Chapter 5 — Conducting Matter

Quote

“It is evident that the electric fluid in conductors may be considered as moveable.”

— Henry Cavendish (1771)

Motivation

Conducting matter excludes static electric fields from its interior: $\mathbf{E}(\mathbf{r})=\rho (\mathbf{r})=0$ for $\mathbf{r}$ inside the conductor volume $V$. All excess charge resides on the surface.

Cavendish rationalized this by arguing that Coulomb forces rearrange charge until the perfect conductor condition is satisfied. Microscopically, the charge density $\rho (\mathbf{r})$ of mobile electrons is associated with delocalized wavefunctions that spread across the sample. An external field distorts these wavefunctions, displacing opposite charges in opposite directions — this is electrostatic induction.

Key Physical Points About Induction

• Quantum mechanically, metals are easier to polarize than dielectrics because their electronic wavefunctions feel a relatively smoother potential energy landscape.
• Electrostatic induction does not involve long-range charge displacement. Equilibrium is established by tiny perturbations of wavefunctions at every point — these Lorentz-average to the macroscopic surface charge density $\sigma ({\mathbf{r}}_S)$.
• This explains why electrostatic equilibrium is reached extremely fast ($\sim 10^{-18}$ s) — classically, electrons would have to traverse macroscopic distances at drift velocity.
• Far away, if the conductor was initially uncharged, the induced charge produces a dipolar field as a first approximation (one sign of charges is pulled and another pushed away by the field): $\mathbf{p}={\int }_S\mathrm{d}S\mathbf{r}\sigma (\mathbf{r})$.

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Thomson’s Theorem

All systems in equilibrium rearrange $\rho$ to minimize total energy, but differ in the degree to which quantum contributions matter compared to electrostatic ones. For a perfect conductor, QM contributes negligibly (except to fix size and shape) because electrons occupy plane-wave-like states — moving them around costs almost no additional QM energy. The origin of electrostatic induction thus stems from minimizing the classical energy alone.

Thomson's Theorem

The electrostatic energy of a body of fixed shape and size is minimized when its charge $Q$ distributes itself to make $\phi$ constant throughout the body — the $\mathbf{E}=0$ condition for a perfect conductor.

Proof Sketch

Let ${\phi }_{\text{ext}}(\mathbf{r})$ be the potential from fixed external charges. The total energy is:

$$U_E[\rho ]=\frac{1}{8\pi {ϵ}_0}{\int }_V{\mathrm{d}}^{3}r{\int }_V{\mathrm{d}}^3{r}^{\prime }\frac{\rho (\mathbf{r})\rho ({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}+{\int }_V{\mathrm{d}}^{3}r\rho (\mathbf{r}){\phi }_{\text{ext}}(\mathbf{r})$$

Minimize $I[\rho ]=U_E[\rho ]-\lambda {\int }_V{\mathrm{d}}^{3}r\rho (\mathbf{r})$ (Lagrange multiplier enforcing $Q=\int \rho {\mathrm{d}}^{3}r$). Both $\rho (\mathbf{r})$ and $\rho ({\mathbf{r}}^{\prime })$ in the self-energy are the same function evaluated at two integration points, so varying $\rho \to \rho +\delta \rho$ produces two first-order terms that are equal upon $\mathbf{r}\leftrightarrow {\mathbf{r}}^{\prime }$ exchange — turning $1/(8\pi {ϵ}_0)$ into $1/(4\pi {ϵ}_0)$. Setting $\delta I=0$ for arbitrary $\delta \rho$:

$$\frac{1}{4\pi {ϵ}_0}{\int }_V{\mathrm{d}}^3{r}^{\prime }\frac{\rho ({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}+{\phi }_{\text{ext}}(\mathbf{r})=\lambda \mathbf{r}\in V$$

The left side is ${\phi }_{\text{tot}}(\mathbf{r})$; the right side is a constant. The neglected $(\delta \rho {)}^2$ terms are positive definite, confirming this is a true minimum.

Boundaries of Applicability

• Perfect conductors only (charges move freely)
• Total charge fixed
• Electrostatics only
• Fixed geometry
• No breakdown or nonlinear effects

Example — Conducting Sphere in Uniform Field

A conducting sphere of radius $R$ in a uniform field ${\mathbf{E}}_0=E_0\hat{\mathbf{z}}$.

By symmetry, $\sigma (\theta )$ is azimuthally symmetric. Inside, $\sigma$ must produce ${\mathbf{E}}_{\text{self}}=-{\mathbf{E}}_0$, so ${\phi }_{\text{self}}(r<R)=E_{0}r\cos \theta$. By continuity at $r=R$, the exterior must be purely $\ell =1$:

$${\phi }_{\text{self}}(r>R)=E_0\frac{R^3}{r^2}\cos \theta$$

This is exactly dipolar (true only for a sphere), with induced dipole moment $\mathbf{p}=\alpha {ϵ}_0{\mathbf{E}}_0$ where the polarizability $\alpha =4\pi R^3$ scales with the volume — a general rule for compact polarizable objects.

Example — Two Distant Spheres Connected by a Wire

Two uncharged spheres (radii $R_1$, $R_2$) connected by a thin wire, separated by distance $\gg R_1,R_2$, in uniform field ${\mathbf{E}}_0$.

Key approximation: At large separation, each sphere’s charge density is approximately uniform, so the potential at each center is the sum of external potential and its own monopole:

$${\phi }_1=-{\mathbf{E}}_0\cdot {\mathbf{r}}_1-\frac{Q}{4\pi {ϵ}_0{R}_1},{\phi }_2=-{\mathbf{E}}_0\cdot {\mathbf{r}}_2+\frac{Q}{4\pi {ϵ}_0{R}_2}$$

Setting ${\phi }_1={\phi }_2$ (conducting wire) fixes $Q$ and gives the coordinate-free dipole moment:

$$\mathbf{p}=Q\mathbf{d}=4\pi {ϵ}_0\frac{R_1{R}_2}{R_1+R_2}({\mathbf{E}}_0\cdot \mathbf{d})\mathbf{d}$$

where $\mathbf{d}={\mathbf{r}}_2-{\mathbf{r}}_1$.

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Surface Charge and Edge Singularities

From the matching conditions and $\mathbf{E}=0$ inside, the tangential component of ${\mathbf{E}}_{\text{out}}$ also vanishes at the surface — the external field is normal to the conductor surface. The normal component gives:

$${\mathbf{E}}_{\text{out}}({\mathbf{r}}_S)=\frac{\sigma ({\mathbf{r}}_S)}{{ϵ}_0}\hat{\mathbf{n}}⟺\sigma ({\mathbf{r}}_S)={ϵ}_0\hat{\mathbf{n}}\cdot {\mathbf{E}}_{\text{out}}({\mathbf{r}}_S)$$

The surface charge density diverges at conductors with sharp, knife-like edges.

Application — Surface Charge on a Conducting Disk

Strategy: Treat the disk as the $b\to 0$ limit of an oblate ellipsoid. A uniformly charged line segment produces equipotential surfaces that are prolate ellipsoids, so superimposing a conductor on any such surface lets us read off $\sigma$ from $\sigma ={ϵ}_0|\nabla \phi {|}_S$.

Key steps:

1. The potential of a line segment of length $2L$ and charge $Q$ is $\phi (u)=\frac{Q/2L}{4\pi {ϵ}_0}\ln \frac{u+L}{u-L}$, where $u=\frac{1}{2}(r_{+}+r_{-})$ defines prolate ellipsoidal surfaces.
2. Compute $|\nabla \phi |$ on the surface $u=b$ using the chain rule through ellipsoidal coordinates.
3. The general result for a conducting ellipsoid with semi-axes $a,b$:

$$\sigma (\rho ,z)=\frac{Q}{4\pi a^{2}b}\frac{1}{\sqrt{{\rho }^2/a^4+z^2/b^4}}$$

4. Taking $b\to 0$ (flattening to a disk of radius $a$):

$$\sigma (\rho )=\frac{Q}{4\pi a\sqrt{a^2-{\rho }^2}}$$

This diverges as $\sigma \sim (a-\rho {)}^{-1/2}$ at the disk edge — the same square-root singularity found in Chapter 7 for conductors with knife-edges. Geometrically, $\sigma (\rho )$ is the projection of a hemisphere with uniform charge density $Q/(4\pi a^2)$ onto its equatorial plane.

Other methods to derive this: Laplace’s equation in ellipsoidal coordinates (Stratton), integral equation methods (Eyges), or purely geometric arguments (Portis).

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Screening and Shielding

A neutral conductor with a vacuum cavity scooped out of its interior has $\mathbf{E}=0$ inside the cavity.

Proof

Suppose $\mathbf{E}\ne 0$ inside the cavity. Any field line would start on positive charge on the cavity wall and end on negative charge elsewhere on the wall. Close this path through the conductor interior (where $\mathbf{E}=0$). The line integral $\oint \mathrm{d}\mathbf{\ell }\cdot \mathbf{E}$ would then be strictly positive (nonnegative integrand, positive on the cavity segment) — contradicting $\nabla \times \mathbf{E}=0$.

The conductor does not screen the field from charge inside the cavity (any Gaussian surface enclosing the conductor also encloses the charge). However, the field outside is independent of the position of charge inside the cavity.

Why the Exterior Field Is Position-Independent

The charge on the inner cavity wall is always $-q$ (by Gauss’s law applied to a surface just inside the conductor). This $-q$ arranges itself so that $\mathbf{E}=0$ inside the conductor body. But the exterior field depends only on the outer surface charge, which is determined by: (i) the total conductor charge $Q_{\text{tot}}$ (which fixes the total outer charge to $Q_{\text{tot}}+q$), and (ii) the condition $\mathbf{E}=0$ inside the conductor. Neither condition involves the position of the interior charge — only its total. Therefore, the exterior field is the same regardless of where $q$ sits inside the cavity.

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Capacitance

Conductors store charge on their easily accessible surfaces. Capacitance quantifies this storage capacity — both in isolation and in the presence of other conductors.

Self-Capacitance

For a single conductor with volume $\Omega$ and surface $S$, introduce the scaled potential $\phi (\mathbf{r})=V\tilde{\phi }(\mathbf{r})$ with $\tilde{\phi }{|}_{\Omega }=1$ and $\tilde{\phi }\to 0$ at infinity:

$$C=\frac{Q}{V}=-{ϵ}_0{\oint }_S\mathrm{d}\mathbf{S}\cdot \nabla \tilde{\phi }$$

This is a purely geometric quantity. For a sphere of radius $R$: $C=4\pi {ϵ}_{0}R$.

For a conducting disk of radius $R$, use the known $\sigma (\rho )$ and the fact that $\phi (0)=V$:

$$\phi (0)=\frac{2}{4\pi {ϵ}_0}{\int }_0^R\frac{\sigma (\rho )2\pi \rho \mathrm{d}\rho }{\sqrt{R^2-{\rho }^2}}=V⟹C=8{ϵ}_{0}R$$

(The factor of 2 accounts for both faces of the disk contributing; upon merging in the infinitesimally thin limit, their charges add.)

Dimensional Analysis for Self-Capacitance

Since all charge is confined to the surface, $C\sim {ϵ}_0\times \text{length}$. A formula $C\approx {ϵ}_0\sqrt{4\pi A}$ (where $A$ is the surface area) turns out to be remarkably accurate for convex bodies. (See Landau & Lifshitz, Electrodynamics of Continuous Media, §2.)

Grounding

The Earth can be modeled as a very large spherical conductor with $C_{\text{Earth}}\to \infty$, so ${\phi }_{\text{Earth}}\to 0$. It acts as a charge reservoir: finite charge transfers don’t change its potential.

Grounding a conductor (connecting it to Earth by a thin wire) allows charge to flow until $\phi =0$. If the conductor initially has net $Q$, charge flows to Earth; if it’s in an external field, charge is drawn from Earth. In both cases, the energy of the system is lowered.

The Capacitance Matrix

For $N$ conductors (e.g., modern integrated circuits with thousands of metallic contacts), we define the capacitance matrix $\mathbf{C}$ via:

$$Q_i=\sum _{j=1}^N{C}_{ij}{\phi }_j$$

Physical interpretation: Fix conductor $k$ at potential ${\phi }_k$ and ground all others. Then $C_{jk}{\phi }_k$ is the charge drawn up from ground onto conductor $j$. By Laplace equation theory, this charge is uniquely determined.

The $C_{ij}$ are purely geometrical — determined by shapes and relative positions of all conductors. The full charge state is obtained by superposition: the charge on each conductor is the sum of contributions from each individual potential.

Connection to Green Functions

The capacitance matrix is encoded directly in the Dirichlet Green function $G_D(\mathbf{r},{\mathbf{r}}^{\prime })$ for the conductor configuration: $C_{ij}$ is essentially the surface integral of ${\partial }^2{G}_D/\partial n\partial n^{\prime }$ between conductors $i$ and $j$. So the Green-function machinery of Chapter 8 computes capacitance matrices for arbitrary geometries.

Properties of the Capacitance Matrix

• Diagonal: $C_{kk}>0$ (not simply the self-capacitance — other grounded conductors influence the potential)
• Symmetry: $C_{kj}=C_{jk}$ (from Green’s reciprocity: ${\sum }_i{\phi }_i{Q}_i^{\prime }={\sum }_i{\phi }_i^{\prime }{Q}_i$)
• Off-diagonal: $C_{kj}\le 0$
• Row sums: ${\sum }_j{C}_{kj}\ge 0$ (equality only when no field lines escape to infinity)

These are the Maxwell inequalities.

Physical Understanding of the Maxwell Inequalities

Set conductor $k$ to unit positive potential, all others grounded. All field lines begin on $k$ and end on other conductors or at infinity (no lines connect other conductor pairs).

• $Q_k=C_{kk}>0$: source of all field lines
• $Q_m=C_{mk}\le 0$: each grounded conductor absorbs field lines (gains negative charge)
• ${\sum }_j{C}_{kj}=Q_{\text{tot}}\ge 0$: total charge is non-negative (since $\phi \ge 0$ everywhere outside, $\phi$ cannot have a minimum in the exterior region, and the minimum boundary value is 0)

The row sum equals zero only when no field lines escape to infinity (complete shielding).

Example — Point Charge Between Grounded Plates (Green's Reciprocity)

A point charge $q$ at height $z_0$ between two infinite grounded plates separated by $d$.

Reciprocity setup: Treat $q$ as an infinitesimal conductor. Choose a comparison system with $q^{\prime }=0$, one plate at $V$, the other grounded. By translational invariance, ${\phi }^{\prime }(z)=Vz/d$ between the plates. Reciprocity gives:

$$Q_L=-\left(1-\frac{z_0}{d}\right)q,Q_R=-\frac{z_0}{d}q$$

The induced charge on each plate is proportional to the distance from $q$ to the other plate. The total induced charge is $-q$ — every field line from $q$ ends on a plate.

Strategy note: For reciprocity problems, start with all primed quantities arbitrary, then choose values that eliminate unknowns and isolate desired quantities.

Historical Example — The Triode

Before transistors, the triode controlled current flow through vacuum between a heated cathode and an anode using a voltage applied to an intermediate metal grid.

From the capacitance matrix: $Q_C=C_{CC}{\phi }_C+C_{CG}{\phi }_G+C_{CA}{\phi }_A$. The grid voltage ${\phi }_G$ modulates the cathode charge $Q_C$ regardless of the cathode–anode potential difference. Since $Q_C\propto E_C$ (the field just outside the cathode), this controls electron acceleration and hence current. Notably, the physical placement of the grid between cathode and anode is irrelevant to this argument — only the capacitance matrix matters.

Coefficients of Potential and Two-Conductor Capacitance

The capacitance matrix is positive definite (from $U_E=\frac{1}{2}{\sum }_{ij}{\phi }_i{C}_{ij}{\phi }_j\ge 0$), so $\mathbf{P}:={\mathbf{C}}^{-1}$ exists:

$${\phi }_i=\sum _j{P}_{ij}{Q}_j$$

Why $C_{kk}>0$ Alone Doesn't Guarantee Invertibility

A diagonal matrix with all positive entries is invertible, but a general matrix with positive diagonal can be singular. For example, $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ has $C_{11}=C_{22}=1>0$ but determinant zero. It is the positive definiteness of $\mathbf{C}$ (from the energy argument) that guarantees invertibility.

For a two-conductor capacitor (charges $\pm Q$, potentials ${\phi }_1,{\phi }_2$):

$$C=\frac{Q}{{\phi }_1-{\phi }_2}=\frac{1}{P_{11}+P_{22}-2P_{12}}=\frac{C_{11}{C}_{22}-C_{12}^2}{C_{11}+C_{22}+2C_{12}}$$

Weak Coupling Limit

When $|C_{12}|\ll C_{11},C_{22}$: $1/C\approx 1/C_{11}+1/C_{22}$ — capacitances in series. This applies when the conductors are far apart and $C_{kk}$ reduce to self-capacitances.

More generally, for any surface $S$ enclosing the positive conductor:

$$C=\frac{{ϵ}_0{\oint }_S\mathrm{d}\mathbf{S}\cdot \mathbf{E}}{{\int }_1^2\mathrm{d}\mathbf{\ell }\cdot \mathbf{E}}$$

where the line integral follows any path from positive to negative conductor.

The most familiar example: a parallel-plate capacitor with $d\ll \sqrt{A}$ gives $C_0={ϵ}_{0}A/d$. Real capacitors have fringing fields that slightly lower $V$ and thus increase $C$ beyond $C_0$.

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Energy of a System of Conductors

Since charge resides on surfaces and conductors are equipotentials:

$$U_E=\frac{1}{2}\sum _{k=1}^N{Q}_k{\phi }_k=\frac{1}{2}\sum _{ij}{\phi }_i{C}_{ij}{\phi }_j=\frac{1}{2}\sum _{ij}{Q}_i{P}_{ij}{Q}_j$$

For a two-conductor capacitor: $U_E=Q^2/(2C)=\frac{1}{2}C({\phi }_1-{\phi }_2{)}^2$.

For a single isolated conductor: $U_E=Q^2/(2C)$.

Application — Coulomb Blockade

A quantum dot (conducting object with $R\sim 10$ nm) separated from an electron reservoir by an insulating barrier. Since $U_E\propto R^{-1}$, the charging energy is large at nanoscale.

With $N$ electrons on the dot and applied potential $\phi$:

$$U_E(N)=\frac{(Ne{)}^2}{2C}-Ne\phi$$

(Here $C\approx C_{11}$, the dot’s self-capacitance, since $C_{22}\gg C_{11},|C_{12}|$.)

Adding one electron ($N\to N+1$) first becomes energetically favorable when $\phi =e(N+\frac{1}{2})/C$. The result is a staircase $Q(\phi )$: horizontal segments where charge is “blocked” by the electrostatic energy cost, with jumps at half-integer values of $\phi C/e$.

At $kT\gg e^2/(2C)$, thermal excitation washes out the staircase, recovering the macroscopic result $Q=C\phi$. This phenomenon enables a capacitance standard by counting individual electrons.

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Forces on Conductors

From $\mathbf{f}=\frac{1}{2}\sigma ({\mathbf{E}}_{\text{in}}+{\mathbf{E}}_{\text{out}})$ and ${\mathbf{E}}_{\text{in}}=0$:

$$\mathbf{F}=\frac{1}{2{ϵ}_0}{\oint }_S\mathrm{d}S{\sigma }^2\hat{\mathbf{n}}=\frac{{ϵ}_0}{2}{\oint }_S\mathrm{d}S{E}_{\text{out}}^2\hat{\mathbf{n}}$$

An outward electrostatic pressure ${\sigma }^2/(2{ϵ}_0)$ acts on every surface element — reflecting the repulsion between infinitesimal bits of same-sign charge (opposed by QM cohesive forces). For an isolated conductor, this of course integrates to zero.

Force via Energy Methods

A subtlety compared to the usual variational virtual-displacement approach: we can hold either charges or potentials of other conductors constant, and carefulness is needed since they are related through $\mathbf{C}$ and $\mathbf{P}$.

Charges held constant: Start from $U_E=\frac{1}{2}{\sum }_k{\phi }_k{Q}_k$ and compute the total differential:

$$\mathrm{d}{U}_E=\frac{1}{2}\sum _k({\phi }_k\mathrm{d}{Q}_k+Q_k\mathrm{d}{\phi }_k)$$

Derivation: Natural Variables of $U_E$

Substitute $\mathrm{d}{\phi }_k={\sum }_j{P}_{kj}\mathrm{d}{Q}_j+{\sum }_j{Q}_j\mathrm{d}{P}_{kj}$ (from ${\phi }_k={\sum }_j{P}_{kj}{Q}_j$):

$$\mathrm{d}{U}_E=\frac{1}{2}\sum _k\left({\phi }_k\mathrm{d}{Q}_k+\sum _j{Q}_k(P_{kj}\mathrm{d}{Q}_j+Q_j\mathrm{d}{P}_{kj})\right)=\sum _k{\phi }_k\mathrm{d}{Q}_k+\frac{1}{2}\sum _{kj}{Q}_k{Q}_j\mathrm{d}{P}_{kj}$$

(using ${\sum }_k{Q}_k{P}_{kj}={\phi }_j$ to combine the first two terms). The $P_{kj}$ depend only on geometry (${\mathbf{R}}_k$), so $\mathrm{d}{P}_{kj}={\sum }_m(\partial P_{kj}/\partial {\mathbf{R}}_m)\cdot \mathrm{d}{\mathbf{R}}_m$. Reading off:

$${\phi }_k=\left(\frac{\partial U_E}{\partial Q_k}\right){|}_{Q^{\prime },\mathbf{R}},{\mathbf{F}}_k=-\left(\frac{\partial U_E}{\partial {\mathbf{R}}_k}\right){|}_{Q,{\mathbf{R}}^{\prime }}=-\frac{1}{2}\sum _{ij}{Q}_i{Q}_j\left(\frac{\partial P_{ij}}{\partial {\mathbf{R}}_k}\right){|}_{Q,{\mathbf{R}}^{\prime }}$$

The result: $Q_k$ and ${\mathbf{R}}_k$ are the natural variables of $U_E$:

$$U_E=U_E(Q_1,\dots ,Q_N;{\mathbf{R}}_1,\dots ,{\mathbf{R}}_N)$$

and:

$${\phi }_k=\left(\frac{\partial U_E}{\partial Q_k}\right){|}_{Q^{\prime },\mathbf{R}},{\mathbf{F}}_k=-\left(\frac{\partial U_E}{\partial {\mathbf{R}}_k}\right){|}_{Q,{\mathbf{R}}^{\prime }}=-\frac{1}{2}\sum _{ij}{Q}_i{Q}_j\left(\frac{\partial P_{ij}}{\partial {\mathbf{R}}_k}\right){|}_{Q,{\mathbf{R}}^{\prime }}$$

where $Q^{\prime }$ and ${\mathbf{R}}^{\prime }$ mean all other charges/positions held fixed.

Thermodynamic Analogy

Just as internal energy $U(S,V)$ has entropy and volume as natural variables, with $T=\partial U/\partial S$ and $-P=\partial U/\partial V$, here ${\phi }_k=\partial U_E/\partial Q_k$ and ${\mathbf{F}}_k=-\partial U_E/\partial {\mathbf{R}}_k$.

Potentials held constant: To switch from charges to potentials as independent variables, perform a Legendre transform ${\hat{U}}_E=U_E-{\sum }_i{Q}_i{\phi }_i$:

The Legendre Transform Across Physics

Whenever the natural variables of an energy function are inconvenient, subtract the product of conjugate pairs to switch:

Domain	Natural function	Transform	New function
Thermodynamics	$U(S,V)$	$F=U-TS$	$F(T,V)$
Classical mechanics	$L(q,\dot{q})$	$H=p\dot{q}-L$	$H(q,p)$
Electrostatics	$U_E(Q_k,{\mathbf{R}}_k)$	${\hat{U}}_E=U_E-{\sum }_i{Q}_i{\phi }_i$	${\hat{U}}_E({\phi }_k,{\mathbf{R}}_k)$

Taking the differential and using $\mathrm{d}{U}_E={\sum }_k{\phi }_k\mathrm{d}{Q}_k-{\sum }_k{\mathbf{F}}_k\cdot \mathrm{d}{\mathbf{R}}_k$:

$$\mathrm{d}{\hat{U}}_E=-\sum _k{Q}_k\mathrm{d}{\phi }_k-\sum _k{\mathbf{F}}_k\cdot \mathrm{d}{\mathbf{R}}_k$$

so ${\hat{U}}_E({\phi }_1,\dots ,{\phi }_N;{\mathbf{R}}_1,\dots ,{\mathbf{R}}_N)$ with:

$$Q_k=-\left(\frac{\partial {\hat{U}}_E}{\partial {\phi }_k}\right){|}_{{\phi }^{\prime },\mathbf{R}},{\mathbf{F}}_k=-\left(\frac{\partial {\hat{U}}_E}{\partial {\mathbf{R}}_k}\right){|}_{\phi ,{\mathbf{R}}^{\prime }}$$

The key observation: ${\hat{U}}_E=U_E-{\sum }_i{Q}_i{\phi }_i=-\frac{1}{2}{\sum }_i{\phi }_i{Q}_i=-U_E$. Using $U_E=\frac{1}{2}{\sum }_{ij}{\phi }_i{C}_{ij}{\phi }_j$ and noting that only $C_{ij}$ varies (potentials held fixed):

$${\mathbf{F}}_k=\frac{1}{2}\sum _{ij}{\phi }_i{\phi }_j\left(\frac{\partial C_{ij}}{\partial {\mathbf{R}}_k}\right){|}_{\phi ,{\mathbf{R}}^{\prime }}$$

Why the Sign Flips

Naively computing $-\partial U_E/\partial {\mathbf{R}}_k$ from $U_E=\frac{1}{2}{\sum }_{ij}{\phi }_i{C}_{ij}{\phi }_j$ gives the wrong sign — because $U_E$ is not a natural function of ${\phi }_k$. Physically: at fixed potentials, the conductors are not a closed system. A reservoir (battery) must supply/extract charge to maintain constant $\phi$ during displacement.

The Battery Argument

Two conductors at fixed potentials maintained by batteries. After displacement $\delta \mathbf{R}$, energy conservation for the isolated system (conductors + batteries):

$$\mathbf{F}\cdot \delta \mathbf{R}+\Delta U_E+\Delta E_{\text{batt}}=0$$

The battery work: $\Delta E_{\text{batt}}=-{\sum }_k{\phi }_k(Q_k-Q_k^{\text{init}})=-2\Delta U_E$.

We also note that at fixed potentials (as is the case here), $\mathrm{F}\cdot \delta \mathrm{R}=-\Delta {\hat{U}}_E$.

Thus, since the total energy change of the conductor subsystem (field energy plus energy exchanged with the reservoir) is precisely the Legendre-transformed energy:

$$\Delta {\hat{U}}_E=\Delta U_E+\Delta E_{\text{batt}}=\Delta U_E-2\Delta U_E=-\Delta U_E$$

Example — Metal Sheet Inserted into a Capacitor

A metal sheet (thickness $t$) inserted distance $x$ into a parallel-plate capacitor (separation $d>t$, voltage $V$, width $w$).

At fixed potential: $F_x=\partial U_E/\partial x$. The field is $V/(d-t)$ above/below the slab and $V/d$ in the empty region. The energy difference:

$$F_x=\frac{1}{2}{ϵ}_0{V}^{2}w\left(\frac{1}{d-t}-\frac{1}{d}\right)\hat{\mathbf{x}}$$

The slab is drawn into the capacitor. The same conclusion from direct force integration requires careful treatment of the non-uniform fringing field at the slab edge.

Example — Two Capacitors: Open vs. Closed Switches

Capacitor $C$ (separation $s$, area $A$, charge $q$) connected to capacitor $C^{\prime }$ by switches.

(a) Switches open: Isolated system. $\delta U_E=-\frac{1}{2}(q/C{)}^2\delta C=\frac{{\sigma }^2}{2{ϵ}_0}A\delta s$. Force is attractive: $F=-{\sigma }^{2}A/(2{ϵ}_0)$.

(b) Switches closed: Charge flows between capacitors ($\delta q=-\delta q^{\prime }$), but since $q/C=q^{\prime }/C^{\prime }=V$ (common voltage), the extra terms cancel:

$$\delta U_E=-\frac{1}{2}{\left(\frac{q}{C}\right)}^2\delta C+\left(\frac{q}{C}-\frac{q^{\prime }}{C^{\prime }}\right)\delta q=-\frac{1}{2}{\left(\frac{q}{C}\right)}^2\delta C$$

Same force. In the reservoir limit $C^{\prime }\to \infty$ (which serves to fix potential on $C$): $\delta {\hat{U}}_E=\frac{1}{2}{V}^2\delta C=-\delta U_E$ ✓.

Real Conductors: Screening Length

Real conductors spread the infinitesimally thin surface layer into a finite one of width $\ell$ (the screening length):

System	$\ell$
Good metals	$\sim 10^{-10}$ m
Biological plasma	$\sim 10^{-8}$ m
Laboratory plasma	$\sim 10^{-5}$ m
Astrophysical plasma	$\sim 10$ m

Toy Model for Screening

A point charge $q$ at the origin in a background of uniform positive density $e{n}_0$ and mobile negative density $-en(\mathbf{r})$:

$${ϵ}_0{\nabla }^2\phi (\mathbf{r})=-e(n_0-n(\mathbf{r}))-q\delta (\mathbf{r})$$

When $q=0$: $n=n_0$, $\phi =0$ (charge neutrality). When $q\ne 0$: the chemical potential $\mu$ controls particle density in near-equilibrium, so $n(\mathbf{r})=n_0(\mu +e\phi (\mathbf{r}))$.

Why $\mu$ Is Constant

In thermodynamic equilibrium, the chemical potential is uniform throughout the system — otherwise particles would flow from high-$\mu$ to low-$\mu$ regions. The spatial variation enters only through the local potential energy $e\phi (\mathbf{r})$, which shifts the effective chemical potential that controls the local density.

Linearizing: $-e(n_0-n)\approx -e^2\phi \partial n_0/\partial \mu$. Defining $1/{\ell }^2:=(e^2/{ϵ}_0)\partial n_0/\partial \mu$:

$${\nabla }^2\phi -\frac{\phi }{{\ell }^2}=-\frac{q}{{ϵ}_0}\delta (\mathbf{r})⟹\phi (r)=\frac{q}{4\pi {ϵ}_{0}r}{e}^{-r/\ell }$$

The induced charge density integrates to $-q$: the mobile charge exactly compensates the impurity. For a sample of scale $D$, the ratio $\ell /D$ measures how nearly perfectly it responds to electrostatic influence.

Physical Origin of Finite Screening Length

A perfect conductor has $\ell =0$ — infinite compressibility (zero energy cost to squeeze charge into an infinitesimally thin layer). Real systems have finite compressibility because the screening charge gains configurational entropy by spreading out: $S_{\text{config}}\propto Nk\ln (V/N)$.

The Helmholtz free energy density $f\sim nkT\ln n$ gives $\mu =kT\ln (n/n_Q)$, so $\partial n/\partial \mu =n/(kT)$. Entropy of a classical gas creates a pressure resisting compression.

Specific Screening Lengths

Debye–Hückel (classical thermal plasma, electrolytes, doped semiconductors):

Using Boltzmann statistics $n_0(\mu )\propto e^{\mu /(kT)}$:

$${\ell }_{DH}=\sqrt{\frac{{ϵ}_{0}kT}{e^2{n}_0}}\sim 10\text{–}100\text{A˚ in electrolytes}$$

Thomas–Fermi (metals at $T=0$):

Using Fermi statistics $n_0(\mu )=\frac{8\pi }{3}(2m\mu /h^2{)}^{3/2}$:

$${\ell }_{TF}=\sqrt{\frac{\pi a_B}{4k_F}},a_B=\frac{4\pi {ϵ}_0{\hslash }^2}{m{e}^2}\sim 1\mathring{\text{A}}\text{ in good metals}$$

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Problems

Screening and Shielding

5.1 — Conductor with a Cavity

A solid conductor has a vacuum cavity of arbitrary shape scooped out of its interior. Use Earnshaw’s theorem to prove that $\mathbf{E}=0$ inside the cavity.

Solution Sketch

The conductor is an equipotential, so the cavity boundary is at a single constant potential $V$. Inside the cavity there is no charge, so Earnshaw’s theorem applies: $\phi$ has no local extrema, meaning both its maximum and minimum are attained on the boundary. But the boundary value is the single constant $V$, so $\phi =V$ everywhere in the cavity, hence $\mathbf{E}=-\nabla \phi =0$.

Note: This is an elegant alternative to the line-integral argument in the main text — Earnshaw’s theorem converts the problem to a trivial boundary-value observation.

Thomson’s Theorem and Induction

5.5 — Charge Distribution Induced on a Neutral Sphere

A point charge $q$ at distance $r>R$ from the center of an uncharged conducting sphere of radius $R$. Express $\sigma (\theta )={\sum }_{\ell =1}^{\infty }{\sigma }_{\ell }{P}_{\ell }(\cos \theta )$. (a) Show that the total electrostatic energy is $U_E=\frac{1}{{ϵ}_0}{\sum }_{\ell }\frac{{\sigma }_{\ell }}{2\ell +1}\left(\frac{R^3{\sigma }_{\ell }}{4\pi (2\ell +1)}+\frac{q{R}^{\ell +2}}{r^{\ell +1}}\right)$. (b) Use Thomson’s theorem to find $\sigma (\theta )$.

Solution Sketch

(a) The total energy $U_E=U_{\text{self}}+U_{\text{interaction}}$. The interaction energy uses the Legendre expansion $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|={\sum }_{\ell }(R/r{)}^{\ell }{P}_{\ell }(\cos \theta )/(r)$ (since $R<r$). The self-energy uses the surface-surface Coulomb integral expanded in spherical harmonics. Legendre orthogonality collapses both to single sums over $\ell$.

(b) Minimize $U_E$ with respect to each ${\sigma }_{\ell }$:

$$\frac{\partial U_E}{\partial {\sigma }_{\ell }}=0⟹{\sigma }_{\ell }=-\frac{(2\ell +1)q{R}^{\ell -1}}{4\pi r^{\ell +1}}$$ $$\sigma (\theta )=-\frac{q}{4\pi }\sum _{\ell =0}^{\infty }(2\ell +1){\left(\frac{R}{r}\right)}^{\ell +1}\frac{P_{\ell }(\cos \theta )}{R^2}$$

Source: C. Donolato, American Journal of Physics 71, 1232 (2003).

5.8 — Don’t Believe Everything You Read in Journals

Three identical conducting spheres at the corners of an equilateral triangle. A voltage $V$ on one sphere allegedly induces rotation in the other two via electrostatic torque. Show the torque is zero.

Solution Sketch

The torque on a spherical conductor with surface charge $\sigma ({\mathbf{r}}_S)$ about its center:

$$\mathbf{N}=\frac{1}{2}{\oint }_S\mathrm{d}S{\mathbf{r}}_S\times {\sigma }^2({\mathbf{r}}_S)\hat{\mathbf{n}}/{ϵ}_0$$

But $\hat{\mathbf{n}}=\hat{\mathbf{r}}$ for a sphere, so ${\mathbf{r}}_S\times \hat{\mathbf{n}}=R\hat{\mathbf{r}}\times \hat{\mathbf{r}}=0$. The torque vanishes identically — the cross product of any radial vector with the (also radial) surface force is always zero on a sphere.

Source: K. Hense, M. Tajmar, and K. Marhold, Journal of Physics A 37, 8747 (2004).

5.9 — Dipole in a Cavity

A point dipole $\mathbf{p}$ at the center of a spherical cavity (radius $R$) in an infinite conductor. (a) Find the induced surface charge density. (b) Show the force on the dipole is zero.

Solution Sketch

(a) The induced $\sigma$ must cancel the dipole field outside the cavity. From Application 4.3, $\sigma (\theta )={\sigma }_0\cos \theta$ on a sphere produces a uniform interior field and a dipole exterior field. Matching:

$$\sigma ({\mathbf{r}}_S)=-\frac{3(\mathbf{p}\cdot \hat{\mathbf{r}})}{4\pi R^3}$$

(b) This $\sigma$ produces a uniform field inside the cavity. Since $\mathbf{F}=(\mathbf{p}\cdot \nabla )\mathbf{E}$ and $\mathbf{E}$ is constant, $\mathbf{F}=0$.

Capacitance and Reciprocity

5.10 — Charge Induction by a Dipole

A point dipole $\mathbf{p}$ at $\mathbf{r}={\mathbf{r}}_0$ outside a grounded conducting sphere of radius $R$. Use Green’s reciprocity to find the charge drawn up from ground.

Solution Sketch

With ${\phi }_C=0$ (grounded sphere) and comparison system ${\rho }^{\prime }=0$ (sphere at potential ${\phi }_C^{\prime }=Q^{\prime }/(4\pi {ϵ}_{0}R)$, producing ${\phi }^{\prime }(\mathbf{r})=Q^{\prime }/(4\pi {ϵ}_{0}r)$ outside):

$$Q{\phi }_C^{\prime }+\int {\mathrm{d}}^{3}r{\rho }_D(\mathbf{r}){\phi }^{\prime }(\mathbf{r})=0$$

Integrating ${\rho }_D=-\mathbf{p}\cdot \nabla \delta (\mathbf{r}-{\mathbf{r}}_0)$ by parts or using the spherically symmetric potential formula, we find:

$$Q=-\frac{R(\mathbf{p}\cdot {\hat{\mathbf{r}}}_0)}{r_0^2}$$

5.11 — Charge Induction by a Potential Patch

A square patch $|x|\le a$, $|y|\le a$ in $z=0$ held at $V$; the rest of $z=0$ grounded; the plane $z=d$ grounded. Find total charge induced on the entire $z=0$ plane.

Solution Sketch

Label the patch (1), rest of $z=0$ (2), and $z=d$ plane (3). We want $Q_1+Q_2$.

Comparison system: Set ${\phi }_1^{\prime }={\phi }_2^{\prime }=:{\phi }^{\prime }$ and ${\phi }_3^{\prime }=0$, making it a parallel-plate capacitor. Reciprocity:

$$V{Q}_3^{\prime }={\phi }^{\prime }(Q_1+Q_2)$$

The comparison system has $C^{\prime }={ϵ}_{0}A/d$ and $Q_3^{\prime }=\frac{(2a{)}^2}{A}{\phi }^{\prime }\frac{{ϵ}_{0}A}{d}$ by the uniform charge density on the lower plate in the primed system. Therefore:

$$Q_1+Q_2=-\frac{4V{ϵ}_0{a}^2}{d}$$

5.12 — Charge Sharing Among Three Metal Balls

Four identical conducting balls on insulating supports: one has charge $Q$ (fixed position), three are uncharged (movable). Describe a procedure using only moving and contact to give the three balls charges $q$, $-q/2$, $-q/2$ (with $q<Q$), leaving the $+Q$ ball unchanged.

Solution Sketch

Method 1: Bring all three uncharged balls into mutual contact and place them near $+Q$. By induction, only the nearest (A) and farthest (C) balls acquire induced charge: $-q$ on A and $+q$ on C, while the middle ball (B) remains neutral (by symmetry). Separate all three while near $Q$. Now touch B to A: being far from $Q$ and nearly isolated, symmetry splits $+q$ equally, giving $-q/2$ each.

Method 2:

1. Move one uncharged ball (C) to $\infty$.
2. Bring the remaining two (A and B) into contact and place them near $+Q$. By induction, the nearer ball (A) acquires $-q$, its touching partner (B) acquires $+q$.
3. Separate A and B while still near $Q$, freezing their charges.
4. Move A (charge $-q$) far from $Q$.
5. Bring C (from $\infty$) into contact with A. Nearly isolated, symmetry splits the charge: each gets $-q/2$.

5.13 — Conducting Disk: Axis Potential and Reciprocity

A conducting disk of radius $R$ at potential $V$. (a) Find $\phi (z)$ on the axis using the known $\sigma (\rho )$. (b) Ground the disk, place charge $q_0$ on the axis at $z=d$. Use reciprocity to find the charge drawn to the disk.

Solution Sketch

(a) Using $\sigma (\rho )=2{ϵ}_{0}V/(\pi \sqrt{R^2-{\rho }^2})$ (both sides contribute; in the finite case bot hare charged, when we bring them together, the density is doubled):

$$\phi (z)=\frac{2V}{\pi }\arctan \frac{R}{|z|}$$

(b) Reciprocity with comparison system = disk at potential $V$, no point charge:

$$Q^{\prime }=-\frac{q_0\phi (d)}{V}=-\frac{2q_0}{\pi }\arctan \frac{R}{d}$$

Source: V.C.A. Ferraro, Electromagnetic Theory (Athlone Press, 1954).

5.14 — Capacitance of Spheres

(a)–(b) Self-capacitance of Earth ($R=6.4\times 10^6$ m) and a nanosphere ($R=10$ nm): compute $C$ in farads and the energy to add one electron. (c) Two spheres (radii $R_A$, $R_B$, separation $R\gg R_A,R_B$, charges $Q_A$, $Q_B$): find $\mathbf{P}$ and $\mathbf{C}$. (d) Compare diagonal $C_{ii}$ with self-capacitances.

Solution Sketch

(a) $C_{\oplus }=4\pi {ϵ}_{0}R\approx 7\times 10^{-4}$ F. Energy to add one electron: $e^2/(2C)\approx 10^{-16}$ eV — negligible.

(b) $C_{\text{nano}}\approx 10^{-18}$ F. Energy: $e^2/(2C)\approx 0.1$ eV — significant at room temperature.

(c) At large separation, each sphere appears as a point charge to the other:

$$\mathbf{P}=\frac{1}{4\pi {ϵ}_0}\left(\begin{array}{cc}{R}_A^{-1}& R^{-1}\\ R^{-1}& R_B^{-1}\end{array}\right),\mathbf{C}=\frac{4\pi {ϵ}_0}{R^2-R_A{R}_B}\left(\begin{array}{cc}{R}_{A}R& -R_A{R}_B\\ -R_A{R}_B& R_{B}R\end{array}\right)$$

(d) As $R\to \infty$: $C_{AA}\to 4\pi {ϵ}_0{R}_A$ and $C_{BB}\to 4\pi {ϵ}_0{R}_B$ — they approach the self-capacitances.

5.15 — Practice with Green’s Reciprocity

(a) Prove the reciprocity theorem ${\sum }_i{q}_i{\tilde{\phi }}_i={\sum }_i{\tilde{q}}_i{\phi }_i$ directly from symmetry of $C_{ij}$. (b) Three identical spheres at equilateral triangle corners. When potentials are $(\phi ,0,0)$, charges are $(q,q_0,q_0)$. Find the charge on each when all potentials equal ${\phi }^{\prime }$. (c) Find potentials when charges are $(q,0,0)$.

Solution Sketch

(a) ${\sum }_i{q}_i{\tilde{\phi }}_i={\sum }_{ij}{C}_{ij}{\phi }_j{\tilde{\phi }}_i={\sum }_{ij}{C}_{ji}{\tilde{\phi }}_j{\phi }_i={\sum }_j{\tilde{q}}_j{\phi }_j$ (using $C_{ij}=C_{ji}$).

(b) Apply reciprocity between $(\phi ,0,0)\leftrightarrow (q,q_0,q_0)$ and $({\phi }^{\prime },{\phi }^{\prime },{\phi }^{\prime })\leftrightarrow (q^{\prime },q^{\prime },q^{\prime })$:

$$q^{\prime }\phi =(q+2q_0){\phi }^{\prime }⟹q^{\prime }=(q+2q_0)\frac{{\phi }^{\prime }}{\phi }$$

(c) By symmetry, potentials are $({\phi }^{\prime \prime },{\phi }_0,{\phi }_0)$. Two applications of reciprocity give two equations for ${\phi }^{\prime \prime }$ and ${\phi }_0$:

$${\phi }_0=\frac{q_0{q}^{\prime \prime }}{(q_0-q)(q+2q_0)}\phi ,{\phi }^{\prime \prime }=\frac{(q_0+q)q^{\prime \prime }}{(q_0-q)(q+2q_0)}\phi$$

Energy and Forces

5.16 — Maxwell Was Not Always Right

A non-conducting square has fixed surface charge. Cut a slice off one side and glue it onto an adjacent side to make a rectangle of equal area and charge. Maxwell argued this proves $C_{\text{rect}}>C_{\text{sq}}$. Pólya called the proof “amazingly fallacious.” (a) Find the logical error. (b) Give a correct physical argument.

Solution Sketch

(a) Maxwell compares four objects: non-conducting square (A), non-conducting rectangle (B), conducting rectangle (C), conducting square (D). He shows $U_E(B)<U_E(A)$ and $U_E(C)<U_E(B)$, hence $U_E(C)<U_E(A)$. But to conclude $C_{\text{rect}}>C_{\text{sq}}$, he needs $U_E(C)<U_E(D)$ — which does not follow from $U_E(C)<U_E(A)$.

(b) Adding charge $\delta Q$ to a conductor costs $\delta U_E=(Q/C)\delta Q$. For a square, the added charge is geometrically closer to all existing charge than for a rectangle of the same area. So $\delta U_E(\text{sq})>\delta U_E(\text{rect})$, meaning $C_{\text{rect}}>C_{\text{sq}}$.

5.18 — Two Pyramidal Conductors

Two pyramid-shaped conductors each carry charge $Q$. (a) Transfer $\delta Q$ from pyramid 2 to pyramid 1. Find the condition on $P_{ij}$ that lowers the energy. (b) Translate to a condition on $C_{ij}$. (c) When can this determine which pyramid is larger?

Solution Sketch

(a) $\delta U_E=Q\delta Q(P_{11}-P_{22})$. Energy decreases ($\delta U_E<0$) when $P_{22}>P_{11}$.

(b) From $\mathbf{P}={\mathbf{C}}^{-1}$: $P_{22}/P_{11}=C_{11}/C_{22}$. So $P_{22}>P_{11}\Rightarrow C_{22}<C_{11}$.

(c) When widely separated, $C_{kk}$ reduce to self-capacitances $\propto$ linear size. So $C_{11}>C_{22}$ means pyramid 1 is larger — charge spreads out more on a larger conductor, lowering the energy.

5.20 — Bounds on Parallel-Plate Capacitance

Capacitor with identical plates of area $A$ separated by $d$. When $d\ll \sqrt{A}$, $C\to C_0={ϵ}_{0}A/d$. (a) Prove: ${\int }_V|\mathbf{E}{|}^2-{\int }_V|{\mathbf{E}}_0{|}^2={\int }_V|\delta \mathbf{E}{|}^2+2{\int }_V{\mathbf{E}}_0\cdot \delta \mathbf{E}$. (b) Show $C>C_0$.

Solution Sketch

(a) Expand $|\mathbf{E}{|}^2=|{\mathbf{E}}_0+\delta \mathbf{E}{|}^2$ and collect terms — direct algebra.

(b) Restrict the exact energy integral to the volume $V$ between the plates (positive integrand → inequality). The first correction term $\int |\delta \mathbf{E}{|}^2\ge 0$. The cross term:

$${\int }_V{\mathbf{E}}_0\cdot \delta \mathbf{E}={\int }_V\nabla {\phi }_0\cdot \nabla (\delta \phi )={\oint }_{\partial V}\mathrm{d}S\delta \phi {\partial }_n{\phi }_0-{\int }_V\delta \phi {\nabla }^2{\phi }_0$$

The volume term vanishes (${\nabla }^2{\phi }_0=0$). The surface term vanishes: $\delta \phi =0$ on plates (same BCs) and $\hat{\mathbf{n}}\cdot \nabla {\phi }_0=0$ on the cylinder walls. Therefore $C\ge C_0$, with equality only for infinite plates.

5.22 — Off-Center Spherical Capacitor

A spherical capacitor (potential difference $V$, capacitance $C$). The inner sphere is displaced by $ϵ$ from center. Show $C^{\prime }=C+O({ϵ}^2)$ via both symmetry and force/energy arguments.

Solution Sketch

Two independent arguments:

(a) Symmetry: The $+ϵ$ capacitor is the $-ϵ$ capacitor rotated by $180°$. So $C^{\prime }=C+Aϵ+B{ϵ}^2+\cdots$ must satisfy $C^{\prime }(ϵ)=C^{\prime }(-ϵ)$, forcing $A=0$.

(b) Force/energy: At $ϵ=0$, the outer sphere produces zero field inside itself (concentric), so $\mathbf{F}=0$ on the inner sphere. But $F\propto \mathrm{d}{C}^{\prime }/\mathrm{d}ϵ$, so $\mathrm{d}{C}^{\prime }/\mathrm{d}ϵ{|}_{ϵ=0}=0$.

5.23 — Force Between Conducting Hemispheres

(a) A spherical shell at potential $V$: cut in half, find the repulsive force. (b) A spherical capacitor (charges $\pm Q$, radii $a<b$): cut in half, find the repulsive force.

Solution Sketch

The electrostatic pressure ${\sigma }^2/(2{ϵ}_0)$ acts outward on each surface element.

(a) $\sigma ={ϵ}_{0}V/R$ (uniform). The net force on one hemisphere (vertical component of pressure integrated):

$$F=\frac{{\sigma }^2}{2{ϵ}_0}\cdot \pi R^2=\frac{\pi {ϵ}_0{V}^2}{2}$$

(b) The field between the shells acts as pressure — pushing the outer shell out and the inner shell in. For the Maxwell stress tensor approach, choose hemispherical surfaces hugging the inner hemisphere from inside and the outer from outside, capturing half of total charge on each shell.

5.24 — Holding a Sphere Together

A conducting shell (radius $R$, charge $Q$) is sawed in half. A point charge $Q^{\prime }$ at the center prevents the halves from flying apart. (a) Find $Q^{\prime }$ that just barely holds the shell together. (b) Does the answer change for a uniformly charged insulating shell?

Solution Sketch

(a) The force per unit area on the shell is $\mathbf{f}=\frac{1}{2}\sigma ({\mathbf{E}}_{\text{in}}+{\mathbf{E}}_{\text{out}})$. For zero net force, ${\mathbf{E}}^{\prime }=-\frac{1}{2}{\mathbf{E}}_{\text{out}}$, giving:

$$\frac{Q^{\prime }}{4\pi {ϵ}_0{R}^2}=\frac{\sigma }{2{ϵ}_0}=\frac{Q}{8\pi {ϵ}_0{R}^2}⟹Q^{\prime }=-Q/2$$

(b) No change — by Gauss’s law, the external field of the shell is the same whether conducting or uniformly charged.

Source: D. Budker, D.P. DeMille, and D.F. Kimball, Atomic Physics (2004).

5.25 — Force Equivalence

Confirm that the constant-$Q$ and constant-$\phi$ force expressions are equivalent using $\mathbf{PC}=\mathbf{I}$.

Solution Sketch

Differentiate $P_{ik}{C}_{km}={\delta }_{im}$:

$$\delta P_{ik}{C}_{km}+P_{ik}\delta C_{km}=0⟹\delta P_{ij}=-P_{ik}\delta C_{km}{P}_{mj}$$

(using $C^{-1}=P$). Multiply by $Q_i$ and $Q_j$, using ${\phi }_k=P_{ki}{Q}_i$:

$$-Q_i\delta P_{ij}{Q}_j={\phi }_k\delta C_{km}{\phi }_m$$

Dividing by $2\delta {\mathbf{R}}_p$ gives the equivalence of the two force formulas ✓.