Chapter 8 — Poisson's Equation

Quote

“It may not be amiss to give a general idea of the method that has enabled us to arrive at results, remarkable in their simplicity and generality, which it would be very hard if not impossible to demonstrate in the ordinary way.”

— George Green (1828)

Motivation

Poisson’s equation,

$${\nabla }^2\phi (\mathbf{r})=-\frac{\rho (\mathbf{r})}{{ϵ}_0},$$

can be solved by direct Coulomb integration when $\rho (\mathbf{r})$ is completely specified. However, in many physically important situations — conducting boundaries, dielectric interfaces, or mobile charges that redistribute in response to the field — some portion of $\rho (\mathbf{r})$ is unknown and must be determined self-consistently. This is the boundary value problem: given the free (volume) charge and conditions on surfaces, find the total potential.

The volume part of $\rho$ is typically specified (external charges we control), while the surface part (induced charges on conductors, polarization charges on dielectrics) is determined by the solution itself. Thus, the task is to find the field produced by charge induced on nearby conductors and/or simple dielectrics.

Strategy: Superposition

Write $\phi (\mathbf{r})$ as a sum of a particular solution of Poisson’s equation (accounting for the known charge) and a general solution of Laplace’s equation (chosen so the total $\phi$ satisfies boundary conditions). This works because solutions to Laplace’s equation form a complete set that can represent any function needed to enforce boundary conditions on $S=\partial V$.

For a charge $q$ fixed at ${\mathbf{r}}_0$ in a volume $V$ with specified $\phi (\mathbf{r})$ on $S$:

$$\phi (\mathbf{r})=\underset{\text{particular sol. of PE}}{\underbrace{\frac{q}{4\pi {ϵ}_0|\mathbf{r}-{\mathbf{r}}_0|}}}+\underset{\text{solution of LE in }V}{\underbrace{\Phi (\mathbf{r})}}$$

Solutions for “nice” geometries (obtained via the method of images) are useful because they can be superposed for more complex configurations — this idea is generalized in the Green function method.

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Method of Images

The method of images can directly attack Poisson’s equation systems in which a planar, spherical, or cylindrical boundary separates space into a volume $V$ and its complement $V^{\prime }$. Given charges in $V$, we place fictitious image charges in $V^{\prime }$ that simulate the effect of the actual induced charge distribution on the boundary.

Rules of the Method

1. Image charges must lie outside the physical volume $V$ — never inside it.
2. Image charges must produce a potential that satisfies Laplace’s equation in $V$ (they are sources only in $V^{\prime }$).
3. The combination (real + image) must satisfy the boundary conditions on $\partial V$.
4. The method provides information about the potential only in $V$ — not in $V^{\prime }$.
5. By uniqueness, the solution so obtained is the solution.

Point Charge and a Grounded Conducting Plane

Charge $q$ at height $z_0$ above an infinite grounded plane at $z=0$. Place image $-q$ at $z=-z_0$:

$$\phi (\rho ,z\ge 0)=\frac{q}{4\pi {ϵ}_0}\left(\frac{1}{\sqrt{{\rho }^2+(z-z_0{)}^2}}-\frac{1}{\sqrt{{\rho }^2+(z+z_0{)}^2}}\right)$$

The first term is the particular solution of PE in $V$; the second solves LE in $V$ and enforces $\phi (z=0)=0$.

The induced surface charge density is:

$$\sigma (\rho )=-{ϵ}_0\hat{\mathbf{z}}\cdot \mathbf{E}(\rho ,z=0^{+})=-\frac{q{z}_0}{2\pi ({\rho }^2+z_0^2{)}^{3/2}}$$

which integrates to $-q$ — every field line leaving $q$ terminates on the plane.

Potential Theory Derivation

We can derive the image solution from potential theory directly: supplement the Coulomb potential with a solution of LE regular at $z\to \infty$ with azimuthal symmetry:

$$\phi (\rho ,z\ge 0)=\frac{q}{4\pi {ϵ}_0}\frac{1}{\sqrt{{\rho }^2+(z-z_0{)}^2}}+{\int }_0^{\infty }\mathrm{d}kA(k)J_0(k\rho )e^{-kz}$$

Using the identity $1/\sqrt{{\rho }^2+z^2}={\int }_0^{\infty }\mathrm{d}k{J}_0(k\rho )e^{-k|z|}$, the condition $\phi (\rho ,0)=0$ fixes $A(k)=-(q/4\pi {ϵ}_0)e^{-k{z}_0}$, reproducing the image solution exactly.

TODO: Can this be done for the sphere image method as well?

Image Energy Is Half the Naïve Answer

The interaction energy is not $q$ in the potential of its image — it is half of that. The image charge is not an independent source; it tracks $q$ as $q$ moves. Always integrate the force, or use $V_E=\frac{1}{2}{ϵ}_0\int |\mathbf{E}{|}^2$ over the physical half-space only. Same trap appears for image charges with spheres, cylinders, and dielectric interfaces.

The Image Energy Subtlety

The force on $q$ is that of interaction with $-q$ at distance $2z_0$: $\mathbf{F}=-q^2\hat{\mathbf{z}}/(16\pi {ϵ}_0{z}_0^2)$. The interaction energy is obtained by integrating this force:

$$V_E(z_0)=-{\int }_{\infty }^{z_0}\mathrm{d}z\hat{\mathbf{z}}\cdot {\mathbf{F}}_q(z)=-\frac{q^2}{16\pi {ϵ}_0{z}_0}$$

This is half the naïve $q$-in-potential-of-$(-q)$ answer ${\tilde{V}}_E=-q^2/(8\pi {ϵ}_0{z}_0)$. The factor of 2 arises because the image charge is not independent — it moves with $q$. In the real system, the charge drawn onto the plane moves across an equipotential (no work done), whereas bringing a real $-q$ from infinity costs additional energy.

Alternatively: The image system’s field is symmetric about $z=0$, so each half-space carries half the total field energy. Since the real system has $\mathbf{E}=0$ for $z<0$, $V_E={\tilde{V}}_E/2$.

Planar Dielectric Interface

Charge $q$ at $z=-d$ in a medium with ${\kappa }_L$; the half-space $z>0$ has ${\kappa }_R$.

Image system: To find $\phi$ in $z<0$: fill all space with ${\kappa }_L$, place $q$ at original position plus image $q_R$ at $z=+d$. To find $\phi$ in $z>0$: fill all space with ${\kappa }_R$, place image $q_L$ at $z=-d$.

The “fill all space” trick works because in a uniform medium, the potential of a point charge is simply Coulomb divided by the permittivity — matching boundary conditions then fixes the image charges. Uniqueness guarantees the result.

$$q_R=\frac{{\kappa }_L-{\kappa }_R}{{\kappa }_L+{\kappa }_R}q,q_L=\frac{2{\kappa }_R}{{\kappa }_L+{\kappa }_R}q$$

The force on $q$ (from the field of $q_R$ alone):

$$\mathbf{F}=-\frac{1}{4\pi {ϵ}_L}\frac{q{q}_R}{4d^2}\hat{\mathbf{z}}$$

which matches the result found in Chapter 6 via direct calculation of ${\sigma }_P$. The stress tensor confirms this is the negative of the force on the interface.

Limits

• ${\kappa }_R\to \infty$: $q_R\to -q$, recovering the grounded conductor result. Field lines are perpendicular to the interface.
• ${\kappa }_R\to {\kappa }_L$: $q_R\to 0$, no interface effects.

TODO: Add picture of field lines.

Multiple Images: Parallel Conducting Planes

Charge $q$ midway between two grounded planes at $z=0$ and $z=d$. Placing $-q$ at $z=-d/2$ and $z=3d/2$ to satisfy each plane individually violates the other’s boundary condition. An infinite sequence of images is required: positive charges at $z=2md+d/2$ and negative charges at $z=2md-d/2$ for all integers $m$.

Asymptotic Screening

For $\rho \gg d$, the potential is dominated by the lowest Fourier mode with periodicity $L=2d$. In this limit ${\nabla }^2\phi \approx {\partial }^2\phi /\partial {\rho }^2+{\partial }^2\phi /\partial z^2=0$ gives:

$$\phi (\rho ,z)\sim e^{-\pi \rho /d}\sin \left(\frac{\pi z}{d}\right),0\le z\le d,\rho \gg d$$

The charges induced on the plates screen the point charge exponentially at large $\rho$ — a 2D analog of Debye screening. This is the same exponential decay found in Chapter 7 for periodic systems: the conducting plates confine the field to a “channel” of width $d$, and laterally the effect dies off on a scale $\sim d/\pi$.

Spherical Boundary: Grounded Sphere

Charge $q$ at distance $s$ from the center of a grounded conducting sphere of radius $R$ (with $s>R$). By symmetry, the image $q^{\prime }$ lies on the line connecting $q$ to the center (choose as $z$ axis). Evaluating $\phi (R,\theta )=0$ at $\theta =0$ and $\theta =\pi /2$:

$$q^{\prime }=-\frac{R}{s}q,b=\frac{R^2}{s}$$

Since $R$ is the geometric mean of $s$ and $b$, the image lies inside the sphere. The potential for $r>R$:

$$\phi (\mathbf{r})=\frac{q}{4\pi {ϵ}_0}\left(\frac{1}{|\mathbf{r}-\mathbf{s}|}-\frac{R/s}{|\mathbf{r}-\hat{\mathbf{s}}{R}^2/s^2|}\right)$$

Since $|q^{\prime }|<q$, not all field lines terminate on the sphere — some escape to infinity (unlike the conducting plane, where the infinite surface captures all lines).

The force on the sphere:

$$\mathbf{F}=-\frac{q^2}{4\pi {ϵ}_0}\frac{Rs}{(s^2-R^2{)}^2}\hat{\mathbf{s}}$$

This force is always attractive and falls off as $1/s^3$ for $s\gg R$ (compared to $1/s^2$ for the plane). The asymptotic difference reflects the geometry: the plane subtends a full half-space of solid angle regardless of distance, while the sphere’s solid angle shrinks as $R^2/s^2$, so its ability to “intercept” field lines diminishes with distance.

Method of Inversion

If $\Phi (r,\theta ,\varphi )$ satisfies LE inside a sphere of radius $R$, then $\Psi (r,\theta ,\varphi )=(R/r)\Phi (R^2/r,\theta ,\varphi )$ satisfies LE outside the sphere. Since $\Psi (R,\theta ,\varphi )=\Phi (R,\theta ,\varphi )$, the unique potential in $r\ge R$ that vanishes at $r=R$ is:

$$\phi (r\ge R,\theta ,\varphi )=\Phi (r,\theta ,\varphi )-\Psi (r,\theta ,\varphi )$$

Choosing $\Phi =q/(4\pi {ϵ}_0|\mathbf{r}-\mathbf{s}|)$ and expanding via Legendre polynomials confirms that $-\Psi$ reproduces the image term.

Generalizations: Fixed Potential and Net Charge

Sphere at fixed potential ${\phi }_0$: Add a charge $q_0=4\pi {ϵ}_{0}R{\phi }_0$ at the center to raise the boundary potential uniformly:

$$\phi (\mathbf{r})=\frac{q}{4\pi {ϵ}_0}\left(\frac{1}{|\mathbf{r}-\mathbf{s}|}-\frac{R/s}{|\mathbf{r}-\hat{\mathbf{s}}{R}^2/s^2|}\right)+\frac{R{\phi }_0}{r},r>R$$

Sphere with net charge $Q$: Add $-q^{\prime }$ at the center to cancel the image, plus $Q$ at the center to set the total charge:

$$\phi (\mathbf{r})=\frac{q}{4\pi {ϵ}_0}\left(\frac{1}{|\mathbf{r}-\mathbf{s}|}-\frac{R/s}{|\mathbf{r}-\hat{\mathbf{s}}{R}^2/s^2|}\right)+\frac{Q+qR/s}{4\pi {ϵ}_{0}r},r>R$$

Force on Neutral Conductors: Attraction Then Repulsion

Let $\rho (\mathbf{r})$ be an arbitrary finite charge distribution. It possesses a family of equipotential surfaces (including $\phi =0$ at infinity). Consider a neutral conductor whose shape matches one of these equipotentials. The interaction energy $V_E$ between conductor and $\rho$ is zero at infinity and zero again when the conductor sits at its shape-matching equipotential. Since the force is attractive at large distance ($Q=0$ means $\rho$ induces a dipole on the conductor, always attractive), $V_E$ must first decrease from zero, reach a minimum, then return to zero. Therefore, the force must become repulsive as the conductor approaches the equipotential surface.

Analogy: A ball rolls into a valley but never rises above sea level — it must roll uphill before reaching the surface.

Line Charge and Conducting Cylinder

Infinite line charge $\lambda$ at distance $s$ from the axis of a conducting cylinder of radius $R$ (with $s>R$). Guess image ${\lambda }^{\prime }$ at position $b$ inside the cylinder:

$$\phi (x,y)=-\frac{1}{4\pi {ϵ}_0}\left[\lambda \ln \left((x-s{)}^2+y^2\right)+{\lambda }^{\prime }\ln \left((x-b{)}^2+y^2\right)\right]$$

For closed, bounded equipotentials to exist, the logarithmic divergence must cancel at large distances: ${\lambda }^{\prime }=-\lambda$. The equipotential condition on the cylinder surface $\rho =R$ then gives:

$$b=\frac{R^2}{s},{\lambda }^{\prime }=-\lambda$$

Same inverse-point relation as the sphere, but now all field lines terminate on the cylinder ($|{\lambda }^{\prime }|=|\lambda |$, unlike the sphere where $|q^{\prime }|<|q|$).

Nonzero Surface Potential

The potential on the cylinder surface is in general nonzero — we have freedom to choose it because the charge distribution is infinite. For the sphere, the finite total charge couples the potential and charge uniquely; here, the infinite line allows independent specification.

Dielectric Cylinder

Line charge $\lambda$ at distance $b$ inside a cylinder of radius $R$ with ${\kappa }_2$, embedded in ${\kappa }_1$.

Interior ($\rho <R$): Fill all space with ${\kappa }_2$, place source $\lambda$ plus image ${\lambda }^{\prime \prime }$ at the inverse point $s=R^2/b$.

Exterior ($\rho >R$): Fill all space with ${\kappa }_1$, place image ${\lambda }^{\prime }$ at the source position plus compensating $\lambda -{\lambda }^{\prime }$ at the origin (net neutrality of the cylinder).

Vector Potential Trick

Matching the normal component of $\mathbf{D}$ via the scalar potential is algebraically awkward. Instead, write $\mathbf{D}=\nabla \times (\psi \hat{\mathbf{z}})$ with $\psi (\varphi )=(\lambda /2\pi )\varphi +K$, giving $\mathbf{D}=(\lambda /2\pi \rho )\hat{\mathbf{\rho }}$. Continuity of ${\mathbf{D}}_{\perp }$ becomes continuity of $\psi$ — much simpler to enforce.

Matching both $\phi$ and $\psi$ at the boundary:

$${\lambda }^{\prime \prime }=\frac{{\kappa }_2-{\kappa }_1}{{\kappa }_2+{\kappa }_1}\lambda ,{\lambda }^{\prime }=\frac{2{\kappa }_1}{{\kappa }_2+{\kappa }_1}\lambda$$

These are identical to the planar dielectric interface ratios — a consequence of the conformal map $w=(z+R)/(z-R)$ that unrolls the cylinder into a plane.

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The Green Function Method

To solve PE ${\nabla }^2\phi (\mathbf{r})=-\rho (\mathbf{r})/{ϵ}_0$ for $\mathbf{r}\in V$ with boundary conditions on $\partial V=S$, introduce a Green function $G(\mathbf{r},{\mathbf{r}}^{\prime })$ satisfying PE for the same volume $V$ with a point source:

$${\nabla }^{2}G(\mathbf{r},{\mathbf{r}}^{\prime })=-\frac{1}{{ϵ}_0}\delta (\mathbf{r}-{\mathbf{r}}^{\prime }),\mathbf{r},{\mathbf{r}}^{\prime }\in V$$

Applying Green’s second identity ${\int }_V{\mathrm{d}}^{3}r(f{\nabla }^{2}g-g{\nabla }^{2}f)={\int }_S\mathrm{d}S\hat{\mathbf{n}}\cdot (f\nabla g-g\nabla f)$ with $f=\phi$ and $g=G$:

$$\phi (\mathbf{r})={\int }_V{\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })G({\mathbf{r}}^{\prime },\mathbf{r})-{ϵ}_0{\int }_S\mathrm{d}{S}^{\prime }\phi ({\mathbf{r}}^{\prime })\frac{\partial G({\mathbf{r}}^{\prime },\mathbf{r})}{\partial n^{\prime }}+{ϵ}_0{\int }_S\mathrm{d}{S}^{\prime }G({\mathbf{r}}^{\prime },\mathbf{r})\frac{\partial \phi ({\mathbf{r}}^{\prime })}{\partial n^{\prime }}$$

This integral equation involves both $\phi$ and $\partial \phi /\partial n$ on $S$. The art is to choose boundary conditions for $G$ that eliminate one, making the formula explicit.

Historical Note: George Green

This theory first appeared in an 1828 memoir by George Green (1793–1841), a self-taught miller from Nottingham who discovered higher mathematics at age 30 through Laplace and Poisson. His privately published 72-page Essay received little attention. William Thomson (Lord Kelvin) found a copy after his own graduation in 1845 and arranged its publication in Germany in 1850. Riemann coined the term “Green function” soon thereafter.

The Dirichlet Green Function

Set $G_D({\mathbf{r}}_S,{\mathbf{r}}^{\prime })=0$ for ${\mathbf{r}}_S\in S$. The third integral vanishes:

$$\phi (\mathbf{r})={\int }_V{\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })G_D({\mathbf{r}}^{\prime },\mathbf{r})-{ϵ}_0{\int }_S\mathrm{d}{S}^{\prime }{\phi }_S({\mathbf{r}}^{\prime })\frac{\partial G_D({\mathbf{r}}^{\prime },\mathbf{r})}{\partial n^{\prime }}$$

Physical interpretation: $G_D(\mathbf{r},{\mathbf{r}}^{\prime })$ is the potential at $\mathbf{r}$ due to a unit point charge at ${\mathbf{r}}^{\prime }$, with $S$ acting as a grounded conductor ($G_D$ vanishes there).

Reciprocity: Using Green’s second identity with $G(\mathbf{x},\mathbf{r})$ and $G(\mathbf{x},{\mathbf{r}}^{\prime })$, integrating over $\mathbf{x}$:

$$G_D(\mathbf{r},{\mathbf{r}}^{\prime })=G_D({\mathbf{r}}^{\prime },\mathbf{r})$$

This symmetry is immensely important: it transforms the formula above into the magic rule:

$$\phi (\mathbf{r})={\int }_V{\mathrm{d}}^3{r}^{\prime }{G}_D(\mathbf{r},{\mathbf{r}}^{\prime })\rho ({\mathbf{r}}^{\prime })-{ϵ}_0{\int }_S\mathrm{d}{S}^{\prime }{\phi }_S({\mathbf{r}}^{\prime })\frac{\partial G_D(\mathbf{r},{\mathbf{r}}^{\prime })}{\partial n^{\prime }}$$

Now $G_D$ is computed as the potential at ${\mathbf{r}}^{\prime }$ for a source at $\mathbf{r}$: we need only calculate one Green function for a given volume, and can then find $\phi (\mathbf{r})$ for arbitrary $\rho$ and ${\phi }_S$.

Free-space Green function ($S$ at infinity): $G_0(\mathbf{r},{\mathbf{r}}^{\prime })=1/(4\pi {ϵ}_0|\mathbf{r}-{\mathbf{r}}^{\prime }|)$.

Interior vs. exterior: For a finite grounded surface, one can define two Green functions — interior ($\mathbf{r},{\mathbf{r}}^{\prime }\in V_{\text{in}}$) and exterior ($\mathbf{r},{\mathbf{r}}^{\prime }\in V_{\text{out}}$). Reciprocity applies to each separately.

Decomposition and the Indirect Effect of Charge

Decompose $G_D=G_0+\Lambda$, where $\Lambda (\mathbf{r},{\mathbf{r}}^{\prime })$ satisfies LE in $V$ and ensures $G_D=0$ on $S$. For a conductor at potential ${\phi }_0$, the surface integral reduces to ${\phi }_0$ and:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}{\int }_V\frac{{\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}+{\int }_V{\mathrm{d}}^3{r}^{\prime }\Lambda (\mathbf{r},{\mathbf{r}}^{\prime })\rho ({\mathbf{r}}^{\prime })+{\phi }_0$$

The first term is the direct Coulomb effect; the second is the indirect effect — the potential at $\mathbf{r}$ due to charge induced on $S$ by a unit source at ${\mathbf{r}}^{\prime }$. This gives the force on a point charge $q$ at $\mathbf{s}$ inside an oddly shaped shell: $\mathbf{F}(\mathbf{s})=-q^2\nabla \Lambda (\mathbf{r},\mathbf{s}){|}_{\mathbf{r}=\mathbf{s}}$ (the charge cannot exert a force on itself).

The Neumann Green Function

We cannot naively set $\partial G_N/\partial n=0$ on $S$ because integrating the defining PE over $V$ gives ${\int }_S\mathrm{d}S\partial G_N/\partial n=-1/{ϵ}_0\ne 0$. Instead:

$$\frac{\partial G_N({\mathbf{r}}_S,{\mathbf{r}}^{\prime })}{\partial n}=-\frac{1}{{ϵ}_{0}A},{\mathbf{r}}_S\in S$$

where $A$ is the area of $S$. Then:

$$\phi (\mathbf{r})=⟨\phi {⟩}_S+{\int }_V{\mathrm{d}}^3{r}^{\prime }{G}_N(\mathbf{r},{\mathbf{r}}^{\prime })\rho ({\mathbf{r}}^{\prime })+{ϵ}_0{\int }_S\mathrm{d}{S}^{\prime }\frac{\partial {\phi }_S({\mathbf{r}}^{\prime })}{\partial n^{\prime }}{G}_N(\mathbf{r},{\mathbf{r}}^{\prime })$$

Neumann conditions do not arise naturally in electrostatics with stationary charge. Although $\sigma ({\mathbf{r}}_S)=-{ϵ}_0\partial \phi /\partial n$ relates the normal derivative to conductor surface charge, $\partial \phi /\partial n$ is determined by the solution, not prescribed in advance. Neumann conditions do appear in steady-current problems and waveguide theory.

Mixed boundary conditions (e.g., a conducting shell occupying only part of $S$, or finite-sized capacitor plates) require more exotic mathematical methods (Sneddon 1966, Fabrikant 1991).

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Computing the Dirichlet Green Function

For planar, cylindrical, and spherical boundaries, the method of images directly gives $G_D$. For other geometries, we present three systematic methods.

Method I: Eigenfunction Expansion

Find eigenfunctions of the Laplacian satisfying homogeneous Dirichlet conditions:

$$-{\nabla }^2{\psi }_n(\mathbf{r})={\lambda }_n{\psi }_n(\mathbf{r}),{\psi }_n({\mathbf{r}}_S)=0$$

By Sturm–Liouville theory, the operator $-{\nabla }^2$ with these BCs is self-adjoint (not the same as Hermitian in general). This guarantees a complete orthonormal set: ${\sum }_n{\psi }_n(\mathbf{r}){\psi }_n^{\ast }({\mathbf{r}}^{\prime })=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$. The Green function is:

$$G_D(\mathbf{r},{\mathbf{r}}^{\prime })=\frac{1}{{ϵ}_0}\sum _n\frac{{\psi }_n(\mathbf{r}){\psi }_n^{\ast }({\mathbf{r}}^{\prime })}{{\lambda }_n}$$

This satisfies the PE (apply the eigenvalue equation) and the BC (${\psi }_n({\mathbf{r}}_S)=0$ for each term). Finite domains give discrete modes; infinite domains give continuous spectra (plane waves in Cartesian, Bessel functions in cylindrical, etc.).

Application — Charge Density to Simulate a Point Charge (Cubical Box)

An insulating cube ($0\le x,y,z\le a$) must carry surface charge $\sigma$ so its external field matches that of a point charge $Q$ at its center. Replace the insulating box by a grounded conducting box with $-Q$ at its center. The charge drawn from ground annuls the exterior field of $-Q$, so its distribution is the desired $\sigma$.

Normalized eigenfunctions:

$${\psi }_{n\ell m}(x,y,z)=\sqrt{\frac{8}{a^3}}\sin \frac{n\pi x}{a}\sin \frac{\ell \pi y}{a}\sin \frac{m\pi z}{a}$$

with ${\lambda }_{n\ell m}=({\pi }^2/a^2)(n^2+{\ell }^2+m^2)$. The surface charge on $z=a$:

$$\sigma (x,y)=-{ϵ}_{0}Q\frac{\partial G_D}{\partial z}{|}_{z=a,{\mathbf{r}}^{\prime }=\text{center}}$$

Method II: Direct Integration

Algorithm:

1. Represent $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ as a product of 1D deltas; use completeness relations for all but one coordinate.
2. Motivate an ansatz for $G$ that reduces the 3D PE to an inhomogeneous ODE in the remaining coordinate.
3. Solve the ODE with continuity of $G$ and a jump condition on $\partial G/\partial (\text{remaining coord.})$ at the source point (from integrating the delta).

Cylindrical Representation of $G_0$

For $G_0(\mathbf{r},{\mathbf{r}}^{\prime })=1/(4\pi {ϵ}_0|\mathbf{r}-{\mathbf{r}}^{\prime }|)$ in cylindrical coordinates, use plane-wave and Fourier expansions:

$$\delta (z-z^{\prime })=\frac{1}{\pi }{\int }_0^{\infty }\mathrm{d}k\cos k(z-z^{\prime }),\delta (\varphi -{\varphi }^{\prime })=\frac{1}{2\pi }\sum _{m=-\infty }^{\infty }{e}^{im(\varphi -{\varphi }^{\prime })}$$

The ansatz $G_0=\frac{1}{2{\pi }^2}{\sum }_m{\int }_0^{\infty }\mathrm{d}k{e}^{im(\varphi -{\varphi }^{\prime })}\cos k(z-z^{\prime })G_m(\rho ,{\rho }^{\prime }|k)$ yields the modified Bessel ODE:

$${ϵ}_0\left[\rho \frac{\mathrm{d}}{\mathrm{d}\rho }\left(\rho \frac{\mathrm{d}{G}_m}{\mathrm{d}\rho }\right)-\left(k^2+\frac{m^2}{{\rho }^2}\right)G_m\right]=-\frac{1}{\rho }\delta (\rho -{\rho }^{\prime })$$

Regularity and continuity at $\rho ={\rho }^{\prime }$ requires $I_m(k{\rho }_{<})K_m(k{\rho }_{>})$ with ${\rho }_{<}=\min (\rho ,{\rho }^{\prime })$, ${\rho }_{>}=\max (\rho ,{\rho }^{\prime })$. The jump condition at $\rho ={\rho }^{\prime }$ (integrating across the delta) fixes the coefficient via the Wronskian $W[K_m,I_m]=1/x$:

$$G_0(\mathbf{r},{\mathbf{r}}^{\prime })=\frac{1}{2{\pi }^2{ϵ}_0}\sum _{m=-\infty }^{\infty }{\int }_0^{\infty }\mathrm{d}k{e}^{im(\varphi -{\varphi }^{\prime })}\cos k(z-z^{\prime })I_m(k{\rho }_{<})K_m(k{\rho }_{>})$$

This is the cylindrical analog of the spherical expansion:

$$\frac{1}{4\pi {ϵ}_0|\mathbf{r}-{\mathbf{r}}^{\prime }|}=\frac{1}{4\pi {ϵ}_0{r}_{>}}\sum _{\ell =0}^{\infty }\frac{4\pi }{2\ell +1}{\left(\frac{r_{<}}{r_{>}}\right)}^{\ell }\sum _{m=-\ell }^{\ell }{Y}_{\ell m}^{\ast }({\hat{\mathbf{r}}}_{<})Y_{\ell m}({\hat{\mathbf{r}}}_{>})$$

Free-Space Spherical Expansion by Direct Integration

Starting from the completeness relation ${\sum }_{\ell m}{Y}_{\ell m}(\hat{\mathbf{r}})Y_{\ell m}^{\ast }({\hat{\mathbf{r}}}^{\prime })=\delta (\hat{\mathbf{r}}-{\hat{\mathbf{r}}}^{\prime })$, the radial ODE is:

$${\partial }_r(r^2{\partial }_r{G}_{\ell })-\ell (\ell +1)G_{\ell }=-\frac{1}{{ϵ}_0}\delta (r-r^{\prime })$$

Regularity at $r=0$ selects $r^{\ell }$ for $r<r^{\prime }$; regularity at $r\to \infty$ selects $r^{-(\ell +1)}$ for $r>r^{\prime }$. Continuity gives $r_{<}^{\ell }/r_{>}^{\ell +1}$. The jump condition $r^{\prime 2}{\partial }_r{G}_{\ell }{|}_{r^{\prime }-\epsilon }^{r^{\prime }+\epsilon }=-1/{ϵ}_0$ gives $A_{\ell }=1/[{ϵ}_0(2\ell +1)]$, reproducing the standard expansion.

Sphere (Exterior) by Direct Integration

For the interior of a sphere (radius $R$), both $\mathbf{r}$ and ${\mathbf{r}}^{\prime }$ are inside, so the outer region ($r^{\prime }<r<R$) has both $r^{\ell }$ and $r^{-(\ell +1)}$ terms. The BCs $G_D(R,r^{\prime })=0$ and regularity at $r=0$ give: $G_D(\mathbf{r},{\mathbf{r}}^{\prime })=\frac{1}{{ϵ}_0}{\sum }_{\ell =0}^{\infty }\frac{1}{2\ell +1}\left(\frac{r_{<}^{\ell }}{r_{>}^{\ell +1}}-\frac{(r_{<}{r}_{>}{)}^{\ell }}{R^{2\ell +1}}\right)P_{\ell }(\hat{\mathbf{r}}\cdot {\hat{\mathbf{r}}}^{\prime })$

Dirichlet Green Function Between Parallel Plates

Two grounded plates at $z=0$ and $z=d$ with a point charge $q$ at $z=z_0$. Use the completeness relation $\frac{2}{d}{\sum }_{n=1}^{\infty }\sin (n\pi z/d)\sin (n\pi z^{\prime }/d)=\delta (z-z^{\prime })$. The resulting radial equation has solutions $I_0$ and $K_0$ with $k=n\pi /d$:

$$\phi (\rho ,z)=\frac{q}{{ϵ}_0\pi d}\sum _{n=1}^{\infty }\sin \frac{n\pi z}{d}\sin \frac{n\pi z_0}{d}{K}_0\left(\frac{n\pi \rho }{d}\right)$$

Since $K_{\alpha }(x\gg 1)\sim \sqrt{\pi /(2x)}{e}^{-x}$, the $n=1$ term dominates for $\rho \gg d$, confirming exponential screening.

The induced charge on the $z=d$ plate: $Q(d)=-z_{0}q/d$, confirming the result from Green’s reciprocity (Chapter 5).

Method III: Splitting

Decompose $G_D=G_0+\Lambda$, where $\Lambda$ satisfies LE in $V$ and is chosen so $G_D=0$ on $S$. This works when $G_0$ is known in the relevant coordinate system.

Exterior Green Function for a Cylinder

Using the cylindrical $G_0$ above, choose:

$$\Lambda (\mathbf{r},{\mathbf{r}}^{\prime })=\frac{1}{2{\pi }^2{ϵ}_0}\sum _m{\int }_0^{\infty }\mathrm{d}k{e}^{im(\varphi -{\varphi }^{\prime })}\cos k(z-z^{\prime })B_m{K}_m(k\rho )K_m(k{\rho }^{\prime })$$

The condition $G_D(\rho =R,{\mathbf{r}}^{\prime })=0$ fixes $B_m=-I_m(kR)/K_m(kR)$:

$$G_D=\frac{1}{2{\pi }^2{ϵ}_0}\sum _m{\int }_0^{\infty }\mathrm{d}k{e}^{im(\varphi -{\varphi }^{\prime })}\cos k(z-z^{\prime })\left[I_m(k{\rho }_{<})-\frac{I_m(kR)}{K_m(kR)}{K}_m(k{\rho }_{<})\right]K_m(k{\rho }_{>})$$

Application — Force on a Charge near a Grounded Tube

Using the exterior cylinder $G_D$, the force on $q$ at distance ${\rho }_0\gg R$:

$$F({\rho }_0)\sim -\frac{q^2}{8\pi {ϵ}_0}\frac{1}{{\rho }_0^2\ln (2{\rho }_0/R)},{\rho }_0\gg R$$

The asymptotic decay is intermediate between the plane ($F\sim {\rho }_0^{-2}$) and the sphere ($F\sim {\rho }_0^{-3}$). Geometrically: the infinite plane subtends a full half-space at any distance; the infinite cylinder’s cross-section shrinks as $R/{\rho }_0$ but extends infinitely along its axis; the sphere’s solid angle shrinks as $R^2/{\rho }_0^2$. The logarithmic correction reflects the marginal nature of 2D screening.

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Complex Potential Methods

The basic machinery — analytic $f(w)=\phi +i\psi$, conformal maps, equipotentials and field lines as level curves — is developed in Chapter 7. Here we apply it to systems with explicit line sources, where the logarithmic singularity of $\ln w$ encodes the line charge.

For 2D problems (line charges), the potential is the real part of an function:

$$f(w)=-\frac{\lambda }{2\pi {ϵ}_0}\ln w=-\frac{\lambda }{2\pi {ϵ}_0}(\ln r+i\theta )$$

analytic everywhere except $w=0$. Superposition builds arbitrary 2D charge distributions.

Line dipole: Taking $\lambda \to \infty$, $a\to 0$ with $p=2\lambda a$ finite:

$$f(w)=-\frac{\lambda }{2\pi {ϵ}_0}\ln \frac{w+a}{w-a}\to -\frac{\lambda a}{\pi {ϵ}_{0}w},\phi (x,y)=\frac{1}{2\pi {ϵ}_0}\frac{\mathbf{p}\cdot \mathbf{r}}{r^2}$$

Application — Wire Chamber

An infinite planar array of parallel lines (spacing $a$, charge density $\lambda$) at height $d$ above a grounded plane. The method of images gives an array of $-\lambda$ lines at $-d$. The complex potential:

$$f(w)=-\frac{\lambda }{2\pi {ϵ}_0}\ln \frac{\sin [(\pi /a)(w-id)]}{\sin [(\pi /a)(w+id)]}$$

gives the physical potential:

$$\phi (x,y)=\frac{\lambda }{4\pi {ϵ}_0}\ln \frac{{\sin }^2(\pi x/a)+{\sinh }^2[\pi (y+d)/a]}{{\sin }^2(\pi x/a)+{\sinh }^2[\pi (y-d)/a]}$$

Interpreting the near-wire equipotentials as wire surfaces (radius $R\ll a$), the capacitance per unit area between wire array and ground plane is:

$$\frac{C}{A}\approx \frac{2\pi {ϵ}_0}{a\ln (a/2\pi R)+2\pi d}$$

which recovers the parallel-plate result $C/A={ϵ}_0/d$ when $a\ll d$. Wire chambers are used in particle physics to infer trajectories of ionizing radiation from the signals induced on the wire array.

Conformal Mapping — Line Charge Between Grounded Plates

Line charge $\lambda$ at $(0,d/2)$ between grounded plates at $y=0$ and $y=d$. The map $g(w)=e^{\pi w/d}$ sends the strip $0\le y\le d$ to the upper half-plane, with the source at $g_0=i$. An image at $g_0^{\ast }=-i$ grounds the real axis:

$$f(g(w))=-\frac{\lambda }{2\pi {ϵ}_0}\ln \frac{e^{\pi w/d}-i}{e^{\pi w/d}+i}$$ $$\phi (x,y)=-\frac{\lambda }{2\pi {ϵ}_0}\ln \sqrt{\frac{e^{2\pi x/d}{\cos }^2(\pi x/d)+[e^{\pi x/d}\sin (\pi y/d)-1{]}^2}{e^{2\pi x/d}{\cos }^2(\pi x/d)+[e^{\pi x/d}\sin (\pi y/d)+1{]}^2}}$$

A single conformal map replaces the infinite sequence of images needed for this geometry.

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The Poisson–Boltzmann Equation

So far, volume charge has been specified and immobile. When mobile charges redistribute in response to the field, $\rho (\mathbf{r})$ is itself determined by $\phi (\mathbf{r})$.

Model: Plane $z=0$ with fixed charge density $\sigma$, embedded in a medium ($ϵ$) containing mobile particles (charge $q$, number density $n_0$) obeying Boltzmann statistics: $n(z)=n_0\exp [-q\phi (z)/kT]$.

The Poisson–Boltzmann equation:

$$\frac{{\mathrm{d}}^2\phi }{\mathrm{d}{z}^2}+\frac{q{n}_0}{ϵ}{e}^{-q\phi /kT}=-\frac{\sigma }{ϵ}\delta (z)$$

Solution Sketch

Multiply-and-integrate trick: Define $\psi =q\phi /kT$ and $\ell =q^2/(ϵkT)$. Multiply the equation by $\mathrm{d}\psi /\mathrm{d}z$ and integrate; using $E(z)\to 0$ as $z\to \infty$:

$$\psi (z)=2\ln \left[\frac{n_0\ell }{2}(z+b)\right],b=\frac{4|q|}{\sigma }$$

The mobile charge density:

$$qn(z)=\frac{2q}{\ell }\frac{1}{(z+b{)}^2}$$

which integrates to $-\sigma /2$ on each side. The electric field:

$$\mathbf{E}(z)=\frac{b}{|z|+b}{\mathbf{E}}_0$$

Algebraic vs. Exponential Screening

Mobile charges screen the field of the charged interface algebraically ($\sim 1/z$). Compare with Thomas–Fermi screening, where a neutral medium of mobile + immobile charges screens a point charge exponentially ($\sim e^{-r/{\lambda }_D}/r$). The difference: here the mobile charges alone must neutralize $\sigma$ — there is no background of fixed opposite charge to set a screening length. The characteristic scale $b=4|q|/\sigma$ is set by the geometry (surface charge density) rather than by a bulk property.

Applications: Colloidal suspensions, biological membranes, semiconductor interfaces (Gouy–Chapman theory), and electrochemistry all involve the Poisson–Boltzmann equation. The nonlinear version is essential when $q\phi \gtrsim kT$.

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Problems

Method of Images

8.1 — The Image Force and Its Limits

A grounded conductor of arbitrary shape has characteristic size $R$. A point charge $q$ is at distance $s\gg R$. Show the force varies as $1/s^3$.

Solution Sketch

Use Green’s reciprocity with system A (charge $q$ near grounded conductor) and system B (no charge, conductor at potential ${\phi }_c$ with charge $q_c=C{\phi }_c$). Reciprocity gives:

$$q{\phi }_A^{(B)}(0)+q_c{\phi }_c=0+q_c{\phi }_c^{(A)}$$

At large $s$: ${\phi }^{(B)}(0)\approx q_c/(4\pi {ϵ}_{0}s)$, so $q_c^{(\text{induced})}=-qC/(4\pi {ϵ}_{0}s)$. Since $C$ depends only on the conductor’s geometry:

$$F\sim \frac{q{q}_c}{4\pi {ϵ}_0{s}^2}\propto \frac{1}{s^3}$$

Key insight: The self-capacitance $C$ — a purely geometric property — controls the asymptotic image force, independent of shape.

8.2 — Point Charge near a Corner

Two grounded semi-infinite conducting planes meet at a right angle. Find the charge induced on each plane when $Q$ is introduced at angle $\alpha$ from one plane.

Solution Sketch

Three image charges complete the $90°$ symmetry: $-Q$ reflected across each plane, and $+Q$ at the diagonally opposite point.

The total solid angle subtended by each plane from $Q$‘s perspective partitions the full $\pi /2$ corner. Through the vertical plane: diagonally opposite $Q$ directly subtends angle $\alpha$, contributing fraction $2\alpha /\pi$ of the field lines. Total induced charge on the vertical plane:

$$Q_{\text{vert}}=Q\frac{2\alpha }{\pi }-Q=Q\left(\frac{2\alpha }{\pi }-1\right)$$

8.4 — A Dielectric Slab Intervenes

Dielectric slab ($\kappa$, thickness $c$) between $z=a$ and $z=b=a+c$. Point charge $q$ at origin. With $\beta =(\kappa -1)/(\kappa +1)$, show:

$$\phi (z>b)=\frac{q(1-{\beta }^2)}{4\pi {ϵ}_0}\sum _{n=0}^{\infty }\frac{{\beta }^{2n}}{\sqrt{(z+2nc{)}^2+{\rho }^2}}$$

Solution Sketch

Three regions with Laplace + point charge structure. Using the Bessel identity $1/\sqrt{{\rho }^2+z^2}={\int }_0^{\infty }\mathrm{d}k{J}_0(k\rho )e^{-k|z|}$, write ansätze:

• Region I ($z<a$): Coulomb + ${\int }_0^{\infty }\mathrm{d}k{J}_0(k\rho )E(k)e^{kz}$
• Region II ($a<z<b$): ${\int }_0^{\infty }\mathrm{d}k{J}_0(k\rho )[B(k)e^{-kz}+C(k)e^{kz}]$
• Region III ($z>b$): ${\int }_0^{\infty }\mathrm{d}k{J}_0(k\rho )A(k)e^{-kz}$

Matching $\phi$ and $\kappa \partial \phi /\partial z$ at $z=a$ and $z=b$ (four equations, four unknowns). Since $1-{\beta }^2=4\kappa /(\kappa +1{)}^2$:

$$A(k)=\frac{q(1-{\beta }^2)}{4\pi {ϵ}_0}\frac{e^{-ka}}{1-{\beta }^2{e}^{-2kc}}$$

Expanding $(1-{\beta }^2{e}^{-2kc}{)}^{-1}={\sum }_n{\beta }^{2n}{e}^{-2nkc}$ and using the Bessel identity in reverse gives the sum over image potentials.

Physical picture: Each reflection at a dielectric boundary picks up a factor $\beta$ — the $n$-th term corresponds to $2n$ internal reflections, analogous to multiple reflections in optics. The coefficient $(1-{\beta }^2)$ is the transmission factor.

If we don’t want to write out the the solution in $z<a$ part as a sum of manually added point charge potential we can split the region at $z=0$ and then have a surface charge density proportional to $\delta (z)$ of magnitude $\sigma (\rho )=q\delta (\rho )/(2\pi \rho )$ and then match at $z=0$ as well.

8.5 — Force on a Dielectric Interface

Point charge $q$ at $z=-d$ in ${\kappa }_L$; half-space $z>0$ has ${\kappa }_R$.

• (a) Show stress tensor gives force on interface equal and opposite to ${\mathbf{F}}_q$.
• (b) Show Coulomb force on ${\sigma }_P$ gives the wrong answer.

Solution Sketch

(a) Integrate the stress tensor on a surface hugging $z=0$ from both sides. Only $D_z$ and $E_{\Vert }$ are continuous; $D_{\Vert }$ is not (since $\nabla \times \mathbf{D}=\nabla \times \mathbf{P}\ne 0$ at the interface). Using $D_z(z=0)=q_{L}d/(4\pi r^3)$ and $E_{\Vert }(z=0)=q_L\rho /(4\pi {\kappa }_L{ϵ}_0{r}^3)$ where $r=\sqrt{{\rho }^2+d^2}$:

$$F_z=\frac{1}{4\pi {ϵ}_L}\frac{q^2}{4d^2}\frac{{\kappa }_L-{\kappa }_R}{{\kappa }_L+{\kappa }_R}$$

Equal and opposite to ${\mathbf{F}}_q$.

(b) The Coulomb force $\mathbf{F}=\int \mathrm{d}S{\sigma }_P({\mathbf{E}}_L+{\mathbf{E}}_R)/2$ gives the correct result divided by ${\kappa }_L$. This is the usual failure of the Coulomb formula for sub-volumes — it misses the field-dependent cohesive forces at the boundary (see Chapter 6).

Note: Naïvely using $\mathbf{D}$ from free charge alone would give $\mathbf{F}=0$ (missing the surface $\nabla \times \mathbf{P}$ contribution to $\mathbf{D}$). The stress tensor via the three image charges — carefully tracking which contributes in which region — is the cleanest approach.

8.7 — Images in Spheres I

Charge $q$ at $2R$ from center of an isolated conducting sphere (radius $R$); force is zero there. Move to $3R$; show force is repulsive with $F=173q^2/(5184\cdot 4\pi {ϵ}_0{R}^2)$.

Solution Sketch

At $s=2R$: the force vanishes, which determines the net charge $Q$ on the sphere. Image charge $q^{\prime }=-qR/s=-q/2$ at $b=R/2$; add $Q-q^{\prime }$ at the center to fix total charge. $F=0$ at $s=2R$ fixes $Q$.

At $s=3R$: new image $q^{\prime }=-q/3$ at $b=R/3$, same $Q$ (isolated sphere, charge unchanged). Compute force from both the image and the central excess charge. The net force is repulsive — consistent with the general argument that neutral conductors can repel at intermediate distances.

8.9 — Debye’s Model for the Work Function

Conducting sphere of radius $R$; image force becomes operative at distance $d$ outside.

• (a) Find $W$.
• (b) Take $R\to \infty$.

Solution Sketch

(a) Removing an electron leaves the sphere with net charge $+e$. The force on the electron has two contributions: the image charge $q^{\prime }=-eR/(R+x)$ at $R^2/(R+x)$, and the Coulomb attraction to the net $+e$ on the sphere. Integrating the total force from $R+d$ to $\infty$:

$$W=\frac{e^2}{8\pi {ϵ}_0}\left(\frac{2}{R+d}-\frac{R}{(R+d{)}^2}+\frac{R}{(R+d{)}^2-R^2}\right)$$

(b) Let $x=d/R$, take $R\to \infty$: $W\to e^2/(16\pi {ϵ}_{0}d)$. The image-force contribution to the work function is understood today to be insignificant — quantum-mechanical effects (exchange-correlation at the surface) dominate.

8.10 — Line Charge and Conducting Cylinder

Line charge $\lambda$ at perpendicular distance $b$ from axis of cylinder $R=b/2$.

Solution Sketch

Image line $-\lambda$ at $b^{\prime }=R^2/b=b/4$ from center, plus line $+\lambda$ at center to fix the cylinder potential.

(a) Surface charge density:

$$\sigma (\varphi )=-\frac{\lambda }{2\pi R}\frac{3}{5-4\cos \varphi }$$

(b) Integrate $\sigma$ to get force. (c) Equivalently, the force per unit length is just the attraction between $\lambda$ and the image $-\lambda$ at distance $b-b^{\prime }=3b/4$:

$$\frac{F}{\ell }=\frac{{\lambda }^2}{2\pi {ϵ}_0}\frac{1}{2(b-b^{\prime })}=\frac{2{\lambda }^2}{3\pi {ϵ}_{0}b}$$

Key simplification: Adding the central image to fix the potential doesn’t affect the force (it’s at the center of the cylinder, so produces no net force on the cylinder by symmetry).

8.11 — Point Dipole in a Grounded Shell

Dipole $\mathbf{p}$ at center of grounded sphere of radius $R$. Show ${\mathbf{E}}_{\text{inside}}={\mathbf{E}}_{\text{dipole}}+\mathbf{p}/(4\pi {ϵ}_0{R}^3)$.

Solution Sketch

Model dipole as $+q$ at $\mathbf{s}$ and $-q$ at $-\mathbf{s}$ with $\mathbf{p}=2q\mathbf{s}$, $|\mathbf{s}|\to 0$. Each charge has an image: $-qR/s$ at $R^2\hat{\mathbf{s}}/s$ and $+qR/s$ at $-R^2\hat{\mathbf{s}}/s$. As $s\to 0$, the image charges shoot to infinity, and the image system approaches two charges $\pm qR/s$ separated by $2R^2/s$, far from the sphere.

In this limit, ${\phi }_{\text{image}}$ at the sphere surface equals that of a uniform field — the same as the field produced by a plane with $\sigma$ given by the stated formula. Using $\sigma (\rho )=-q{z}_0/[2\pi ({\rho }^2+z_0^2{)}^{3/2}]$ with $s^{\prime }=R^2/s\to \infty$:

$$E_{\text{ind}}=\frac{q^{\prime }}{2\pi {ϵ}_0{s}^{\prime 2}}=\frac{qR/s}{2\pi {ϵ}_0(R^2/s{)}^2}=\frac{qs}{2\pi {ϵ}_0{R}^3}=\frac{p}{4\pi {ϵ}_0{R}^3}$$

directed along $\mathbf{p}$.

8.12 — Inversion in a Cylinder

Solution Sketch

(a) If $\Phi (\rho ,\varphi )$ satisfies LE for $\rho <R$, then $\Psi (\rho ,\varphi )=\Phi (R^2/\rho ,\varphi )$ satisfies LE for $\rho >R$ (substitute $u=R^2/\rho$, use chain rule).

(b) For a line charge at $s>R$: $\Phi$ satisfies LE inside the cylinder. Since $\Psi (R,\varphi )=\Phi (R,\varphi )$, the function $\phi =\Phi -\Psi$ vanishes at $\rho =R$ and satisfies PE outside (with the line source).

(c) For a dielectric cylinder (${\kappa }_1$ inside, ${\kappa }_2$ outside, source at $s>R$): try ${\phi }_{\text{out}}=\Phi +S\Psi$ and ${\phi }_{\text{in}}=T\Phi$. Matching $\phi$ and $\kappa \partial \phi /\partial \rho$ at $\rho =R$:

$$S=\frac{{\kappa }_2-{\kappa }_1}{{\kappa }_2+{\kappa }_1},T=\frac{2{\kappa }_2}{{\kappa }_2+{\kappa }_1}$$

(d) $\Psi (\rho ,\varphi )=\Phi (R^2/\rho ,\varphi )$ is the potential of a source at the inverse point $R^2/s$ — used only in the region opposite to the real source. This is precisely the image construction.

Green Functions

8.14 — Green Function Inequalities

Solution Sketch

(a) $G_D=G_0+\Lambda$ where $\Lambda$ is the potential of the negative charge drawn from ground onto $S$. This potential is negative everywhere in $V$, so $\Lambda <0$ and $G_D<G_0=1/(4\pi {ϵ}_0|\mathbf{r}-{\mathbf{r}}^{\prime }|)$.

(b) Let $\Omega =V$ minus an infinitesimal sphere around ${\mathbf{r}}^{\prime }$. In $\Omega$, ${\nabla }^2{G}_D=0$, so by Earnshaw’s theorem $G_D$ has no local extrema. On $\partial V$: $G_D=0$. On the small sphere: $G_D>0$ (dominated by $G_0$). Since $G_D$ cannot have a minimum in $\Omega$, it cannot be negative anywhere — hence $G_D\ge 0$. (Strict positivity $G_D>0$ in the interior follows from the strong maximum principle for harmonic functions: if function attains a minimum value at any interior point, then it’s constant throughout the domain, which we know it is not.)

8.15 — Potential of a Voltage Patch

Grounded plane $z=0$ except for area $S_0$ at potential ${\phi }_0$. Show:

$$\phi (x,y,z)=\frac{{\phi }_0|z|}{2\pi }{\int }_{S_0}\frac{{\mathrm{d}}^2{r}^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}$$

Solution Sketch

No charge in the volume, so only the surface term of the magic rule contributes. The Dirichlet Green function for the half-space $z>0$ is the method-of-images result. Taking the normal derivative at $z^{\prime }=0$:

$${ϵ}_0\frac{\partial G_D}{\partial z^{\prime }}{|}_{z^{\prime }=0}=\frac{z}{2\pi [(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2+z^2{]}^{3/2}}$$

Substituting into the magic rule with ${\phi }_S={\phi }_0$ on $S_0$ and ${\phi }_S=0$ elsewhere gives the result. The symmetry $\phi (x,y,z)=\phi (x,y,-z)$ extends it to $|z|$.

8.17 — Free-Space Green Functions by Eigenfunction Expansion

Find $G_0^{(d)}(\mathbf{r},{\mathbf{r}}^{\prime })$ in $d=1,2,3$ dimensions.

Solution Sketch

The Laplacian in free space commutes with translations; its eigenfunctions are plane waves $e^{i\mathbf{k}\cdot \mathbf{r}}$ with eigenvalue $k^2$. Physically, a plane wave has uniform curvature everywhere, so ${\nabla }^2$ can only multiply it by a constant.

$d=3$: $A(\mathbf{k})=1/({ϵ}_0{k}^2)$, Fourier transform gives $G_0^{(3)}=1/(4\pi {ϵ}_0|\mathbf{r}-{\mathbf{r}}^{\prime }|)$.

$d=1$: Direct integration of ${\partial }_{xx}G=-\delta (x-x^{\prime })/{ϵ}_0$ using $\int \mathrm{d}x\delta (x)=\frac{1}{2}(1+\text{sgn}(x))+C$ and $\int \mathrm{d}x\text{sgn}(x)=|x|+C$ gives $G_0^{(1)}=-|x-x^{\prime }|/(2{ϵ}_0)$ (respecting reciprocity and dropping terms not symmetric in $x-x^{\prime }$).

$d=2$: $A(\mathbf{k})=1/(4{\pi }^2{ϵ}_0{k}^2)$. In polar coordinates with regularization $k^{-1}={\lim }_{\eta \to 0}k/(k^2+{\eta }^2)$: ${\int }_0^{\infty }\mathrm{d}kk/(k^2+{\eta }^2)J_0(k\rho )=K_0(\eta \rho )\to -\ln \rho$ as $\eta \to 0$. So $G_0^{(2)}=-\ln |\mathbf{\rho }-{\mathbf{\rho }}^{\prime }|/(2\pi {ϵ}_0)$.

General $d$: Starting from $G_0^{(d)}=\frac{1}{{ϵ}_0(2\pi {)}^d}\int \frac{{\mathrm{d}}^{d}k}{k^2}{e}^{i\mathbf{k}\cdot \mathbf{r}}$, write ${\mathrm{d}}^{d}k=k^{d-1}\mathrm{d}k\mathrm{d}{\Omega }_d$. The angular integral over the $d$-sphere gives:

$$\int \mathrm{d}{\Omega }_d{e}^{ikr\cos \theta }=\frac{(2\pi {)}^{d/2}}{(kr{)}^{d/2-1}}{J}_{d/2-1}(kr)$$

The remaining radial integral is evaluated via:

$${\int }_0^{\infty }\mathrm{d}k{k}^{\nu -1}{J}_{\nu }(kr)=\frac{2^{\nu -1}\Gamma (\nu )}{r^{\nu }}$$

with $\nu =d/2-1$. Combining:

$$G_0^{(d)}(\mathbf{r})=\frac{\Gamma \left(\frac{d-2}{2}\right)}{4{\pi }^{d/2}{ϵ}_0{r}^{d-2}}$$

This breaks at $d=2$ where $\Gamma (0)$ diverges, reflecting the logarithmic singularity.

Physical interpretation of singularity scaling: In $d$ dimensions, flux through a $(d-1)$-sphere (area $S_{d-1}=2{\pi }^{d/2}/\Gamma (d/2)$, scaling as $r^{d-1}$) must equal the enclosed charge. As $r\to 0$: in $d=1$ the “sphere” is two points (constant area, $E\sim \text{const}$, $G\sim r$); in $d=2$ it’s a circle ($E\sim 1/r$, $G\sim \ln r$); in $d=3$ it’s a sphere ($E\sim 1/r^2$, $G\sim 1/r$).

Symmetry: $G_0^{(d)}(\mathbf{r},{\mathbf{r}}^{\prime })=G_0^{(d)}(|\mathbf{r}-{\mathbf{r}}^{\prime }|)$ because the defining equation and BCs are translationally and rotationally invariant.

8.19 — Cube Green Function

Interior Dirichlet Green function for cube $-a\le x,y,z\le a$. Find $\sigma (0,0)$ on $z=a$ face for $Q$ at center.

Solution Sketch

Subtlety: On $[-a,a]$, odd cosines alone miss the odd-function subspace. Use shifted sines ${\psi }_m(x)=\sqrt{1/a}\sin (m\pi (x+a)/(2a))$ which form a complete orthonormal set vanishing at $x=\pm a$. The ansatz:

$$G(\mathbf{r},{\mathbf{r}}^{\prime })=\frac{1}{{ϵ}_0{a}^2}\sum _{m,n}{G}_{mn}(z,z^{\prime }){\psi }_m(x){\psi }_m(x^{\prime }){\psi }_n(y){\psi }_n(y^{\prime })$$

reduces to the ODE $[{\partial }_z^2-{\kappa }^2]G_{mn}=-(4/{ϵ}_0)\delta (z-z^{\prime })$ with ${\kappa }^2={\pi }^2(m^2+n^2)/(4a^2)$.

Continuity at $z=z^{\prime }$, vanishing at $z=\pm a$, and the jump condition fix:

$$G_{mn}(z,z^{\prime })=\frac{\sinh [\kappa (a+z_{<})]\sinh [\kappa (a-z_{>})]}{\kappa \sinh (2\kappa a)}$$

The surface charge on $z=a$:

$$\sigma (0,0)=\frac{Q}{2a^2}\sum _{m,n\text{odd}}\frac{1}{\cosh \left(\frac{\pi }{2}\sqrt{m^2+n^2}\right)}\approx 0.1233\frac{Q}{a^2}$$

8.22 — Green Function for a Dented Beer Can

Cylinder of radius $R$, height $h$, with a wedge removed: $0\le \varphi \le 2\pi /p$.

Solution Sketch

The ansatz uses two completeness relations: particle-in-a-box for $0<z<h$ and $0<\varphi <2\pi /p$:

$$G_D=\sum _{n,m\ge 1}\frac{2p}{{ϵ}_0\pi h}\sin \frac{n\pi z}{h}\sin \frac{n\pi z^{\prime }}{h}\sin \frac{mp\varphi }{2}\sin \frac{mp{\varphi }^{\prime }}{2}{g}_{\lambda }(\rho ,{\rho }^{\prime }|\kappa )$$

with $\kappa =n\pi /h$ and $\lambda =mp/2$. The radial ODE:

$${ϵ}_0\left\{\frac{1}{\rho }{\partial }_{\rho }(\rho {\partial }_{\rho })-\left({\kappa }^2+\frac{{\lambda }^2}{{\rho }^2}\right)\right\}{g}_{\lambda }=-\frac{1}{\rho }\delta (\rho -{\rho }^{\prime })$$

Regularity at $\rho =0$, vanishing at $\rho =R$, and the jump condition give:

$$g_{\lambda }(\rho ,{\rho }^{\prime }|\kappa )=-\frac{I_{\lambda }(\kappa {\rho }_{<})}{{ϵ}_0{I}_{\lambda }(\kappa R)}\left[I_{\lambda }(\kappa R)K_{\lambda }(\kappa {\rho }_{>})-K_{\lambda }(\kappa R)I_{\lambda }(\kappa {\rho }_{>})\right]$$

TODO: Can we get something interesting from this, like the field pattern of a charge inside a dented can?

8.23 — Weyl’s Formula

Derive $G_0(\mathbf{r},{\mathbf{r}}^{\prime })=\frac{1}{2{ϵ}_0}\int \frac{{\mathrm{d}}^2{k}_{\perp }}{(2\pi {)}^2}\frac{e^{i{\mathbf{k}}_{\perp }\cdot ({\mathbf{r}}_{\perp }-{\mathbf{r}}_{\perp }^{\prime })}}{k_{\perp }}{e}^{-k_{\perp }|z-z^{\prime }|}$.

Solution Sketch

Write $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })=\delta ({\mathbf{r}}_{\perp }-{\mathbf{r}}_{\perp }^{\prime })\delta (z-z^{\prime })$ with $\delta ({\mathbf{r}}_{\perp }-{\mathbf{r}}_{\perp }^{\prime })=\int {\mathrm{d}}^2{k}_{\perp }/(2\pi {)}^2{e}^{i{\mathbf{k}}_{\perp }\cdot ({\mathbf{r}}_{\perp }-{\mathbf{r}}_{\perp }^{\prime })}$. Ansatz: $G_0=\int {\mathrm{d}}^2{k}_{\perp }/(2\pi {)}^2{e}^{i{\mathbf{k}}_{\perp }\cdot ({\mathbf{r}}_{\perp }-{\mathbf{r}}_{\perp }^{\prime })}A(k_{\perp },z,z^{\prime })$. Substituting into PE: $({\partial }_z^2-k_{\perp }^2)A=-\delta (z-z^{\prime })/{ϵ}_0$. Solution regular at $|z|\to \infty$: $A=e^{-k_{\perp }|z-z^{\prime }|}/(2{ϵ}_0{k}_{\perp })$.

8.24 — Electrostatics of a Cosmic String

A cosmic string reduces the angular range to $0\le \varphi <2\pi /p$ where $p^{-1}=1-4G\mu /c^2$.

Solution Sketch

(a) $G_0^{(2)}=-\ln |\mathbf{\rho }-{\mathbf{\rho }}^{\prime }|/(2\pi {ϵ}_0)$ (from 8.17).

(b) Adapted delta function: $\delta (\varphi -{\varphi }^{\prime })=(p/2\pi ){\sum }_m{e}^{imp(\varphi -{\varphi }^{\prime })}$.

(c) Direct integration gives:

$$G_0^p=\frac{1}{2\pi }\sum _{m=1}^{\infty }\frac{\cos [mp(\varphi -{\varphi }^{\prime })]}{m}{\left(\frac{{\rho }_{<}}{{\rho }_{>}}\right)}^{mp}-\frac{p}{2\pi }\ln {\rho }_{>}$$

(d) Summing the series in closed form and checking $p=1$ recovers the standard result.

(e) The self-force: $\mathbf{F}=(p-1)q^2\hat{\mathbf{\rho }}/(4\pi {ϵ}_0\rho )$, attractive toward the string.

Physical insight: This is a purely topological effect. The missing wedge means field lines wrap around a slightly smaller angular range, arriving back at the charge asymmetrically. The charge is pulled toward the apex by its own distorted field. For integer $p$, unrolling the cone into flat space shows $p$ copies of the charge (like multiple images in a wedge of mirrors), all pulling toward the apex. The wedge contribution is $G_0^p-G_0^1$ — the free-space self-energy is subtracted, leaving only the topological part.

Analogy: Shout in flat space — sound leaves symmetrically and never returns. Cut a wedge, glue into a cone — some sound converges at the apex and reflects back.

8.25 — Complex Potential for a Line Array

Show $f(z)=-(\lambda /2\pi {ϵ}_0)\ln \tan (\pi z/a)$ is the complex potential for alternating line charges at $x=na$ and $x=(n+\frac{1}{2})a$.

Solution Sketch

Near $z=na+z_0$ with $|z_0|\ll a$: $\tan (\pi z/a)\approx (-1{)}^n\pi z_0/a$, so $f\sim -\lambda \ln z_0$ (positive line charge). Near $z=(n+\frac{1}{2})a$: $\tan \to \cot$, so $f\sim +\lambda \ln z_0$ (negative line). The physical potential:

$$\phi (x,y)=\frac{\lambda }{4\pi {ϵ}_0}\ln \frac{\cosh (2\pi y/a)-\cos (2\pi x/a)}{\cosh (2\pi y/a)+\cos (2\pi x/a)}$$

For $|y|\gg a$: $\cosh (2\pi y/a)\gg 1$, so $\ln (1\pm \cos /\cosh )\approx \pm 2\cos /\cosh$, giving $\phi \sim e^{-2\pi |y|/a}\cos (2\pi x/a)$ — exponential screening from the alternating charges, with decay length $a/(2\pi )$.

Physical picture: The alternating array is charge-neutral on scales $\gg a$. Its far field decays exponentially because positive and negative contributions cancel to leading order — only the lowest Fourier mode with period $a$ survives, and it decays on the scale $a/(2\pi )$.