }_t\mathbf{A}=K{\mathbf{p}}^{\prime \prime }/(c^{2}r)$. Thus the electric field $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ isFor the magnetic field, use the curl product rule and the chain rule:Consequently the fields of a time-dependent electric dipole includeThese formulas hold for $r>0$. The $r^{-3}$ electric field is the retarded electric dipole field, the $r^{-2}$ terms are induction contributions, and the $r^{-1}$ terms describe electric dipole radiation. Retaining the dependence of $t_0$ on position in every spatial derivative is what produces both the induction and radiation contributions.

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.

Take zero electric potential at infinity and put $K=(4\pi {\epsilon }_0{)}^{-1}$. By spherical symmetry and Gauss's law, the electric field is radial, with magnitude determined by the total enclosed electric charge. Thus, away from the idealized shell surfaces,Integrating $E_r=-d\Phi /dr$ from infinity, or adding the three spherical-shell potentials, givesThe expressions agree at each interface, so the potential is continuous everywhere, including the origin. The electric field has different interior and exterior limits at each infinitely thin charged surface, with jump equal to surface charge density divided by ${\epsilon }_0$. An ideal sheet has no single classical field value on the sheet itself; the displayed adjacent limits specify the field there. In conducting material of nonzero thickness the internal field is zero.

The electrostatic energy is the integral of ${\epsilon }_0|\mathbf{E}{|}^2/2$. There is no contribution from $r<a$, and spherical integration givesThe dependence on $b$ cancels because the magnitude of enclosed electric charge is $|q|$ on both sides of the middle shell, even though the field direction changes.

In a steady state, Maxwell's equations imply $\nabla \times \mathbf{E}=0$, since the magnetic field is time independent. The divergence of the steady Ampère equation gives $\nabla \cdot \mathbf{j}=0$. With uniform electrical conductivity, Ohm's law is $\mathbf{j}=\sigma \mathbf{E}$, soFor the usual straight-wire model, the cylindrical side is insulating and the two flat end faces are equipotentials. Put $\mathbf{E}=-\nabla V$. The electric potential solves Laplace's equation, with constant end values and zero normal derivative on the side. The linear axial potential $V(z)=V(0)-\Delta Vz/l$ satisfies all these conditions. It is unique: the difference $h$ of two solutions has zero end values and zero side normal derivative, and integration by parts gives $\int |\nabla h{|}^{2}dV=0$. ThusThis proves uniform current in a straight homogeneous wire with the usual contact conditions. Integrating the electric current density over a cross-section and using Ohm's law then givesThe second expression is the rate of Ohmic heating.

Strictly, steady current and uniform conductivity alone do not force uniformity; the contact conditions are implicit in the wire model. For example, let $\chi$ be a nonconstant transverse Neumann eigenfunction on the disk, satisfying ${\Delta }_{\perp }\chi =-{\lambda }^2\chi$ and zero normal derivative at its edge. Then $V=-E_{0}z+C\chi \cosh (\lambda (z-l/2))$ is harmonic and preserves an insulating cylindrical side, but its nonconstant end potentials produce nonuniform current. Its added axial current has zero cross-sectional mean. This supplies a counterexample if arbitrary contacts are permitted.

A bend generally redistributes the electric current density: the inner side of the bend carries more current than the outer side, as illustrated by $j_{\theta }\propto 1/r$ in an annular conductor. Exact resistance and local Ohmic heating can therefore change. For a slender, gently curved wire the straight-wire formulas remain the leading approximation; the identity $\dot{Q}=I^{2}R$ remains valid for the total steady dissipation.

For the charge-generated electric field, use the electric potentialThe vector potential is time independent, so its usual additional contribution $-{\partial }_t\mathbf{A}$ is zero. The continuity equation now givesThus the second term in part (iii) is exactly the displacement current contribution ${\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$, andThis is the Ampère-Maxwell equation. The constructed field also has zero curl, consistent with the time-independent magnetic field and Faraday's law. A static addition to the electric field does not change this conclusion; the integral construction selects the fields sourced by the given densities rather than an independently added radiation field.

Take the electric dipole moment along the polar axis, $\mathbf{p}=p\hat{\mathbf{z}}$. The electric field is minus the gradient of the electric potential; in spherical polar coordinates,Since $\mathbf{p}=p(\cos \theta {\hat{\mathbf{e}}}_r-\sin \theta {\hat{\mathbf{e}}}_{\theta })$ and $\mathbf{p}\cdot {\hat{\mathbf{e}}}_r=p\cos \theta$, the radial and angular components combine asThe origin is excluded because the point-dipole potential is singular there.

Spherical symmetry and the integral Gauss law give $4\pi r^2{E}_r(r)=Q/{ϵ}_0$ for $r>R$. Together with the stipulated zero interior field,The electric field jumps across the ideal charged surface; its value exactly on that surface is not needed for the volume integral. The electrostatic energy of a uniformly charged spherical shell isAs a check, the shell has constant electric potential $Q/(4\pi {ϵ}_{0}R)$, with zero reference at infinity, so $U=Q\varphi (R)/2$ gives the same result.

The differential Gauss law is $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. On a large ball $B_R$, the product rule and $\nabla \varphi =-\mathbf{E}$ giveThe divergence theorem consequently expresses the electrostatic energy asFor a localized charge distribution with the electric potential chosen to vanish at infinity, $\varphi =O(R^{-1})$ and $\mathbf{E}=O(R^{-2})$, so the surface term is $O(R^{-1})$ and tends to zero. More generally, it suffices to assume that this boundary integral vanishes. Passing to all space givesThe physical decay and potential reference are relevant: $\mathbf{E}\to 0$ by itself does not specify the additive constant in $\varphi$. Adding a constant $C$ changes the charge-based expression by $CQ/2$ when the total charge is $Q$, while leaving the field unchanged. The zero-at-infinity potential is the usual reference implicit in this all-space electrostatic energy formula.

Write $b=|{\mathbf{r}}_0|$, $r=|\mathbf{r}|$, $\mathbf{P}=\mathbf{r}-{\mathbf{r}}_0$ and $P=|\mathbf{P}|$. Initially assume $0<b<a$ and a nonzero current dipole. The possible singularities of the continued formulas occur when $r=0$, $P=0$ orBy the Cauchy-Schwarz inequality, $D\ge 0$. Away from the two endpoints it vanishes exactly when $\mathbf{P}$ is antiparallel to $\mathbf{r}$, namely on the open segment $\mathbf{r}=s{\mathbf{r}}_0$, $0<s<1$. Thus both formulas are regular in the physical exterior $r>a$; the singularities requested here belong to their analytic expressions continued into the sphere.

To determine the order on the segment, let $\mathbf{e}={\mathbf{r}}_0/b$ and approach an interior point as $\mathbf{r}=z\mathbf{e}+\rho {\mathbf{e}}_{\rho }$, with $0<z<b$ and ${\mathbf{e}}_{\rho }\perp \mathbf{e}$ a unit transverse direction. Expansion givesConsequently the leading singular terms in the electric potential and magnetic scalar potential areThus a dipole with a nonzero transverse moment has first-order line singularities on the open segment: their generic growth is $1/\rho$, the inverse distance to that segment. Angular coefficients can vanish along particular approach directions without removing the line singularity.

At the source point put $\mathbf{r}={\mathbf{r}}_0+P\mathbf{n}$ with a fixed direction $\mathbf{n}\ne -\mathbf{e}$. The explicit dipole term givesThe source therefore has a second-order electric dipole singularity and, for a nonradial moment, a first-order magnetic singularity. Approaches tangent to the singular segment need not have these fixed-angle scalings.

At the origin put $\mathbf{r}=r\mathbf{n}$, with $\mathbf{n}\ne \mathbf{e}$. The leading terms areThe electric expression has a first-order point singularity there. The magnetic expression has a generally direction-dependent finite radial limit, so the origin is a nonremovable endpoint singularity of radial order zero, rather than an isolated $1/r$ pole. It can still be unbounded along approaches whose angle to the segment also tends to zero. Stating the approach matters for this nonisolated singularity.

For a radial current dipole, $\mathbf{Q}=q\mathbf{e}$, the transverse line terms cancel and the electric line singularity is removable. Its only electric singularities are the first-order term $-q/(4\pi \sigma br)$ at the origin and the second-order dipole at ${\mathbf{r}}_0$. The magnetic potential is identically zero because $\mathbf{Q}\times {\mathbf{r}}_0=0$. Finally, if ${\mathbf{r}}_0=0$, the two terms in the given exterior electric expression combine toThere is then just a second-order electric singularity at the origin. A zero dipole moment makes both potentials zero.