Electromagnetism course of the University of Oxford (B7.2)

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.

Let $\mathbf{n}=\mathbf{r}/a$ be the unit normal from the electrical conductor into the vacuum. The electrostatic boundary conditions at a conductor, together with the vanishing interior electric field, giveWrite $u=\cos \theta$ in spherical coordinates, $D=|\mathbf{r}-{\mathbf{r}}_0|$ on $r=a$, and $z^{\star }=a^2/r_0$. Differentiating the electric potential at fixed angles givesHere the distance identity in the preceding answer was used before simplifying. Hence the induced surface charge density isFor positive $q$ this surface charge density is negative everywhere, with largest magnitude nearest the point charge.

Integrate the surface charge density using the surface integral in spherical coordinates:Since $r_0>a$, the elementary integral isSubstitution yields the total induced electric chargeThe equality with the image charge also follows by applying Gauss's law to the exterior electric field: the physical and fictitious descriptions give the same electric flux through a sphere enclosing the conductor but excluding the physical point charge.

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.

Choose a spherical Gaussian surface of radius $s$ with $a<s<r_0$. Its physical enclosed electric charge is the total induced electric charge $Q_a$ on the inner sphere. The method of images produces the same electric field on that Gaussian surface, so Gauss's law identifies $Q_a$ with the sum of the enclosed image charges. These are the first family with $n\le -1$ and the second family with $n\le 0$:Summing these geometric series givesThe electric field in the empty central cavity vanishes by electrostatic shielding, so it contributes no additional inner-face surface charge density. The formula tends to $-q$ as $r_0\downarrow a$ and to zero as $r_0\uparrow b$, as expected when the point charge approaches one grounded boundary or the other.

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.

Write the constant spherical harmonic $Y_{00}=C$. Its orthonormality gives $4\pi |C{|}^2=1$, independently of its phase convention. Orthogonality to $Y_{00}$ givesOnly the $\ell =m=0$ term contributes to the surface integral of the surface charge density. Multiplying by the surface-area factor $a^2$ givesThus the spherical harmonic calculation reproduces the total induced electric charge:

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.

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 $\mathbf{H}$.

Allow time dependence while keeping the constitutive relations linear, isotropic, nondispersive, and independent of time. The Maxwell equations in matter giveFor the Poynting vector $\mathcal{P}=\mathbf{E}\times \mathbf{H}$, the divergence identity yieldsThe first two terms are exact time derivatives. Taking the zero-field energy density as zero gives the electromagnetic energy densityConsequently the Poynting theorem is the continuity equationIn integral form,The left side is the rate of increase of stored electromagnetic energy plus outward Poynting flux. The right side accounts for work delivered to the free electric current; in an Ohmic medium ${\mathbf{J}}_f\cdot \mathbf{E}$ is the local rate of Joule heating.

Work in cylindrical coordinates and use the assumed independence of $\phi$ and $z$. In each material region, the zero curl of the static electric field givesThe tangential electric field is zero at a perfect conductor. The dielectric-interface boundary conditions make its tangential components continuous at each material interface, so $E_z=E_{\phi }=0$ throughout the gap. They also vanish in the empty interior and exterior for the regular solution with no applied external electric field. HenceThis specifies an axis of direction: the sign of $E_{\rho }$ depends on the sign of $V$.

For the static magnetic field, Gauss's law for magnetism impliesThus $\rho B_{\rho }$ is constant in each layer. Continuity of the normal magnetic field across every interface makes that constant global, and regularity at the axis makes it zero. Next use $\nabla \times \mathbf{H}=0$ away from the conducting surfaces. Its $\phi$ component gives $d{H}_z/d\rho =0$. There is no azimuthal surface current density, so $H_z$ is continuous across the conducting surfaces and the material interface. The decay of the exterior magnetic field at infinity therefore forces this global constant to zero. The only remaining component isThe magnetostatic Ampère law fixes its sign: positive inner electric current in the $+z$ direction produces positive $B_{\phi }$ for positive magnetic permeability.

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.

Since ${\mathbf{e}}_{\rho }\times {\mathbf{e}}_{\phi }={\mathbf{e}}_z$, the Poynting vector is axial. The magnetic permeability cancels from $\mathbf{E}\times \mathbf{B}/\mu$ in each material layer:The signed Poynting flux through the plane with unit normal $+{\mathbf{e}}_z$ is its surface integral with area element $\rho d\rho d\phi$:ThusThe transmitted electromagnetic energy per unit time is the circuit's terminal power. For $VI>0$, the Poynting flux points in the direction of the inner electric current. The electromagnetic energy travels through the insulating gap and reaches a load at the end of the coaxial cable. The radii and material constants redistribute the electromagnetic energy density and the local Poynting vector, while their integrated Poynting flux depends only on the terminal electric potential difference and electric current.

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.

Take the ideal electric dipole limit with a fixed smooth electric dipole moment $\mathbf{p}(t)$, so that ${\mathbf{r}}_0(t)=\mathbf{p}(t)/(2q)$; its required time derivatives are bounded on the time interval under consideration. Fix $r>0$ and put $\mathbf{n}=\mathbf{r}/r$, $t_0=t-r/c$. For $s=+1$ or $-1$, the source position is $s\mathbf{p}(t)/(2q)$. A Taylor expansion of the distance givesSubstitute this into the defining equation for retarded time:Boundedness first implies $t_r^s-t_0=O(q^{-1})$. The Taylor expansion $\mathbf{p}(t_r^s)=\mathbf{p}(t_0)+O(q^{-1})$ then changes the displayed correction only by $O(q^{-2})$. HenceThe nearer positive displacement reduces the propagation distance and consequently increases the retarded time. The Big O notation here is for $q\to \infty$ at a fixed observation event away from the origin. Keeping a fixed smooth $\mathbf{p}$ is essential: bounded amplitude alone, with arbitrarily increasing time frequency as $q$ varies, would not justify this Taylor expansion.

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.