Chapter 9 — Steady Current

Quote

“The endless circulation of the electric fluid may appear paradoxical and even inexplicable, but it is no less true.”

— Alessandro Volta (1800)

Motivation

By local charge conservation, the current $I={\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{j}$ and charge density $\rho (\mathbf{r},t)$ satisfy the continuity equation:

$$\frac{\partial \rho (\mathbf{r},t)}{\partial t}+\nabla \cdot \mathbf{j}(\mathbf{r},t)=0$$

When the charge density has no explicit time dependence, we arrive at the steady current condition:

$$\nabla \cdot \mathbf{j}(\mathbf{r})=0$$

Borrowing intuition from Gauss’s law, this tells us that steady current lines either close on themselves or begin and end at infinity — there are no sources or sinks. Combined with $\nabla \times \mathbf{E}=0$ (the static limit), the electric and magnetic sectors decouple completely. This chapter focuses on the electric side: given $\nabla \cdot \mathbf{j}=0$ and $\nabla \times \mathbf{E}=0$, what determines the current flow through ohmic matter?

The key results: the Drude model gives a microscopic motivation for Ohm’s law $\mathbf{j}=\sigma \mathbf{E}$; closing Maxwell’s equations with this constitutive relation reduces the problem to Laplace’s equation in uniform conductors, making all of Chapter 7 and Chapter 8 directly applicable; Joule heating is unavoidable and ohmic currents minimize it (a variational principle); maintaining steady current requires non-electrostatic energy sources (EMF); and drift-diffusion connects ohmic transport to screening via the Einstein relation.

The fluid analogy — $\mathbf{j}(\mathbf{r})=\rho (\mathbf{r})\mathbf{v}(\mathbf{r})$ for convection current — is intuitive and often useful, but breaks down for phenomena that depend on inter-particle forces. As J.J. Thomson noted in 1937: “The service of the electric fluid concept to the science of electricity, by suggesting and coordinating research, can hardly be overestimated.” The limitations become apparent when we need quantum mechanics to describe real transport (e.g., band structure, Pauli exclusion).

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The Child–Langmuir Law

Physical Setup

A parallel-plate vacuum diode: heat one plate (the cathode) so electrons are ejected thermionically (we assume with speed zero, so just barely). As the current increases, the space charge of ejected electrons screens the applied field, imposing a maximum current — a self-limiting process.

Initially $\phi (x)=Vx/L$, but as temperature rises and more electrons are emitted, we must solve Poisson’s equation. Equilibrium is reached when $\mathrm{d}\phi /\mathrm{d}x{|}_{x=0}=0$ — at that point the cathode field vanishes and no additional electrons can cross the gap.

From $\nabla \cdot \mathbf{j}=0$ in 1D, $\rho (x)v(x)$ is constant. With $v(0)=0$, energy conservation gives $\frac{1}{2}m{v}^2(x)=e\phi (x)$. Poisson’s equation becomes:

$$\frac{{\mathrm{d}}^2\phi }{\mathrm{d}{x}^2}=-\frac{\rho }{{ϵ}_0}=\frac{|j|}{{ϵ}_0}\sqrt{\frac{m}{2e\phi }}$$

Derivation

Multiply both sides by $\mathrm{d}\phi /\mathrm{d}x$ and integrate using BCs $\phi (0)=0$, $\mathrm{d}\phi /\mathrm{d}x{|}_{x=0}=0$:

$${\left(\frac{\mathrm{d}\phi }{\mathrm{d}x}\right)}^2=\frac{4|j|}{{ϵ}_0}\sqrt{\frac{m}{2e}}\sqrt{\phi }$$

Take the square root and integrate again with $\phi (0)=0$, then impose $\phi (L)=V$:

$$|j|=\frac{4}{9}\sqrt{\frac{2e}{m}}\frac{{ϵ}_0{V}^{3/2}}{L^2}$$

The $V^{3/2}$ scaling (not $V$) is the hallmark of space-charge-limited transport. The extra half-power comes from energy conservation coupling carrier velocity to potential ($v\propto \sqrt{\phi }$): the applied voltage drives both the carrier speed and the charge density needed to sustain the current, so doubling $V$ more than doubles $j$. A quick order-of-magnitude check: $I\sim Q/\Delta t\sim (CV)/(L/\sqrt{2eV/m})\sim ({ϵ}_{0}A/L)\cdot V\cdot \sqrt{2eV/m}/L$, which reproduces the $V^{3/2}/L^2$ dependence.

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The Drude Model and Ohm’s Law

To fully describe current in matter requires quantum mechanics and statistical physics — conduction is an intrinsically non-classical phenomenon. Phenomenologically, however, many systems obey Ohm’s law:

$$\mathbf{j}=\sigma \mathbf{E}$$

This is a constitutive relation describing the many-particle system’s linear response to an applied stimulus.

The Drude model provides a simple classical motivation. Charge carriers (electrons in metals, ions in plasmas) experience an external field and collisions with heavier, nearly immobile scatterers. If $\tau$ is the mean time between collisions:

$$m\dot{\mathbf{v}}=q\mathbf{E}-\frac{m\mathbf{v}}{\tau }$$

The Friction Term Is Statistical

The $-m\mathbf{v}/\tau$ term encodes a statistical statement: $\Delta \mathbf{p}=0-m\mathbf{v}$. Just before a collision the carrier has drift momentum $m\mathbf{v}$; just after, the velocity direction is completely randomized, so the average post-collision drift is zero. The carrier still moves fast (thermal velocity $v_{\text{th}}\gg v_d$), but in a random direction.

Individual collisions may be elastic or inelastic — what matters is that each one destroys the ordered drift momentum. Over many carriers and many collisions, this is a systematic transfer of directed kinetic energy into disordered thermal motion (heat). The friction term is thus dissipative from the drift perspective even though the Drude model does not track the carrier’s thermal energy.

In steady state $\dot{\mathbf{v}}=0$, giving the drift velocity ${\mathbf{v}}_d=q\tau \mathbf{E}/m$. With carrier density $n$, this yields $\mathbf{j}=nq{\mathbf{v}}_d=\sigma \mathbf{E}$ with:

$$\sigma =\frac{n{q}^2\tau }{m}$$

Quantum Mechanical Remark

The quantum result for metals has the same algebraic form, but $m\to m^{\ast }$ (effective mass from band structure), $\tau \to {\tau }_{\text{tr}}$ (transport scattering time from Fermi surface), and $n$ counts only carriers near the Fermi level. The Drude form is more robust than the model deserves.

Technical note: For thin filamentary wires where $\mathrm{d}\mathbf{\ell }$ is parallel to $\mathbf{j}$ everywhere, we can make the replacement:

$$\int {\mathrm{d}}^{3}r\mathbf{j}\times \mathbf{C}\to I\int \mathrm{d}\mathbf{\ell }\times \mathbf{C}$$

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Ohmic Media and Laplace’s Equation

With $\mathbf{j}=\sigma \mathbf{E}=-\sigma \nabla \phi$ and $\nabla \cdot \mathbf{j}=0$, a uniform conductor ($\sigma$ space-independent) satisfies:

$${\nabla }^2\phi (\mathbf{r})=0$$

in its interior. This is the crucial bridge: all the potential-theory machinery of Chapter 7 and Chapter 8 applies directly to steady currents in homogeneous ohmic media.

Matching Conditions

At an interface between media with conductivities ${\sigma }_1$ and ${\sigma }_2$:

Condition	Origin
${\hat{\mathbf{n}}}_2\cdot ({\mathbf{j}}_1-{\mathbf{j}}_2)=0$	$\nabla \cdot \mathbf{j}=0$ (no charge accumulation)
${\hat{\mathbf{n}}}_2\times ({\mathbf{E}}_1-{\mathbf{E}}_2)=0$	$\nabla \times \mathbf{E}=0$ (irrotational)

Equivalently: ${\sigma }_1\partial {\phi }_1/\partial n{|}_S={\sigma }_2\partial {\phi }_2/\partial n{|}_S$ and ${\phi }_1{|}_S={\phi }_2{|}_S$.

An important consequence: at a boundary between an ohmic medium and an insulator ($\sigma \to 0$), no current flows through, giving a natural Neumann BC: $\partial \phi /\partial n{|}_S=0$.

Resistance

For two conducting electrodes at potential difference $V$ driving steady current $I$ through an ohmic sample:

$$R:=\frac{V}{I}=\frac{1}{\sigma }\frac{{\int }_A^B\mathrm{d}\mathbf{\ell }\cdot \mathbf{E}}{{\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{E}}$$

Uniform Current Density in a Wire

For a wire of arbitrary cross section $A$, uniform over its length $L$, the potential $\phi =-Vz/L$ satisfies Laplace’s equation, gives the correct electrode potentials, and has $\partial \phi /\partial n=0$ on the insulating sides. By uniqueness, it is the solution. The current density $\mathbf{j}=\sigma V/L\hat{\mathbf{z}}$ is therefore uniform, giving:

$$R=\frac{\rho L}{A},\rho :=1/\sigma$$

This applies to, e.g., a washer of height $h$ with inner/outer radii $a,b$: current flows radially, so $\mathrm{d}R=\rho \mathrm{d}r/(2\pi rh)$ and $R=(\rho /2\pi h)\ln (b/a)$.

The $RC={ϵ}_0/\sigma$ Relation

Writing out $R$ and $C$ explicitly for the same electrode geometry embedded in a uniform medium:

$$RC=\frac{1}{\sigma }\frac{{\int }_A^B\mathrm{d}\mathbf{\ell }\cdot \mathbf{E}}{{\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{E}}\cdot \frac{{ϵ}_0{\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{E}}{{\int }_A^B\mathrm{d}\mathbf{\ell }\cdot \mathbf{E}}=\frac{{ϵ}_0}{\sigma }$$

The field distributions cancel because both ${\phi }_{\text{ohmic}}$ and ${\phi }_{\text{vacuum}}$ satisfy Laplace’s equation with identical BCs. This holds for any electrode geometry in a uniform ohmic medium.

This is extremely useful: whenever one of $R$ or $C$ is known for a given geometry, the other follows immediately.

Application: Contact Resistance

Physical setup: Current flows through a single circular hole of radius $a$ in an insulating planar baffle that divides an infinite ohmic medium.

Strategy: Exploit $RC={ϵ}_0/\sigma$. Replace the hole with a negatively charged conducting disk, remove the baffle, and note that $\mathbf{E}$ from the disk reproduces the hole’s field pattern (with reversed arrows on one side). The relevant capacitance is half the disk’s self-capacitance (charge accumulates on both sides of a disk, but the hole problem has current entering from one side only):

$$C_{\text{eff}}=\frac{C_{\text{self}}}{2}=\frac{8{ϵ}_{0}a}{2}=4{ϵ}_{0}a$$

The entry resistance is $R_{\text{in}}={ϵ}_0/(\sigma C_{\text{eff}})=1/(4\sigma a)$. Total contact resistance (entry + exit in series):

$$R=\frac{1}{2\sigma a}$$

Real contacts and constriction resistance. The intrinsic roughness of material surfaces ensures that when two macroscopic conductors are pressed together, current flows only at the tiny asperities where the surfaces truly touch — not over the entire nominal contact area. Each asperity acts as a small orifice of effective radius $a_i$ through an insulating gap (the air/vacuum between the non-touching regions). The total contact resistance is then the parallel combination of many single-orifice resistances:

$$R_{\text{contact}}={\left(\sum _i\frac{1}{R_i}\right)}^{-1}={\left(\sum _{i}2\sigma a_i\right)}^{-1}$$

This is often called constriction resistance — current lines are forced to constrict through a small area before fanning out again, and the bottleneck geometry (not the bulk resistivity) dominates the resistance. In practice, the number and size of asperities depend on surface finish, applied pressure, and material hardness — increasing the clamping force deforms asperities, enlarges effective contact radii, and lowers $R_{\text{contact}}$.

This physics is central to: electrical connectors and switches (where contact resistance causes heating at joints and limits current ratings); spot welding (where the constriction resistance at the interface intentionally concentrates Joule heating to fuse the workpieces); probe-based measurements like the four-point probe; and microelectronics, where shrinking interconnects push contact areas toward the nanoscale.

At nanometer scales, the contact radius $a$ approaches the electron mean free path $\lambda$. When $a\lesssim \lambda$, electrons traverse the constriction ballistically (no scattering inside the contact), and transport crosses over from the ohmic $R\propto 1/a$ regime to the Sharvin (ballistic) regime $R\propto 1/a^2$, where resistance is set by the quantum of conductance and the contact cross section rather than by bulk resistivity.

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Joule Heating

Power Dissipation

The dissipative nature of the Drude friction term is visible directly:

$$P={\mathbf{F}}_{\text{friction}}\cdot \mathbf{v}=-\frac{m{v}_d^2}{\tau }<0$$

Per unit volume (with carrier density $n$): $\mathrm{d}u/\mathrm{d}t=-nm{v}_d^2/\tau =-\sigma E^2=-\mathbf{j}\cdot \mathbf{E}$.

Model-Independent Derivation

The result $\mathrm{d}W/\mathrm{d}t=\int {\mathrm{d}}^{3}r\mathbf{j}\cdot \mathbf{E}$ follows from the Lorentz force alone, without assuming any transport model. The magnetic force does no work ($\mathbf{v}\times \mathbf{B}\cdot \mathbf{v}=0$), so only $\mathbf{E}$ contributes.

For a single charge $q_k$ at ${\mathbf{r}}_k(t)$ in a potential $\phi (\mathbf{r},t)$, the energy is $W_k=q_k\phi ({\mathbf{r}}_k(t),t)$. Taking the total (convective) derivative (cf. Chapter 2):

$$\frac{\mathrm{d}{W}_k}{\mathrm{d}t}=q_k\frac{\partial \phi }{\partial t}{|}_{{\mathbf{r}}_k}+q_k({\mathbf{v}}_k\cdot \nabla )\phi {|}_{{\mathbf{r}}_k}$$

The first term tracks explicit time dependence of the potential at the particle’s location; the second tracks the particle moving through a spatially varying potential. In the steady current regime, all fields are time-independent, so $\partial \phi /\partial t=0$ and:

$$\frac{\mathrm{d}{W}_k}{\mathrm{d}t}=q_k{\mathbf{v}}_k\cdot \nabla \phi =-q_k{\mathbf{v}}_k\cdot \mathbf{E}({\mathbf{r}}_k)$$

Summing over all charges and passing to the continuum:

$$\frac{\mathrm{d}W}{\mathrm{d}t}=-\sum _k{q}_k{\mathbf{v}}_k\cdot \mathbf{E}({\mathbf{r}}_k)\to -\int {\mathrm{d}}^{3}r\rho (\mathbf{r})\mathbf{v}(\mathbf{r})\cdot \mathbf{E}(\mathbf{r})=-\int {\mathrm{d}}^{3}r\mathbf{j}\cdot \mathbf{E}$$

The sign: $\mathrm{d}W/\mathrm{d}t<0$ means the charges lose potential energy; that energy goes into Joule heat (no kinetic energy change at constant drift velocity). In steady state, $|\mathrm{d}W/\mathrm{d}t|=\int {\mathrm{d}}^{3}r\mathbf{j}\cdot \mathbf{E}$ is the Joule heating rate.

For a sample of resistance $R$ carrying total current $I$, using $\nabla \cdot \mathbf{j}=0$ and the divergence theorem:

$${\int }_V{\mathrm{d}}^{3}r\mathbf{j}\cdot \mathbf{E}=-{\int }_V{\mathrm{d}}^{3}r\nabla \cdot (\mathbf{j}\phi )=-{\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{j}\phi =({\phi }_A-{\phi }_B)I=I^{2}R$$

Variational Principle: Minimum Dissipation

Key Result

Among all divergenceless current densities carrying a fixed total current $I$ through specified electrodes, the ohmic current $\mathbf{j}=\sigma \mathbf{E}$ minimizes the total Joule heating $\int {\mathrm{d}}^{3}r|\mathbf{j}{|}^2/\sigma =\int {\mathrm{d}}^{3}r\mathbf{j}\cdot \mathbf{E}$. Ohm’s law is not assumed — it emerges from the minimization; $\nabla \times \mathbf{E}=0$ is not assumed either.

Proof via Constrained Variation

Minimize $|\mathbf{j}{|}^2/\sigma$ subject to:

• $\nabla \cdot \mathbf{j}=0$ everywhere, enforced by a Lagrange function $\psi (\mathbf{r})$;
• $I=\pm {\int }_{S_{\alpha }}\mathrm{d}{\mathbf{S}}_{\alpha }\cdot \mathbf{j}$ at each electrode, enforced by Lagrange constants ${\lambda }_{\alpha }$.

The functional:

$$F[\mathbf{j}]=\frac{1}{2}{\int }_V{\mathrm{d}}^{3}r\frac{|\mathbf{j}{|}^2}{\sigma }-{\int }_V{\mathrm{d}}^{3}r\psi (\mathbf{r})\nabla \cdot \mathbf{j}+\sum _{\alpha }{\lambda }_{\alpha }{\int }_{S_{\alpha }}\mathrm{d}{\mathbf{S}}_{\alpha }\cdot \mathbf{j}$$

Integrate the $\psi \nabla \cdot \mathbf{j}$ term by parts. Since current flows only through the electrodes (not the insulating boundary), the surface terms reduce to electrode contributions. The variation gives:

$$\delta F={\int }_V{\mathrm{d}}^{3}r\delta \mathbf{j}\cdot \left(\frac{\mathbf{j}}{\sigma }+\nabla \psi \right)-\sum _{\alpha }{\int }_{S_{\alpha }}\mathrm{d}{\mathbf{S}}_{\alpha }\cdot \delta \mathbf{j}(\psi -{\lambda }_{\alpha })$$

The extremum requires $\mathbf{j}=-\sigma \nabla \psi$ with $\psi$ constant on each electrode — so $\psi$ is the electrostatic potential and Ohm’s law emerges. The second-order term $\frac{1}{2}\int {\mathrm{d}}^{3}r|\delta \mathbf{j}{|}^2/\sigma >0$ confirms this is a minimum.

Physically: current concentrates in regions of highest $\sigma$ to minimize $|\mathbf{j}{|}^2/\sigma$.

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Electromotive Force

Since $\nabla \times \mathbf{E}=0$, the field $\mathbf{E}=\mathbf{j}/\sigma$ inside a conductor does zero work around any closed loop (by Stokes’ theorem). A steady current flowing in a closed circuit therefore requires a non-electrostatic energy source — chemical, thermal, gravitational, nuclear, or electromagnetic (but not static). By historical convention, such sources provide electromotive force (EMF).

The Fictitious Field ${\mathbf{E}}^{\prime }$

We represent the effect of any EMF source using a fictitious field ${\mathbf{E}}^{\prime }$ that modifies Ohm’s law:

$$\mathbf{j}(\mathbf{r})=\sigma [\mathbf{E}(\mathbf{r})+{\mathbf{E}}^{\prime }(\mathbf{r})]$$

${\mathbf{E}}^{\prime }$ is localized (e.g., between battery terminals), specified once and for all, and generally does not satisfy Maxwell’s equations. It acts as a source term, analogous to $\rho$ in Poisson’s equation. (The exception is Faraday’s law, where ${\mathbf{E}}^{\prime }$ is a true electric field — handled in later chapters.)

Circuit Relations

For two points 1 and 2 in a generic circuit:

Quantity	Definition
Voltage difference	$V_1-V_2={\int }_1^2\mathrm{d}\mathbf{\ell }\cdot \mathbf{E}$
EMF	${\mathcal{E}}_{12}={\int }_1^2\mathrm{d}\mathbf{\ell }\cdot {\mathbf{E}}^{\prime }$
Resistance	$R_{12}=\frac{1}{\sigma I}{\int }_1^2\mathrm{d}\mathbf{\ell }\cdot \mathbf{j}=\frac{L_{12}}{A\sigma }$

From $\mathbf{j}=\sigma (\mathbf{E}+{\mathbf{E}}^{\prime })$:

$$I{R}_{12}={\phi }_1-{\phi }_2+{\mathcal{E}}_{12}$$

(This uses $\nabla \times \mathbf{E}=0$, so $\int \mathrm{d}\mathbf{\ell }\cdot \mathbf{E}=-\Delta \phi$ is path-independent. With time-varying fields, an extra $-{\int }_1^2\mathrm{d}\mathbf{\ell }\cdot \partial \mathbf{A}/\partial t$ appears.)

TODO: Blog post with interactive demo for slowly varying (AC) circuits.

Closing the circuit ($1$ and $2$ identification):

$$\mathcal{E}=IR$$

where $R$ is the total circuit resistance (including the EMF source’s internal resistance), and:

$$\mathcal{E}=\oint \mathrm{d}\mathbf{\ell }\cdot {\mathbf{E}}^{\prime }=\frac{1}{q}\oint \mathrm{d}\mathbf{\ell }\cdot \mathbf{F}$$

where we write ${\mathbf{E}}^{\prime }=\mathbf{F}/q$ to express the EMF as the work done per unit charge by the non-electrostatic “force” $\mathbf{F}$ around the circuit.

Power Budget of an EMF Source

Setting points $1$ and $2$ just outside the terminals of the EMF source:

$$P=I{\mathcal{E}}_{12}=I({\phi }_1-{\phi }_2)+I^2{R}_{12}⟹\mathrm{d}U=\mathrm{d}q{\mathcal{E}}_{12}=\mathrm{d}q({\phi }_1-{\phi }_2)+I^2{R}_{12}\mathrm{d}t$$

The EMF power splits into:

• establishing the terminal potential difference ${\phi }_1-{\phi }_2$, and
• Joule losses in the internal resistance $R_{12}$.

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Kirchhoff’s Laws

Restating $\nabla \cdot \mathbf{j}=0$ and $\nabla \times \mathbf{E}=0$ for filamentary-wire circuits:

Current law (at each node): charge does not accumulate, so

$$\sum _k{I}_k=0$$

Voltage law (around each closed loop): $\oint \mathrm{d}\mathbf{\ell }\cdot \mathbf{E}=0$, giving

$$\sum _k{\mathcal{E}}_k=\sum _n{I}_n{R}_n$$

(EMF positive if the loop passes through from $-$ to $+$; voltage drop positive if loop direction matches $I_n$.)

This yields a system of $b$ linear equations for $b$ unknown branch currents — $n-1$ from the current law (one per independent node) and $b-n+1$ from the voltage law (one per independent loop) — which is always uniquely solvable for a connected circuit. Negative solutions simply indicate the actual current direction is opposite to the assumed one.

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Surface Charges on Current-Carrying Wires

What maintains the static field $\mathbf{E}=\mathbf{j}/\sigma$ inside a wire, especially far from the EMF source? It cannot be volume charge: $\rho \propto \nabla \cdot \mathbf{E}\propto \nabla \cdot \mathbf{j}=0$. The source must be charge on the conductor surface.

Physical Picture

The electric field that drives the current is produced by surface charges, not (only) by the battery’s EMF directly. These charges accumulate at bends and interfaces, redirecting the field to follow the conductor shape. Energy transport is via the Poynting vector through fields outside the wire, not through the wire itself.

The surface charges change identity as carriers flow (individual electrons enter and leave the surface layer), but their macroscopic magnitude and distribution remain constant — the “static” regime is static only in the macroscopic average sense.

Order-of-Magnitude Estimate

At a $90°$ bend in a wire with conductivity $\sigma$: by symmetry, charge patches on the vertical and horizontal surfaces produce a normal field $E_n$ that redirects the current. As a first approximation, $E_n\sim E$ (the normal field must be comparable to the longitudinal field it redirects), so $E\sim I/(A\sigma )$ and $Q/(A{ϵ}_0)\sim E$:

$$Q\sim \frac{{ϵ}_{0}I}{\sigma }$$

For 1 A through copper ($\sigma =6\times 10^7{\Omega }^{-1}{\text{m}}^{-1}$): $Q\sim 10^{-19}$ C — roughly one electron’s worth. The fact that such a tiny charge suffices to steer macroscopic currents reflects the enormous strength of the Coulomb interaction at atomic scales.

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Current Source Distributions

Taking the divergence of $\mathbf{j}=\sigma (\mathbf{E}+{\mathbf{E}}^{\prime })$ with $\nabla \cdot \mathbf{j}=0$:

$$\sigma {\nabla }^2\phi (\mathbf{r})=\sigma \nabla \cdot {\mathbf{E}}^{\prime }(\mathbf{r})=:-f(\mathbf{r})$$

where $f(\mathbf{r})$ represents the spatial distribution of current sources/sinks. This is Poisson’s equation, so all the machinery of Chapter 8 applies.

Point Sources

A point current source $f(\mathbf{r})=I\delta (\mathbf{r})$ at the origin of an infinite conducting medium:

$$\phi (\mathbf{r})=\frac{I}{4\pi \sigma r}$$

A hemispherical electrode on a conducting surface (all current flows into a half-space):

$$\phi (\mathbf{r})=\frac{I}{2\pi \sigma r}$$

Both satisfy $\partial \phi /\partial n=0$ on the surface of the medium.

Neumann Green Function Approach

For a finite conducting sample with insulating boundaries ($\partial \phi /\partial n{|}_S=0$):

$$\phi (\mathbf{r}\in V)=⟨\phi {⟩}_S+\frac{1}{\sigma }{\int }_V{\mathrm{d}}^3{r}^{\prime }{G}_N(\mathbf{r},{\mathbf{r}}^{\prime })f({\mathbf{r}}^{\prime })$$

up to the irrelevant surface-average constant. The Neumann BC constraint ${\int }_S\mathrm{d}S\partial G_N/\partial n^{\prime }=-1/{ϵ}_0$ may complicate analytical work but poses little difficulty numerically.

Application — Four-Point Resistance Probe

Setup: Four uniformly spaced leads (separation $s$) contact the surface of a sample with thickness $h$ and resistivity $\rho =1/\sigma$. Current enters through the outer leads; voltage is measured between the inner leads.

Method: Model contacts as infinitesimal hemispheres. For a semi-infinite medium, superpose ${\phi }_{\pm }(\mathbf{r})=\pm \rho I/(2\pi r_{\pm })$. For finite thickness $h$, the BC $\partial \phi /\partial z{|}_{z=-h}=0$ is enforced by method of images — an infinite array of image point currents at $z=\pm 2nh$.

Result:

$$R=\frac{\rho }{2\pi }\left[\frac{1}{s}+\frac{2}{h}\sum _{n=1}^{\infty }\left(\frac{1}{\sqrt{(s/2h{)}^2+n^2}}-\frac{1}{\sqrt{(s/h{)}^2+n^2}}\right)\right]$$

Limits:

• $h/s\gg 1$ (thick sample): images negligible, $R=\rho /(2\pi s)$.
• $h/s\ll 1$ (thin film): Euler–Maclaurin gives $R=(\rho /\pi h)\ln 2$. Cross-check: replace point sources with line sources in 2D, ${\phi }_{\pm }(\mathbf{r})=\pm (\rho {\ell }^{\prime }/2\pi )\ln r_{\pm }$, and the same result follows directly.

Surface Potential and Internal Sources

Physical Insight

The surface potential of a conductor encodes the internal source distribution: measuring $\phi$ on $S$ determines the first moment of $f(\mathbf{r})$ exactly, without knowing the detailed source structure inside $V$. This is the principle behind organ activity monitoring (brain, heart) — surface measurements recover enough about the source distribution for diagnostic purposes.

Proof

Define the dipole moment of the source distribution: ${\pi }_k:={\int }_V{\mathrm{d}}^{3}rf(\mathbf{r})r_k$.

Starting from $\sigma {\nabla }^2\phi =-f$:

$${\pi }_k=-\sigma {\int }_V{\mathrm{d}}^{3}r{r}_k{\nabla }^2\phi =-\sigma {\int }_V{\mathrm{d}}^{3}r\nabla \cdot (r_k\nabla \phi )+\sigma {\int }_V{\mathrm{d}}^{3}r\nabla \phi \cdot \nabla r_k$$

Apply the divergence theorem to the first term: the insulating BC $\partial \phi /\partial n{|}_S=0$ kills it. For the second, use ${\nabla }^2{r}_k=0$ and the identity $\nabla \phi \cdot \nabla r_k=\nabla \cdot (\phi \nabla r_k)$:

$${\pi }_k=\sigma {\int }_S\mathrm{d}\mathbf{S}\cdot \phi \nabla r_k=\sigma {\int }_S\mathrm{d}{S}_k\phi$$

The splitting $\mathbf{r}\to r_k$ (to be able to apply convenient vector calculus identities) and the identity $\mathbf{V}\cdot \nabla r_k=V_k$ are the key technical steps.

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Drift-Diffusion and the Einstein Relation

If the number density $n(\mathbf{r})$ of carriers varies in space, kinetic theory adds a diffusion current — Fick’s law:

$${\mathbf{j}}_n=-D\nabla n$$

with diffusion constant $D>0$. For carriers of charge $q$, the drift-diffusion (Nernst–Planck) equation is:

$$\mathbf{j}(\mathbf{r})=\sigma \mathbf{E}(\mathbf{r})-qD\nabla n(\mathbf{r})$$

Using ${ϵ}_0\nabla \cdot \mathbf{E}=qn$ and $\nabla (\nabla \cdot \mathbf{E})={\nabla }^2\mathbf{E}$ (from $\nabla \times \mathbf{E}=0$):

$$\mathbf{j}(\mathbf{r})=\sigma \mathbf{E}(\mathbf{r})-{ϵ}_{0}D{\nabla }^2\mathbf{E}(\mathbf{r})$$

In equilibrium ($\mathbf{j}=0$), this gives ${\nabla }^2\mathbf{E}-(\sigma /{ϵ}_{0}D)\mathbf{E}=0$, identifying a characteristic screening length:

$$\ell =\sqrt{\frac{{ϵ}_{0}D}{\sigma }}$$

Comparing with the screening length from Chapter 5:

$$\frac{\sigma }{{ϵ}_{0}D}=\frac{1}{{\ell }^2}$$

This is the Einstein relation — it connects the screening length (a purely equilibrium quantity) to the ratio of two transport coefficients ($\sigma$ and $D$). For good metals, ${\ell }_{\text{TF}}$ is microscopic, so diffusion corrections to Ohm’s law arise only within atomic distances of surfaces or interfaces. For thermal plasmas, ${\ell }_{\text{DH}}$ can be macroscopically large, making diffusion currents significant over large distances from boundary layers in the ionosphere, doped semiconductors, and living cells.

Application — The Resting Potential of a Cell

A potential difference of $\sim 100$ mV exists across nerve and muscle cell walls (interior lower). The cell wall separates two compositionally different conducting plasmas: high K${}^{+}$ concentration inside, low outside.

Fick’s law drives a K${}^{+}$ diffusion current outward, destroying local charge neutrality and forming an electric double layer (like a parallel-plate capacitor). The resulting field drives a conduction current of K${}^{+}$ back inward. In equilibrium, the two currents balance.

Defining mobility $\hat{\mu }=\sigma /(nq)$ so $v_d=\hat{\mu }\mathbf{E}$, the Nernst–Planck equation $\mathbf{j}=qn\hat{\mu }\mathbf{E}-qD\nabla n=0$ gives $\hat{\mu }\mathrm{d}\phi =-D\mathrm{d}n/n$. Integrating across the membrane:

$$\Delta \phi ={\phi }_{\text{out}}-{\phi }_{\text{in}}=\frac{D}{\hat{\mu }}\ln \frac{n_{\text{in}}}{n_{\text{out}}}=\frac{k_{B}T}{q}\ln \frac{n_{\text{in}}}{n_{\text{out}}}$$

(The last step uses the Einstein relation for the Debye–Hückel regime.) For a quiescent neuron: $n_{\text{in}}\approx 120$ mmol/L, $n_{\text{out}}\approx 5$ mmol/L, $T=310$ K:

$$\Delta \phi \approx 86\text{mV}$$

in reasonable agreement with experiment.

TODO: Blog post on cell resting potential with interactive visualization.

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Problems

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9.1 — A Power Theorem

A steady current density $\mathbf{j}$ is confined to a volume $V$. Prove that the rate at which work is done on these charges by a static electric field (from charges not necessarily in $V$) is zero.

Solution Sketch

$\mathcal{P}={\int }_V{\mathrm{d}}^{3}r\mathbf{j}\cdot \mathbf{E}$. For a static field $\mathbf{E}=-\nabla \phi$:

$$\mathcal{P}=-{\int }_V{\mathrm{d}}^{3}r\mathbf{j}\cdot \nabla \phi ={\int }_V{\mathrm{d}}^{3}r\phi \nabla \cdot \mathbf{j}-{\int }_V{\mathrm{d}}^{3}r\nabla \cdot (\mathbf{j}\phi )$$

First term vanishes by $\nabla \cdot \mathbf{j}=0$. Second gives ${\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{j}\phi =0$ because confinement implies $\hat{\mathbf{n}}\cdot \mathbf{j}=0$ on $S$.

Physical insight: A conservative field cannot do net work on charges circulating in closed loops. Any net energy transfer to steady currents requires a non-conservative (EMF) source.

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9.2 — A Salt-Water Tank

A battery maintains potential difference $V$ between the two halves of the top cover ($x<L/2$ at $+V$, $x>L/2$ at $0$) of a tank ($L\times \infty \times h$) filled with salty water of conductivity $\sigma$. Find $\mathbf{j}(x,y,z)$.

Solution Sketch

Laplace’s equation with Neumann BCs on $x=0$, $x=L$, $z=0$, and the split Dirichlet BC at $z=h$. Separate variables:

$$\phi (x,z)=\sum _{n=0}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)\cosh \left(\frac{n\pi z}{h}\right)$$

Orthogonality on the top BC gives $A_0=V/2$ and:

$$A_n=\frac{2V\sin (n\pi /2)}{n\pi \cosh (n\pi )}$$

Only odd $n$ contribute. Then $\mathbf{j}=-\sigma \nabla \phi$.

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9.3 — Radial Hall Effect

An infinitely long cylindrical conductor carries current density $j_z(r)$.

• (a) Find $E_r(r)$ ensuring zero radial Lorentz force on every electron.
• (b) Show ${\rho }_c=-{\rho }_{+}/(1-v^2/c^2)$.
• (c) Estimate $\Delta \phi$ for a 1 cm${}^2$ copper wire at 1 A.

Solution Sketch

(a) Ampère gives $B_{\varphi }(r)=({\mu }_0/r){\int }_0^r\mathrm{d}sj(s)s$. Zero radial Lorentz force requires:

$$E_r(r)=\frac{{\mu }_{0}v}{r}{\int }_0^r\mathrm{d}sj(s)s$$

(b) Gauss gives $E_r(r)=(1/{ϵ}_{0}r){\int }_0^r\mathrm{d}s[{\rho }_c(s)+{\rho }_{+}]s$. Equating with (a) using $j={\rho }_{c}v$ and ${\mu }_0{ϵ}_0=1/c^2$:

$${\rho }_c=-\frac{{\rho }_{+}}{1-v^2/c^2}$$

The mobile charge density exceeds $|{\rho }_{+}|$ by a tiny relativistic correction. To preserve neutrality, electrons squeeze radially inward, leaving a thin positive surface layer.

(c) For copper: ${\rho }_{+}\approx 1.3\times 10^{10}$ C/m${}^3$, $v\approx 8\times 10^{-5}$ m/s, $v^2/c^2\sim 10^{-25}$. The radial field $E_r\approx \frac{1}{2}{\mu }_0{v}^2{\rho }_{+}r$ gives $\Delta \phi \sim 10^{-25}$ V — completely negligible.

Physical insight: This is magnetism as a relativistic correction to electrostatics. The magnetic pinching force on current carriers is balanced by an electric field from a minute charge imbalance (${\rho }_c\ne -{\rho }_{+}$). Every current-carrying wire is relativistic in this sense — a theme developed fully in the relativity chapters.

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9.4 — Acceleration EMF

A rod with cross section $A$ and conductivity $\sigma$ accelerates at $\mathbf{a}$ along its length.

• (a) Show $I=A\sigma ma/e$.
• (b) Generalize to $\mathcal{E}=\oint \mathrm{d}\mathbf{\ell }\cdot (m\mathbf{a}/e)$.
• (c) Estimate the current for a copper ring ($r=1$ cm, $A=1$ mm${}^2$) oscillating at 500 Hz with amplitude $1°$.

Solution Sketch

(a) In the rod’s frame, the Drude equation with no external field: $m(\mathbf{a}+\dot{\mathbf{v}})+m\mathbf{v}/\tau =0$. Steady state: $\mathbf{v}=-\mathbf{a}\tau$, giving $\mathbf{j}=\sigma m\mathbf{a}/e$ and $I=A\sigma ma/e$.

(b) Effective EMF field: ${\mathbf{E}}^{\prime }=m\mathbf{a}/e$, so $\mathcal{E}=\oint \mathrm{d}\mathbf{\ell }\cdot (m\mathbf{a}/e)$.

(c) For angular acceleration $\dot{\Omega }$, $a=r\dot{\Omega }$ and ring area $S=\pi r^2$:

$$I=\frac{2mS\dot{\Omega }}{eR}$$

Numerically $I\approx 15$ nA — extraordinarily small. The inertial EMF $\sim ma/e$ is negligible because the electron mass is so small: mechanical accelerations achievable in the lab produce pseudo-forces on electrons that are dwarfed by typical electromagnetic forces.

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9.5 — Membrane Boundary Conditions

A thin membrane (conductivity ${\sigma }^{\prime }$, thickness $\delta$) separates two regions of conductivity $\sigma$ with uniform current in the $z$-direction. Derive the effective across-the-membrane matching conditions for $\phi (z)$.

Solution Sketch

Laplace’s equation in three regions gives piecewise linear $\phi =A_{i}z+B_i$. Continuity of $\phi$ and $\sigma {\phi }^{\prime }$ at each interface ($z=0$, $z=\delta$) yields:

$$\phi ({\delta }^{+})-\phi (0^{-})=\frac{\delta \sigma }{{\sigma }^{\prime }}{\frac{\mathrm{d}\phi }{\mathrm{d}z}|}_{z=0^{-}}$$ $${\frac{\mathrm{d}\phi }{\mathrm{d}z}|}_{z={\delta }^{+}}-{\frac{\mathrm{d}\phi }{\mathrm{d}z}|}_{z=0^{-}}=0$$

The first says the potential jump is proportional to the normal current; the second says the normal derivative is continuous. The thin-membrane limit replaces a three-region problem with effective two-region matching conditions.

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9.6 — Current Flow to a Bump

A voltage $V_0$ drives steady current through an ohmic medium ($\sigma$) between parallel plates separated by $d$. The lower plate has a hemispherical bump of radius $R\ll d$. Find the current $I$ into the bump.

Solution Sketch

For $d\gg R$, the unperturbed field is $E_0\approx V_0/d$. The bump acts as a grounded hemisphere in a uniform field, which acquires surface charge ${\sigma }_{\text{surf}}=3{ϵ}_0{E}_0\cos \theta$. Integrating the normal current:

$$I=\frac{3\pi \sigma V_0{R}^2}{d}$$

This is three times the current through the flat disk area $\pi R^2$ — the hemisphere “attracts” current by concentrating field lines.

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9.8 — Spherical Child–Langmuir Problem

A spherical vacuum diode with cathode radius $a$, anode radius $b>a$ at potential $V$. Find the maximum thermionic current. Electrons have zero initial velocity.

Solution Sketch

Spherical symmetry: $j=\rho v=I/(4\pi r^2)$ and $\frac{1}{2}m{v}^2=e\phi$. Substitute $r=a{e}^t$ and $\phi =y(t)(I/4\pi {ϵ}_0{)}^{2/3}(m/2e{)}^{1/3}$:

$$y^{\prime \prime }+y^{\prime }=y^{-1/2},y(0)=0,y^{\prime }(0)=0$$

No closed-form solution, but $y(t)$ defines the current implicitly:

$$I=\sqrt{\frac{2e}{m}}4\pi {ϵ}_0\frac{V^{3/2}}{[y(x){]}^{3/2}},x=\ln (b/a)$$

Limit $b\gg a$: Asymptotic solution $y(t)\approx (9t/2{)}^{2/3}$ gives:

$$I\approx \sqrt{\frac{2e}{m}}\frac{8\pi {ϵ}_0{V}^{3/2}}{9[\ln (b/a){]}^2}$$

The change of variable $r=a{e}^t$ is the key technique — it absorbs the $1/r^2$ geometry into a simpler ODE.

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9.9 — Honeycomb Resistor Network

An infinite 2D honeycomb network with edge resistance $r$ has one edge removed. Find the resistance between the two endpoints $A,B$ of the missing edge.

Solution Sketch

Superposition trick. With the edge present: by three-fold symmetry, current $I$ entering $A$ splits equally ($I/3$ per edge), and similarly (another system) for $I$ leaving at $B$. Superposing: $2I/3$ flows through the $AB$ edge, giving $R_{\text{full}}=2r/3$.

The full network is the missing-edge resistance $X$ in parallel with $r$:

$$\frac{1}{X}+\frac{1}{r}=\frac{3}{2r}⟹X=2r$$

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9.10 — Refraction of Current Density

Show that current density obeys a “law of refraction” at the flat boundary between two ohmic media with conductivities ${\sigma }_1$ and ${\sigma }_2$.

Solution Sketch

From matching conditions (${\sigma }_1{E}_{1n}={\sigma }_2{E}_{2n}$ normal, $E_{1t}=E_{2t}$ tangential):

$$\frac{\tan {\theta }_1}{{\sigma }_1}=\frac{\tan {\theta }_2}{{\sigma }_2}$$

Current bends toward the normal in the more conducting medium — analogous to optical refraction with $\sigma$ as refractive index. Each region satisfies Laplace’s equation individually since $\sigma$ is piecewise constant.

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9.12 — A Separation-Independent Resistance

Two highly conducting spheres (radii $a_1,a_2$) inject and extract current deep inside a tank of weakly conducting fluid (conductivity $\sigma$). Show $R$ is nearly independent of their separation $d$ when $d\gg a_1,a_2$.

Solution Sketch

Direct: Model each electrode as a point source/sink $\mathbf{j}=\pm I\hat{\mathbf{r}}/(4\pi r^2)$ (superposition). Integrating potential differences:

$$R\approx \frac{1}{4\pi \sigma }\left(\frac{1}{a_1}+\frac{1}{a_2}-\frac{2}{d}+\mathcal{O}\left(\frac{1}{d^2}\right)\right)\stackrel{d\to \infty }{\to }\frac{1}{4\pi \sigma }\left(\frac{1}{a_1}+\frac{1}{a_2}\right)$$

Via $RC={ϵ}_0/\sigma$: Two-conductor capacitance $C=1/(P_{11}+P_{22}-2P_{12})$ with $P_{ij}=1/(4\pi {ϵ}_0{r}_{ij})$. Same result — the $1/d$ cross-terms vanish for large separation, so each electrode contributes independently.

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9.13 — Inhomogeneous Conductivity

Steady current flows in the $x$-direction in an infinite strip $|y|<L$ with $\sigma (x)={\sigma }_0(1+a\cos kx)$, $a<1$. Find $\mathbf{E}$ everywhere.

Solution Sketch

$|\mathbf{j}|$ is constant, so $E_x=j/\sigma (x)$. Inside the strip:

$${\phi }_{\text{in}}(x,y)=-\frac{j}{{\sigma }_0}x+\frac{ja}{{\sigma }_{0}k}\sin kx$$

The $\sin kx$ term creates nonzero $E_y$ at $|y|=L$. Matching continuity and decay at infinity:

$$\phi (x,y)=-\frac{j}{{\sigma }_0}x+\frac{ja}{{\sigma }_{0}k}\sin kx{e}^{\pm k(y\mp L)},|y|>L$$

So the conductivity modulation “leaks” field into the exterior, decaying exponentially on scale $1/k$.

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9.15 — Resistance of an Ohmic Sphere

Current $I$ enters and exits an ohmic sphere (radius $R$, conductivity $\sigma$) through small polar electrodes ($\theta \le \alpha$, $\pi -\alpha \le \theta$, with $\alpha \ll 1$). Find the resistance.

Solution Sketch

Expand ${\phi }_{\text{in}}=\sum A_{\ell }{r}^{\ell }{P}_{\ell }(\cos \theta )$. By antisymmetry, only odd $\ell$ contribute. Matching $-\partial \phi /\partial r{|}_{r=R}$ to the electrode current (there is current only in small regions of azimuthal angle $\alpha$) and using the hint $\int \mathrm{d}\theta \sin \theta P_{\ell }=[P_{\ell +1}-P_{\ell -1}]$:

$$\phi (r,\theta )=\frac{I}{\sigma \pi (R\alpha {)}^2}\sum _{k=0}^{\infty }\frac{1}{2k+1}\frac{r^{2k+1}}{R^{2k}}[P_{2k+2}(\cos \alpha )-P_{2k}(\cos \alpha )]P_{2k+1}(\cos \theta )$$

The resistance scales as $R_{\text{elec}}\sim 1/(\sigma \pi R{\alpha }^2)$ and diverges as $\alpha \to 0$ — finite current through an infinitesimal area requires infinite voltage.

Technical point: To isolate individual Legendre modes, one must integrate over the full angular range $[0,\pi ]$. Partial-range integration couples infinitely many modes.

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9.16 — Space-Charge-Limited Current in Matter (Mott–Gurney Law)

Replace the vacuum diode with a poor conductor ($v=\hat{\mu }E$, permittivity $ϵ$). Find the maximum current density.

Solution Sketch

With $v=\hat{\mu }E$, Poisson’s equation becomes $(\mathrm{d}\phi /\mathrm{d}x)({\mathrm{d}}^2\phi /\mathrm{d}{x}^2)=j/(\hat{\mu }ϵ)$, i.e. $\frac{\mathrm{d}}{\mathrm{d}x}({\phi }^{\prime }{)}^2=2j/(\hat{\mu }ϵ)$. Integrating twice with $\phi (0)={\phi }^{\prime }(0)=0$:

$$j=\frac{9}{8}\hat{\mu }ϵ\frac{V^2}{L^3}$$

This is the Mott–Gurney law. The $V^2/L^3$ scaling (vs. Child–Langmuir’s $V^{3/2}/L^2$) reflects $v\propto E$ (mobility) rather than $v\propto \sqrt{\phi }$ (energy conservation). Governs charge transport in organic semiconductors and ionic conductors.

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9.17 — van der Pauw’s Formula

An ohmic film (conductivity $\sigma$, thickness $d$, semi-infinite width). Current $I$ enters at $A$, exits at $B$. Contact separations $a,b,c$.

• (a) Find ${\phi }_{AC}$.
• (b) Prove $e^{-\pi d\sigma R_{AB,CD}}+e^{-\pi d\sigma R_{BC,DA}}=1$.

Solution Sketch

(a) Line source: $\phi (\rho )=-(I/\pi d\sigma )\ln \rho$ (from $I=\int \mathrm{d}\mathbf{S}\cdot \mathbf{j}$ over a half-cylinder). Distance $A\to C$ is $a+b$:

$${\phi }_{AC}=-\frac{I}{\pi d\sigma }\ln (a+b)$$

(b) Superposing both current leads:

$$R_{AB,CD}=\frac{1}{\pi d\sigma }\ln \frac{(b+c)(a+b)}{b(a+b+c)},R_{BC,DA}=\frac{1}{\pi d\sigma }\ln \frac{(b+c)(a+b)}{ca}$$

Using $b(a+b+c)=(a+b)(b+c)-ac$, the exponentials add to 1.

Significance: This identity allows resistivity measurement of a flat sample of arbitrary shape — the geometry drops out entirely. It is the foundation of the van der Pauw technique, ubiquitous in semiconductor characterization.

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9.18 — Rayleigh–Carson Reciprocity

An ohmic sample of arbitrary shape: current $I_A$ enters/exits at one pair of points, $I_B$ at another. If $I_A=I_B$, prove $V_A=V_B$ for two systems by evaluating $\int {\mathrm{d}}^{3}r[\nabla \cdot ({\phi }_A{\mathbf{j}}_B)-\nabla \cdot ({\phi }_B{\mathbf{j}}_A)]$.

Solution Sketch

Configurations $A,B$ satisfy $\sigma {\nabla }^2{\phi }_{A,B}=I_{A,B}[\delta (\mathbf{r}-{\mathbf{r}}_1)-\delta (\mathbf{r}-{\mathbf{r}}_2)]$. With $\mathbf{j}=-\sigma \nabla \phi$, expand:

$$M=\int {\mathrm{d}}^{3}r[\nabla {\phi }_A\cdot {\mathbf{j}}_B-{\phi }_A\nabla \cdot {\mathbf{j}}_B-\nabla {\phi }_B\cdot {\mathbf{j}}_A+{\phi }_B\nabla \cdot {\mathbf{j}}_A]$$

The $\nabla \phi \cdot \mathbf{j}$ terms cancel (both give $-\sigma \nabla {\phi }_A\cdot \nabla {\phi }_B$). The delta functions give $M=\sigma (I_B{V}_A-I_A{V}_B)$. By the divergence theorem $M=0$ (no current at infinity). So $V_A=V_B$ when $I_A=I_B$.

TODO: Add sketch.

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9.19 — The Electric Field of an Ohmic Tube

Roll an ohmic sheet into a cylinder of radius $a$, insert a thin line EMF at the slot so $\phi (a,\varphi )=V_0\varphi /(2\pi )$ for $-\pi <\varphi <\pi$. Find $\phi$ everywhere, sum the series, compute ${\sigma }_{\text{surf}}(\alpha )$ and $\mathbf{E}$.

An useful exercise in technical sense with nice visualizations.

Solution Sketch

(a) The BC is a sawtooth — Fourier series: $\varphi /(2\pi )={\sum }_{k=1}^{\infty }(-1{)}^{k-1}\sin (k\varphi )/(k\pi )$. Matching regularity at $\rho =0$ and $\rho =\infty$:

$$\phi (\rho ,\varphi )=\frac{V_0}{\pi }\sum _{k=1}^{\infty }\frac{(-1{)}^{k-1}}{k}\{\begin{array}{ll}(\rho /a{)}^k\sin k\varphi ,& \rho <a\\ (a/\rho {)}^k\sin k\varphi ,& \rho >a\end{array}$$

Why no linear $\varphi$ term: A term $\phi \propto \varphi$ is forbidden in the interior (actually fine on the boundary as it excludes a single point due to EMF source): $\varphi$ is multivalued (jumps by $2\pi$ around the full circle). The Fourier series represents the sawtooth on the open interval $⟨-\pi ,\pi ⟩$ using single-valued functions.

(b) Using $\sum b^k\sin k\varphi /k=\arctan [b\sin \varphi /(1-b\cos \varphi )]$: ${\phi }_{\text{in}}=V_0\alpha /\pi$ where $\alpha =\arctan [\rho \sin \varphi /(a+\rho \cos \varphi )]$ is the polar angle from the slot. Equipotentials inside are straight lines from the EMF seat.

(c) At large $\rho$ only $k=1$ survives: ${\phi }_{\text{out}}\sim V_{0}a\sin \varphi /(\pi \rho )$ — a dipole field perpendicular to the slot.

(d) ${\sigma }_{\text{surf}}(\alpha )=({ϵ}_0{V}_0/2\pi a)\tan \alpha$.

(e) ${\mathbf{E}}_{\text{in}}=-(V_0/\pi s)\hat{\mathbf{\alpha }}$, where $s$ is distance from the slot. Diverges at the EMF seat.

TODO: Add plots of equipotentials and field lines.

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9.20 — Current Density in a Curved Wire

A potential difference $V$ drives current through a wire with a circular-arc bend (inner radius $R_1$, outer radius $R_2$). Find $j(r)$.

Solution Sketch

The potential drop $\Delta \phi =E(r)r\Delta \theta$ between two cross sections is independent of $r$ (constant voltage across each cross section). Therefore $E\propto 1/r$ and:

$$j=\sigma E\propto \frac{1}{r}$$

Current density is highest at the inner radius. For a narrow wire ($R_2-R_1\ll R_1$), $j\approx \text{const}$ — the usual result.

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9.22 — Joule Heating of a Shell

Current flows on a spherical shell (radius $R$, conductivity $\sigma$) with BCs $\phi (\alpha )=V\cos n\varphi$ and $\phi (\pi -\alpha )=-V\cos n\varphi$. Find the Joule heating rate.

Important as it illustrates both the 2D Laplace approach (as the conductor here is a 2D space so we are solving Laplace in the shell material, not inside and outside it as usual) + non-vanishing of the tangential field.

Solution Sketch

2D Laplace on the sphere: the substitution $y=\ln [\tan (\theta /2)]$ maps $\theta \in (0,\pi )$ to $y\in (-\infty ,\infty )$, giving ${\partial }^2\phi /\partial y^2+{\partial }^2\phi /\partial {\varphi }^2=0$ — flat-space Laplace in $(y,\varphi )$.

The solution with $\cos n\varphi$ dependence:

$$\phi (\theta ,\varphi )=[A{\tan }^n(\theta /2)+B{\cot }^n(\theta /2)]\cos n\varphi$$

with $A=V/[{\tan }^n(\alpha /2)-{\cot }^n(\alpha /2)]$ and $B=-A$.

Computing $\mathbf{K}=\sigma {\mathbf{E}}_{\text{tang}}$ and integrating $\mathcal{P}=\int \mathrm{d}S|\mathbf{K}{|}^2/\sigma$. For $n=1$:

$$\mathcal{P}=\frac{2\pi \sigma V^2}{\cos \alpha }$$

Subtlety: $E_r$ need not vanish — with no field inside the shell, ${ϵ}_0{E}_r^{\text{out}}={\sigma }_{\text{surf}}$. These surface charges are implicit in the $\delta \to 0$ shell limit.

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9.23 — Resistance of a Shell

A spherical shell (radius $a$) has conductivity $\sigma$ for ${\alpha }_1<\theta <\pi -{\alpha }_2$ and is perfectly conducting otherwise. Potential difference $V$ between poles. Find the resistance.

Illustrates both the 2D Laplace approach as well as building up the total resistance from infinitesimal parts.

Solution Sketch

(a) Azimuthally symmetric 2D Laplace on the sphere reduces (via $y=\ln [\tan (\theta /2)]$) to ${\phi }^{\prime \prime }=0$:

$$\phi =A+B\ln [\tan (\theta /2)]$$

BCs give $B=V/\ln [\tan ({\alpha }_1/2)\tan ({\alpha }_2/2)]$. Surface current $K=-\sigma B/(a\sin \theta )$, total current $I=2\pi \sigma B$:

$$R=\frac{1}{2\pi \sigma }\ln [\tan ({\alpha }_1/2)\tan ({\alpha }_2/2)]$$

(b) Series approach: each thin ring at angle $\theta$ has radius $a\sin \theta$ and width $a\mathrm{d}\theta$, giving $\mathrm{d}R=\mathrm{d}\theta /(2\pi \sigma \sin \theta )$. Integrating from ${\alpha }_1$ to $\pi -{\alpha }_2$ confirms the result.

Technical tricks: The series approach also follows directly from the Laplace solution: the potential drop across a single ring is $\mathrm{d}\phi =B\mathrm{d}y=B\mathrm{d}\theta /\sin \theta$, since

$$\ln [\tan ((\theta +\mathrm{d}\theta )/2)]-\ln [\tan (\theta /2)]\approx \mathrm{d}\theta /\sin \theta$$

and the current through it is $I=2\pi \sigma B$, so $\mathrm{d}R=\mathrm{d}\phi /I=\mathrm{d}\theta /(2\pi \sigma \sin \theta )$. The substitution $y=\ln [\tan (\theta /2)]$ converts $\int \mathrm{d}\theta /\sin \theta \to \int \mathrm{d}y$, making the integral elementary — a general technique for 2D Laplace on the sphere with azimuthal symmetry.

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9.24 — Resistance of the Atmosphere

The atmosphere’s conductivity increases with height as $\sigma (r)={\sigma }_0+A(r-r_0{)}^2$ with ${\sigma }_0=3\times 10^{-14}$ S/m, $A=0.5\times 10^{-20}$ S/m${}^3$. At $H=50$ km the atmosphere becomes effectively a perfect conductor. The surface field is $E_0\approx 100$ V/m downward. Estimate the atmospheric resistance.

Solution Sketch

Current conservation: $j(r)=j(r_0)r_0^2/r^2$ with $j(r_0)={\sigma }_0{E}_0$. The voltage:

$$V=j(r_0)r_0^2{\int }_{r_0}^{r_0+H}\frac{\mathrm{d}r}{r^2[{\sigma }_0+A(r-r_0{)}^2]}$$

Partial fractions yield an arctan. Numerically, $V\approx 370$ kV, giving:

$$R\approx 250\Omega$$

Physical context: This is the resistance of the global atmospheric electric circuit. The Earth and the ionosphere form the electrodes. Thunderstorms maintain the $\sim 370$ kV potential difference, driving a global fair-weather current of $\sim 1500$ A through this $\sim 250\Omega$ resistance.

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9.25 — Ohmic Loss in an Infinite Circuit

An infinite ladder network built from a repeating $R_2$–$R_1$–$R_2$ motif. Determine $R_1/R_2$ such that all $R_1$ resistors together dissipate a fraction $\alpha$ of the total power. Find the maximum $\alpha$.

TODO: Add circuit diagram in TikZ.

Solution Sketch

Total resistance by the self-similar argument ($R_1$ in parallel with $R$, that in series with $2R_2$):

$$R=2R_2+\frac{R_{1}R}{R_1+R}⟹R=R_2+\sqrt{R_2^2+2R_1{R}_2}$$

Power in all $R_1$s: At each stage, the current $I_n$ entering the $n$-th cell splits: fraction $R/(R_1+R)$ through $R_1$ and $R_1/(R_1+R)$ into the rest. The current reaching stage $n$ is $I_n=I\cdot (R_1/(R_1+R){)}^n$, so the power dissipated in the $n$-th $R_1$ is $I_n^2{R}^2{R}_1/(R_1+R{)}^2$. Summing the geometric series:

$$P_{R_1}^{\text{total}}=\frac{I^2{R}^2{R}_1}{(R_1+R{)}^2}\sum _{n=0}^{\infty }{\left(\frac{R_1}{R_1+R}\right)}^{2n}=\frac{I^2{R}^2{R}_1}{(R_1+R{)}^2-R_1^2}=I^{2}R\frac{R_1}{2R_1+R}$$

so $\alpha =P_{R_1}^{\text{total}}/(I^{2}R)=R_1/(2R_1+R)$. Setting $x=R_1/R_2$:

$$x=\frac{2\alpha (1-\alpha )}{(1-2\alpha {)}^2},\alpha =\frac{1}{2+x+\sqrt{x^2+2x}}$$

Since $x>0$: $\alpha <1/2$ always. As $R_1/R_2\to \infty$, $\alpha \to 1/2$ from below — the series $R_2$ resistors always dissipate at least half the total power because all current must pass through them.