Chapter 3 — Electrostatics

Quote

The subject I am going to recommend to your attention almost terrifies me. The variety it presents is immense, and the enumeration of facts serves to confound rather than to inform. The subject I mean is electricity.

— Leonhard Euler (1761)

Motivation

The Maxwell equations for time-independent $\rho (\mathbf{r})$ — with $\partial /\partial t=0$ and $\mathbf{j}=0$ — make it convenient to develop exact approaches for force $\mathbf{F}$, torque $\mathbf{N}$, and electrostatic energies.

Sommerfeld (1952) divides electrostatics into summation problems (given $\rho$, find $\mathbf{E}$) and boundary value problems (given potentials/charges on conductors, find $\mathbf{E}$). This chapter deals primarily with summation; Chapter 7 and Chapter 8 treat boundary value problems.

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The Electrostatic Field

The Maxwell equations specify $\nabla \cdot \mathbf{E}$ and $\nabla \times \mathbf{E}$, which by the Helmholtz theorem uniquely determine $\mathbf{E}$. The curl vanishes; the divergence gives $\rho /{ϵ}_0$. Pulling $\nabla$ inside the integral:

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

The Coulomb force between two distributions follows by superposition:

$$\mathbf{F}=\frac{1}{4\pi {ϵ}_0}\int {\mathrm{d}}^{3}r\int {\mathrm{d}}^3{r}^{\prime }{\rho }^{\ast }(\mathbf{r})\rho ({\mathbf{r}}^{\prime })\frac{\mathbf{r}-{\mathbf{r}}^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}$$

Self-Force Vanishes

The antisymmetry of $(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ under $\mathbf{r}\leftrightarrow {\mathbf{r}}^{\prime }$ immediately shows that the force of ${\rho }^{\ast }$ on $\rho$ equals minus the force of $\rho$ on ${\rho }^{\ast }$. Setting $\rho ={\rho }^{\ast }$: no charge distribution exerts a net force on itself (though it can exert a force on a part of itself).

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The Scalar Potential

Manipulating vectors is harder than manipulating scalars. From the Helmholtz decomposition — before pushing $\nabla$ inside the integral — we identify the electrostatic scalar potential:

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

so that $\mathbf{E}(\mathbf{r})=-\nabla \phi (\mathbf{r})$.

Surface and line charge versions:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \mathrm{d}S\frac{\sigma ({\mathbf{r}}_S)}{|\mathbf{r}-{\mathbf{r}}_S|}\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \mathrm{d}\ell \frac{\lambda (\ell )}{|\mathbf{r}-\mathbf{\ell }|}$$

(Example of a line charge: DNA in vivo.)

Combining Gauss’s law with $\mathbf{E}=-\nabla \phi$ gives Poisson’s equation:

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

When Poisson's Equation Is a Necessity

When $\rho (\mathbf{r})$ is specified once and for all, we can use Coulomb’s integral directly — Poisson’s equation is just an alternative. But when polarizable matter is present, $\rho$ itself depends on $\mathbf{E}$ (and hence on $\phi$). The charge density is no longer an input but part of the unknown. Then Poisson’s equation — a self-consistent PDE for $\phi$ — is the only way forward (Chapter 7 and Chapter 8).

Conservative Nature

The Coulomb force $\mathbf{F}=q\mathbf{E}$ is conservative: the work done is path-independent. This follows from Stokes’ theorem:

$$W_{\text{cycle}}=q{\oint }_C\mathrm{d}\mathbf{\ell }\cdot \mathbf{E}=q{\int }_S\mathrm{d}\mathbf{S}\cdot \nabla \times \mathbf{E}=0$$

Splitting the cycle into two paths between the same endpoints gives $W_1=-W_2$; reversing one shows $W_1=W_2$. This also proves $\mathbf{E}=-\nabla \phi$ without invoking the Helmholtz theorem.

The work done moving a charge from reference point ${\mathbf{r}}_0$ to $\mathbf{r}$:

$$\phi (\mathbf{r})=-{\int }_{{\mathbf{r}}_0}^{\mathbf{r}}\mathrm{d}\mathbf{\ell }\cdot \mathbf{E}$$

We usually set $\phi (\infty )=0$ (possible for any finite charge distribution; fails for infinite geometries like a charged plane).

Matching Conditions for $\phi$

From the matching conditions for $\mathbf{E}$ with ${\hat{\mathbf{n}}}_2$ outward from region 2:

$$\frac{\partial {\phi }_2}{\partial n_2}-\frac{\partial {\phi }_1}{\partial n_2}=\frac{\sigma }{{ϵ}_0}$$

Since $\mathbf{E}$ is at most discontinuous, the integral of $\mathbf{E}$ across the boundary vanishes as the path shrinks to zero:

$${\phi }_1({\mathbf{r}}_S)={\phi }_2({\mathbf{r}}_S)$$

Exception: Dipole Layers

A dipole layer perpendicular to the surface produces a discontinuity in $\phi$. The distribution contains a derivative of $\delta (z)$, so one partial integration yields a surviving $\delta$ contribution.

The continuity of $\phi$ carries the same information as ${\hat{\mathbf{n}}}_2\times ({\mathbf{E}}_1-{\mathbf{E}}_2)=0$ — both encode $\nabla \times \mathbf{E}=0$.

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Earnshaw’s Theorem

Earnshaw's Theorem

The scalar potential $\phi (\mathbf{r})$ in a finite, charge-free region $\mathcal{R}$ takes its maximum and minimum values on the boundary of $\mathcal{R}$.

Consequence: No collection of point charges can be held in stable equilibrium by electrostatic forces alone.

Proof

Suppose $\phi (\mathbf{r})$ has a local minimum at point $P\in \mathcal{R}$. Then on a small surface $S$ enclosing $P$:

$${\oint }_S\mathrm{d}\mathbf{S}\cdot \nabla \phi >0⟹{\int }_V{\mathrm{d}}^{3}r\nabla \cdot \mathbf{E}<0$$

But $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0=0$ everywhere in $V$ (charge-free region). Contradiction. The same argument applies to a maximum with the inequality reversed.

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Equipotentials and Field Lines

Equipotentials: Surfaces defined by $\phi (\mathbf{r})=\text{const.}$ No work is done moving a charge along them — $\mathbf{E}$ is either zero or perpendicular to the equipotential at every point.

Electric field lines: Curves that follow $\mathbf{E}$ at every point, defined by $\mathrm{d}\mathbf{\ell }=\lambda \mathbf{E}$ for some parameter $\lambda$. In Cartesian coordinates: $dx/E_x=dy/E_y=dz/E_z$. These have rare analytic solutions.

Quote

We cannot afford to despise the humbler method of actually drawing tentative figures on paper and selecting that which appears least unlike the figure we require. — James Clerk Maxwell (1891)

Two field lines cannot cross (except at null points where $\mathbf{E}=0$), because $\mathbf{E}$ has a unique direction at every point.

Electric flux: The net number of lines through an arbitrary surface $S$ is proportional to:

$${\Phi }_E={\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{E}$$

For a closed surface, the divergence theorem gives ${\Phi }_E=Q_V/{ϵ}_0$. If $Q_V=0$, every field line that enters $V$ must also leave it.

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The Charged Line Segment

An instructive example (used in Chapter 5 for the conducting disk): a uniform charge per unit length $\lambda$ on $|z|<L$.

Direct integration:

$$\phi (z,\rho )=\frac{\lambda }{4\pi {ϵ}_0}\ln \left[\frac{\sqrt{(L-z{)}^2+{\rho }^2}+L-z}{\sqrt{(L+z{)}^2+{\rho }^2}-L-z}\right]$$

Ellipsoidal Coordinates

Define $r_{\pm }=\sqrt{{\rho }^2+(L\pm z{)}^2}$ (distances from the segment endpoints) and introduce:

$$u=\frac{1}{2}(r_{-}+r_{+}),t=\frac{1}{2}(r_{-}-r_{+})$$

Motivation: We’re looking for coordinates in which $\phi$ depends on only one variable. The identity $ut=-zL$ eliminates both $z$ and $t$ from the potential, giving:

$$\phi =\frac{\lambda }{4\pi {ϵ}_0}\ln \frac{u+L}{u-L}$$

This depends on $u$ alone — the equipotentials are surfaces of constant $u$, i.e., $r_{+}+r_{-}=\text{const.}$, which is the geometric definition of a prolate ellipsoid with foci at the segment endpoints. The curves of constant $t$ (hyperbolae with the same foci) are the field lines.

Limiting cases:

Far Field ( $R\gg L$)

With $R=\sqrt{{\rho }^2+z^2}$ and $Q=2L\lambda$:

$$\phi \approx \frac{Q}{4\pi {ϵ}_{0}R}$$

Any finite charged object looks like a point charge from sufficiently far away. The precise meaning of “sufficiently far” is made clear by the multipole expansion.

Near Field ( $\rho \ll L$, $z\ll L$)

$$\phi (\rho )\approx -\frac{\lambda }{2\pi {ϵ}_0}\ln \rho +\frac{\lambda }{2\pi {ϵ}_0}\ln (2L)$$

The first term is the potential of an infinite line charge — to a nearby observer, the finite segment looks infinite. The divergent constant as $L\to \infty$ has no observable consequences; it drops out of $\mathbf{E}=-\nabla \phi =\frac{\lambda }{2\pi {ϵ}_0\rho }\hat{\mathbf{\rho }}$.

TODO: Add equipotential plot (generate in C++).

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Gauss’s Law in Practice

For direct calculations of field using ${ϵ}_0{\oint }_S\mathrm{d}\mathbf{S}\cdot \mathbf{E}=Q_V$, symmetries are essential — they determine the form of $\mathbf{E}$ and the choice of Gaussian surface.

Symmetry	$\mathbf{E}$ form	Gaussian surface	Result
Spherical: $\rho =\rho (r)$	$E(r)\hat{\mathbf{r}}$	Sphere	$E(r)=\frac{Q(r)}{4\pi {ϵ}_0{r}^2}$
Cylindrical: $\rho =\rho (\rho )$	$E(\rho )\hat{\mathbf{\rho }}$	Cylinder	$E(\rho )=\frac{\lambda (\rho )}{2\pi {ϵ}_0\rho }$
Planar: $\rho =\rho (z)$	$E(z)\hat{\mathbf{z}}$	Pillbox	$E(z)=\frac{\sigma (z)}{2{ϵ}_0}\frac{z}{z}$

The last two cases give potentials with infinite additive constants — artifacts of the infinite geometry.

Uniqueness for Infinite Geometries

The Gauss’s law solution is unique by the Helmholtz theorem for densities that vanish fast enough at infinity. For infinite lines/planes, uniqueness can be established as the limit of a sequence of finite distributions.

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Solid Angle

Definition

The solid angle ${\Omega }_S(\mathbf{r})$ subtended by a surface $S$ at observation point $O$ is the area of the projection of $S$ onto a unit sphere centered at $O$:

$${\Omega }_S(\mathbf{r})={\int }_S\mathrm{d}{\Omega }_S(\mathbf{r})={\int }_S\mathrm{d}\mathbf{S}\cdot \frac{{\mathbf{r}}_S-\mathbf{r}}{|{\mathbf{r}}_S-\mathbf{r}{|}^3}$$

${\Omega }_S$ changes sign when $\mathbf{r}$ passes through $S$ — because ${\mathbf{r}}_S-\mathbf{r}$ reverses direction relative to $\mathrm{d}\mathbf{S}$.

For a closed surface enclosing volume $V$:

$${\Omega }_S=\{\begin{array}{ll}4\pi & O\in V\\ 0& O\notin V\end{array}$$

Why ${\Omega }_S=0$ Outside and $4\pi$ Inside

If $O$ is outside $V$, every ray from $O$ intersects $S$ an even number of times. Adjacent intersections project onto the unit sphere with opposite signs (because $\hat{\mathbf{n}}\cdot {\hat{\mathbf{r}}}_S$ flips), so they cancel pairwise.

If $O$ is inside $V$, every ray makes an odd number of intersections, and the set closest to $O$ projects onto the complete unit sphere: area $4\pi$.

Combined with Coulomb’s field of a point charge, this is exactly Gauss’s law:

$${\Phi }_E(S)={\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{E}=\frac{q}{4\pi {ϵ}_0}{\Omega }_S$$

Warning

This derivation relies critically on the inverse-square nature of the Coulomb force. Gauss’s law does not hold for other radial force laws.

TODO: Add solid angle diagrams.

Application: Field Lines of a Point Charge in a Uniform Field

For a point charge $q>0$ in $\mathbf{E}=E\hat{\mathbf{z}}$, define polar coordinates $(r,\theta )$ centered on the charge. A field line passes through point $P$ at angle $\theta$.

The electric flux through the disk $S$ (perpendicular to $z$, bounded by the circle that $P$ traces under azimuthal symmetry) has two contributions:

$${\Phi }_E(S)=\frac{q{\Omega }_S}{4\pi {ϵ}_0}+E\pi r^2{\sin }^2\theta$$

The solid angle subtended by the disk equals that of a spherical cap:

$${\Omega }_S=2\pi (1-\cos \theta )$$

Now take another point $P^{\prime }$ on the same field line and its corresponding disk $S^{\prime }$. The closed Gaussian surface $G$ formed by $S$, $S^{\prime }$, and the surface of revolution $S_R$ swept by the field line segment $P{P}^{\prime }$ encloses no charge and ${\Phi }_E$ through $S_R$ is zero by construction. Therefore:

$${\Phi }_E(S)+{\Phi }_E(S^{\prime })=0$$

Since $P^{\prime }$ is arbitrary, ${\Phi }_E(S^{\prime })$ is constant along a field line. Letting $a=2\pi {ϵ}_{0}E/q$ and $b=1+2{ϵ}_0{\Phi }_E(S^{\prime })/q$, the field line equation is:

$$a{r}^2{\sin }^2\theta -\cos \theta =-b$$

Each value of $b$ gives a distinct field line.

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The Electric Stress Tensor

If $\mathbf{E}$ is the field produced by $\rho$ occupying $\Omega$, the force on $\rho (\mathbf{r}\in V)$ due to $\rho (\mathbf{r}\in \Omega \setminus V)$ is $\mathbf{F}={\int }_V{\mathrm{d}}^{3}r\rho (\mathbf{r})\mathbf{E}(\mathbf{r})$.

If we eliminate $\rho$ using ${ϵ}_0\nabla \cdot \mathbf{E}=\rho$:

$$F_j={ϵ}_0{\int }_V{\mathrm{d}}^{3}r{E}_j{\partial }_i{E}_i={ϵ}_0{\int }_V{\mathrm{d}}^{3}r[{\partial }_i(E_j{E}_i)-E_i{\partial }_i{E}_j]$$

Since $\nabla \times \mathbf{E}=0$ implies ${\partial }_i{E}_j={\partial }_j{E}_i$:

$$F_j={ϵ}_0{\int }_V{\mathrm{d}}^{3}r{\partial }_i\left(E_i{E}_j-\frac{1}{2}{\delta }_{ij}{E}^2\right)$$

This motivates the Maxwell electric stress tensor:

$$T_{ij}(\mathbf{E})={ϵ}_0\left(E_i{E}_j-\frac{1}{2}{\delta }_{ij}{E}^2\right)$$

The two terms have distinct mechanical meaning: ${ϵ}_0{E}_i{E}_j$ is a tension along field lines (like rubber bands in tension pulling the surface outward along $\mathbf{E}$), and $-\frac{1}{2}{ϵ}_0{E}^2{\delta }_{ij}$ is an isotropic pressure transverse to them (field lines repel each other sideways).

The force becomes a surface integral:

$$\mathbf{F}={\oint }_S\mathrm{d}\mathbf{S}\hat{\mathbf{n}}\cdot \mathsf{T}(\mathbf{E})={ϵ}_0{\oint }_S\mathrm{d}\mathbf{S}\left[(\hat{\mathbf{n}}\cdot \mathbf{E})\mathbf{E}-\frac{1}{2}{E}^2\hat{\mathbf{n}}\right]$$

Why This Is Useful

The stress tensor replaces a volume integral over charge with a surface integral over the field — and the surface $S$ need not coincide with the charge boundary. A clever choice of $S$ can dramatically simplify the calculation.

Physically: the net force on charge inside $V$ is “transmitted” through each surface element $\hat{\mathbf{n}}\mathrm{d}S$ by a vector force density $f_j={\hat{n}}_i{T}_{ij}$. That this surface can be chosen arbitrarily in vacuum led Faraday and Maxwell (and their contemporaries) to view the vacuum as an elastic medium (“luminiferous aether”) capable of supporting stresses.

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Force on a Charged Surface

The force per unit area on a charged surface is not obvious: the normal component of $\mathbf{E}$ is discontinuous there, jumping by $\sigma /{ϵ}_0$ across the surface. So which value of $\mathbf{E}$ do we use — the field just above, just below, or something else?

The resolution is to split the surface into a small disk around ${\mathbf{r}}_S$ plus the rest (surface with hole). In the limit where the disk is infinitesimally close to ${\mathbf{r}}_S$, it looks like an infinite plane with density $\sigma ({\mathbf{r}}_S)$, contributing $\pm \sigma /(2{ϵ}_0)\hat{\mathbf{n}}$ on either side. The remaining field ${\mathbf{E}}_S$ from the surface-with-hole is smooth and continuous through the hole (this is the crucial point — removing the local source removes the discontinuity). So on either side:

$${\mathbf{E}}_1={\mathbf{E}}_S+\frac{\sigma }{2{ϵ}_0}{\hat{\mathbf{n}}}_1,{\mathbf{E}}_2={\mathbf{E}}_S+\frac{\sigma }{2{ϵ}_0}{\hat{\mathbf{n}}}_2$$

No surface element exerts a force on itself, so only ${\mathbf{E}}_S$ drives the force per unit area at ${\mathbf{r}}_S$:

$$\mathbf{f}=\sigma {\mathbf{E}}_S=\frac{\sigma }{2}({\mathbf{E}}_1+{\mathbf{E}}_2)$$

${\mathbf{E}}_S$ is the simple average of the fields on either side — an exact result, not an approximation.

Matching Condition for Free

Taking the difference of the two equations above and dotting with ${\hat{\mathbf{n}}}_2$ immediately recovers the standard matching condition ${\hat{\mathbf{n}}}_2\cdot ({\mathbf{E}}_1-{\mathbf{E}}_2)=\sigma /{ϵ}_0$. So this decomposition simultaneously solves two problems: it identifies the correct field for computing surface forces and derives the boundary condition.

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Electrostatic Potential Energy

Potential Energy of a Charge in an External Field

From $\mathbf{F}=q\mathbf{E}=-q\nabla \phi$, identify:

$$V_E=q\phi (\mathbf{r})$$

By superposition, for a distribution ${\rho }_2(\mathbf{r})$ in the potential ${\phi }_1(\mathbf{r})$ produced by other sources:

$$V_E=\int {\mathrm{d}}^{3}r{\rho }_2(\mathbf{r}){\phi }_1(\mathbf{r})$$

Earnshaw Revisited

Mechanical equilibrium of $q$ requires $q\phi (\mathbf{r})$ to have a local minimum at ${\mathbf{r}}_q$, but Earnshaw says no such minimum exists in charge-free space. Therefore, no classical model of matter composed of positive nuclei and negative point electrons can be stable — quantum mechanics is essential for the stability of matter.

Virtual Displacement and Force

Classical Virtual Displacement

A virtual displacement is an imagined, infinitesimal shift $\delta \mathbf{s}$ that is consistent with the constraints, occurs at frozen time ($\delta t=0$), and is used to probe the system’s response.

Displace ${\rho }_2$ by $\delta \mathbf{s}$ with ${\phi }_1$ fixed:

$$\delta {\rho }_2(\mathbf{r})={\rho }_2(\mathbf{r}-\delta \mathbf{s})-{\rho }_2(\mathbf{r})=-\delta \mathbf{s}\cdot \nabla {\rho }_2$$ $$\delta V_E=-\int {\mathrm{d}}^{3}r{\phi }_1(\mathbf{r})\delta \mathbf{s}\cdot \nabla {\rho }_2(\mathbf{r})$$

Integration by parts moves $\nabla$ onto ${\phi }_1$:

$$\delta V_E=\int {\mathrm{d}}^{3}r{\rho }_2(\mathbf{r}){\mathbf{E}}_1(\mathbf{r})\cdot \delta \mathbf{s}$$

recovering the Coulomb force $\mathbf{F}=-\nabla V_E$.

Green’s Reciprocity

Using the Coulomb expression for $V_E$:

$$\int {\mathrm{d}}^{3}r{\rho }_2(\mathbf{r}){\phi }_1(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int {\mathrm{d}}^{3}r\int {\mathrm{d}}^3{r}^{\prime }\frac{{\rho }_2(\mathbf{r}){\rho }_1({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}=\int {\mathrm{d}}^3{r}^{\prime }{\phi }_2({\mathbf{r}}^{\prime }){\rho }_1({\mathbf{r}}^{\prime })$$

Often useful via clever choices of “system 1” and “system 2” to solve problems that are difficult to approach directly.

Key Application: Non-Overlapping Spherical Distributions

For two non-overlapping, spherically symmetric distributions with charges $Q_1,Q_2$ separated by $R$: the potential outside any spherical distribution is that of a point charge:

$$V_E=\int {\mathrm{d}}^{3}r{\rho }_1(\mathbf{r}){\phi }_2(\mathbf{r})=\int {\mathrm{d}}^{3}r{\rho }_1(\mathbf{r}){\phi }_{\text{point},2}(\mathbf{r})$$

(since ${\rho }_1$ lives entirely outside ${\rho }_2$, where ${\phi }_2={\phi }_{\text{point},2}$). Now apply reciprocity — replace the true ${\rho }_1$ with a point charge $Q_1\delta (\mathbf{r}-{\mathbf{r}}_1)$:

$$=\int {\mathrm{d}}^{3}r{\rho }_{\text{point},1}(\mathbf{r}){\phi }_2(\mathbf{r})=Q_1{\phi }_{\text{point},2}({\mathbf{r}}_1)=\frac{Q_1{Q}_2}{4\pi {ϵ}_{0}R}$$

Each step uses only that the potential outside a sphere is Coulombic. The interaction energy — and thus the force — is the same as if each distribution were a point charge.

Application: Mean Value Theorem for $\phi$

$$\frac{1}{4\pi R^2}{\oint }_S\mathrm{d}S\phi (\mathbf{r})=\phi (0)$$

Setup: System 1: point charge $q$ at the origin (produces ${\phi }_1=q/(4\pi {ϵ}_{0}r)$). System 2: no charges inside the sphere, but ${\phi }_2(\mathbf{r})$ specified on $S$.

The trick: Choose ${\rho }_2\sim \delta (r-R)$ — charge only on the sphere surface. The reciprocity integral $\int {\mathrm{d}}^{3}r{\rho }_2{\phi }_1$ picks out ${\phi }_1(R)=q/(4\pi {ϵ}_{0}R)$ times the total charge on $S$. The other integral $\int {\mathrm{d}}^{3}r{\rho }_1{\phi }_2$ picks out $q{\phi }_2(0)$. Equating and dividing by $q$ gives the mean value theorem.

Key principle: Place charge only where you want to know the potential.

Total Electrostatic Energy

The total energy $U_E$ is the work to assemble a charge distribution from infinity, performed quasistatically (ensuring thermodynamic reversibility). $U_E$ achieves its minimum for the ground state.

$$U_E=\frac{1}{2}\int {\mathrm{d}}^{3}r\rho (\mathbf{r})\phi (\mathbf{r})$$

The factor $1/2$ prevents double counting — bringing chunk 1 into the potential of chunk 2 and vice versa.

Derivation via $\lambda$-Scaling

Let the charge distribution at any intermediate stage be $\lambda \rho (\mathbf{r})$; by linearity the potential is $\lambda \phi (\mathbf{r})$. Adding charge $(\mathrm{d}\lambda )\rho (\mathbf{r})$ to any volume element costs $\mathrm{d}\lambda \cdot \lambda \int {\mathrm{d}}^{3}r\rho \phi$. Integrating from $\lambda =0$ to $1$:

$$U_E={\int }_0^1\mathrm{d}\lambda \lambda \int {\mathrm{d}}^{3}r\rho \phi =\frac{1}{2}\int {\mathrm{d}}^{3}r\rho \phi$$

Example — Uniformly Charged Sphere: Assemble by adding spherical shells. At intermediate radius $r$, the potential at the surface is ${\phi }_S=\rho r^2/(3{ϵ}_0)$. Adding a shell of charge $\mathrm{d}q=4\pi r^2\mathrm{d}r\rho$:

$$U_E=\frac{3}{5}\frac{Q^2}{4\pi {ϵ}_{0}R}$$

Point-Charge Divergence

The $R\to 0$ limit diverges — the self-energy of a point charge is infinite. This is the price of using delta functions. The pathology is remedied only in QED (cf. Section 23.6.3).

Energy in Terms of the Field

Eliminating $\rho$ using Gauss’s law and integrating by parts:

$$U_E=\frac{{ϵ}_0}{2}\int {\mathrm{d}}^{3}r|\mathbf{E}{|}^2\ge 0$$

Apparent Contradiction

This is manifestly non-negative, but $U_E=\frac{1}{2}\int \rho \phi$ can be negative (e.g., two opposite point charges). The resolution: the field form includes the divergent self-energies, while the $\rho \phi$ form for point charges excludes them. Care is required when mixing the two.

The naive identification $u_E=\frac{1}{2}{ϵ}_0|\mathbf{E}{|}^2$ as an energy density is correct (Chapter 15), though the same identification for $\frac{1}{2}\rho \phi$ gives a different quantity.

In matter, $\frac{1}{2}{ϵ}_0|\mathbf{E}{|}^2$ generalizes to $\frac{1}{2}\mathbf{E}\cdot \mathbf{D}$, and a second concept appears — the energy change $\Delta U_E$ on inserting a sample into a pre-existing field. See Chapter 6.

Interaction Energy

For $\rho ={\rho }_1+{\rho }_2$:

$$U_E[{\rho }_1+{\rho }_2]=U_E[{\rho }_1]+U_E[{\rho }_2]+V_E$$

where the interaction energy is:

$$V_E=\frac{1}{4\pi {ϵ}_0}\int {\mathrm{d}}^{3}r\int {\mathrm{d}}^3{r}^{\prime }\frac{{\rho }_1(\mathbf{r}){\rho }_2({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

When ${\rho }_2$ creates an unspecified external potential ${\phi }_{\text{ext}}$, the self-energy $U_E[{\rho }_2]$ is absent from the problem.

Three Energies, Easy to Confuse

• Total energy $U_E[\rho ]$: the work to assemble the entire charge distribution from infinity. Includes self-energies of every piece.
• Self-energy $U_E[{\rho }_k]$: the total energy of one piece in isolation. Diverges for point charges; subtracted away in most physical answers.
• Interaction energy $V_E$: the cross term — work to bring the assembled pieces together, holding each rigid. Has no self-energy and so is finite even for point charges.

Common slip: writing $U_E$ when you really mean $V_E$ (or vice versa) for a system of point charges. Rule of thumb: if the answer involves only inter-particle distances and is finite, it’s $V_E$; if it would diverge for point sources, it’s $U_E$.

Application: Ionization Potential of a Metal Cluster (Bréchignac et al., 1989)

Experiments show $I_N=U_E^{\text{final}}-U_E^{\text{initial}}=W+a{N}^{-1/3}$ for a cluster of $N$ monovalent atoms. Model: cluster as a sphere of radius $R=R_S{N}^{1/3}$ (with $\frac{4}{3}\pi R_S^3=n^{-1}$, $n$ the valence electron density) each atom in the cluster by a sphere of radius $R_S<R$. Conservation of volume: $R=R_S{N}^{1/3}$. The volume of each atomic sphere is chosen as $\frac{4}{3}\pi R_S^3=n^{-1}$ where n is the valence electron density of the metal.

Initial state: A single atom (sphere of radius $R_S$) with uniform positive charge density ${\rho }_{+}=|e|n$ and a point electron $-e$ at its center. By Gauss’s law inside/outside the sphere:

$$U_E^{\text{initial}}=\frac{1}{2}{ϵ}_0\int {\mathrm{d}}^{3}r|{\mathbf{E}}_i{|}^2+\int {\mathrm{d}}^{3}r{\rho }_{+}{\phi }_{-e}=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{R_S}\left[\frac{3}{5}-\frac{3}{2}\right]$$

Final state: The electron is gone; charge $+e$ distributed uniformly over the cluster surface (radius $R$). By Gauss’s law:

$$U_E^{\text{final}}=\frac{e^2}{8\pi {ϵ}_{0}R}$$

Using $R=R_S{N}^{1/3}$:

$$I_N\simeq \frac{e^2}{4\pi {ϵ}_0{R}_S}\left[\frac{9}{10}+\frac{1}{2}{N}^{-1/3}\right]$$

reproducing the experimental $W+a{N}^{-1/3}$ scaling.

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Problems

Fields and Potentials

3.3 — Spherical Shell and Volume by Superposition

Find $\mathbf{E}(\mathbf{r})$ inside and outside (a) a uniformly charged spherical shell by superposing ring fields, and (b) a uniformly charged sphere by superposing disk fields.

Solution Sketch

(a): Let $R$ be the shell radius with $\sigma =Q/(4\pi R^2)$. Each ring at angle $\theta$ contributes charge $\mathrm{d}q=\sigma 2\pi R^2\sin \theta \mathrm{d}\theta$ at distance $x-R\cos \theta$ from the observation point on the $x$-axis.

For $x>R$: all rings are to the left, integration gives $E(x)=Q/(4\pi {ϵ}_0{x}^2)$.

For $x<R$: rings on either side of the observation point contribute with opposite signs. The two sign flips (direction and distance) cancel, and the calculation is identical except $|x-R|=R-x$, giving $E=0$ inside.

(b): Same approach with disks instead of rings, using $\mathrm{d}\sigma =\rho R\sin \theta \mathrm{d}\theta$. Outside: $Q/(4\pi {ϵ}_0{r}^2)$. Inside: $\rho r/(3{ϵ}_0)$.

Both recover the standard Gauss’s law results by pure superposition.

3.7 — Field Lines of a Non-Uniformly Charged Disk

A disk of radius $a$ carries uniform $\sigma <0$ on its face and uniform $\lambda >0$ on its rim, with net positive charge. Sketch the field line topology.

Solution Sketch

Three limiting regimes dictate the topology:

1. Far from disk: resembles a positive point charge
2. Near the face (away from rim): resembles an infinite sheet of negative charge (lines pointing inward)
3. Near the rim: resembles a positive line charge (lines pointing outward)

These incompatible local patterns must connect, forcing null points where $\mathbf{E}=0$ and field lines cross. By Earnshaw’s theorem, these null points are saddle points of $\phi$ — $\mathbf{E}$ converges from two directions and diverges in the perpendicular directions.

3.8 — Charged Slab and Sheet

(b) Find $\mathbf{E}(\mathbf{r})$ for $\rho (x,y,z)={\sigma }_0\delta (x)+{\rho }_0\Theta (x)-{\rho }_0\Theta (x-b)$. (c) Show that this distribution exerts no net force on itself.

Solution Sketch

A useful drill in superposition. Apply Gauss’s law to each component (infinite sheet + finite slab), sum the fields. For self-force: the antisymmetry argument — the integrand in the self-force integral is odd under the exchange that maps $\rho (\mathbf{r})\rho ({\mathbf{r}}^{\prime })(\mathbf{r}-{\mathbf{r}}^{\prime })$ to its negative, although direct calculation is instructive.

3.9 — Electric Flux Through a Plane

A charge distribution $\rho (\mathbf{r})$ with total charge $Q$ occupies a finite volume in $z<0$. Show that ${\int }_{z=0}\mathrm{d}\mathbf{S}\cdot \mathbf{E}=Q/(2{ϵ}_0)$.

Solution Sketch

For each point charge $q_k$ in the distribution, exactly half of its flux lines pass through $z=0$ (by the solid angle argument: the plane subtends $\Omega =2\pi$ at any point below it). By superposition, the same holds for any localized $\rho$.

Theorems and Identities

3.6 — General Electrostatic Torque

Show that the torque of ${\rho }^{\prime }$ on $\rho$ is:

$$\mathbf{N}=-\frac{1}{4\pi {ϵ}_0}\int {\mathrm{d}}^{3}r\int {\mathrm{d}}^3{r}^{\prime }\frac{\mathbf{r}\times {\mathbf{r}}^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}\rho (\mathbf{r}){\rho }^{\prime }({\mathbf{r}}^{\prime })$$

Solution Sketch

Start from $\mathbf{N}=\int {\mathrm{d}}^{3}r\mathbf{r}\times \rho (\mathbf{r}){\mathbf{E}}^{\prime }(\mathbf{r})$. Write ${\mathbf{E}}^{\prime }=-\nabla {\phi }^{\prime }$ and integrate by parts — the gradient moves from ${\phi }^{\prime }$ to $\mathbf{r}\times \rho$, generating the $(\mathbf{r}-{\mathbf{r}}^{\prime })$ kernel. Since $\mathbf{r}\times \mathbf{r}=0$, only the $\mathbf{r}\times {\mathbf{r}}^{\prime }$ cross term survives.

Note: ${\mathbf{N}}^{\prime }=-\mathbf{N}$ — in an isolated system, total torque vanishes.

Source: P.C. Clemmow, An Introduction to Electromagnetic Theory (Cambridge, 1973).

3.10 — Two Electrostatic Theorems

(a) Use Green’s second identity with $f=\phi$, $g=1/|\mathbf{r}|$ to prove $\phi (0)=⟨\phi {⟩}_S$ (mean value theorem) in a charge-free sphere. (b) Use part (a) to give an alternative proof of Earnshaw’s theorem.

Solution Sketch

(a): ${\nabla }^2(1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-4\pi \delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ picks out $\phi (0)$ from the volume integral. The ${\nabla }^2\phi =0$ term vanishes (charge-free). The surface terms give $⟨\phi {⟩}_S$.

(b): Shrink the sphere to arbitrarily small size. The average $⟨\phi {⟩}_S$ cannot exceed the maximum of $\phi$ on $S$, nor fall below the minimum. Therefore, $\phi$ at the center is bounded by its surface values — it cannot be a local extremum. This is Earnshaw’s theorem.

Energy and Stress

3.11 — Potential, Field, and Energy of a Charged Disk

A disk of radius $R$ with uniform $\sigma >0$. (a) Potential on the symmetry axis. (b) Potential at the rim. (c) Sketch field lines. (d) Total energy $U_E$.

Solution Sketch

(a): Simple by direct integration of rings.

(b): Use a point on the rim as origin. In polar coordinates centered there, the maximum radius for a given angle is $r_{\text{max}}=2R\cos \theta$ (this results in intertwined integrals, i.e., bounds of each one depend on the other). Integrate $r$ first (easier), then $\theta$ from $0$ to $\pi /2$:

$$\phi (0)=\frac{\sigma R}{\pi {ϵ}_0}$$

(d): Assemble in circular annuli. At intermediate radius $r$, the rim potential is $\sigma r/(\pi {ϵ}_0)$. Adding an annulus $\mathrm{d}q=\sigma 2\pi r\mathrm{d}r$:

$$U_E=\frac{2{\sigma }^2{R}^3}{3{ϵ}_0}$$

Source: O.D. Jefimenko, Electricity and Magnetism (1966).

3.12 — Spherical Shell with a Hole

A circular hole of radius $b$ is bored through a shell of radius $R$ with uniform $\sigma$. Show that at the center of the hole:

$$E(P)=\frac{\sigma }{2{ϵ}_0}[1-\sin ({\theta }_0/2)]$$

where ${\theta }_0$ is the cone opening angle.

Solution Sketch

(a): By symmetry, $\mathbf{E}$ is radial at $P$. Each annular ring at angle $\theta$ on the shell contributes charge $\mathrm{d}q=\sigma 2\pi R^2\sin \theta \mathrm{d}\theta$ at distance $s=2R\sin (\theta /2)$. The radial projection factor is $\cos \alpha =\sin (\theta /2)$, so:

$$E(P)=\frac{\sigma }{2{ϵ}_0}{\int }_{{\theta }_0}^{\pi }\cos (\theta /2)\mathrm{d}\theta =\frac{\sigma }{2{ϵ}_0}[1-\sin ({\theta }_0/2)]$$

(b): When ${\theta }_0\ll 1$, the shell is essentially complete. By Gauss’s law: $E_{\text{in}}=0$, $E_{\text{out}}=\sigma /{ϵ}_0$. The field on the shell is the average: $\sigma /(2{ϵ}_0)$. This matches the ${\theta }_0\to 0$ limit.

Source: E.M. Purcell, Electricity and Magnetism, 2nd ed. (1985).

3.13 — Uniformly Charged Cube

Find the ratio ${\phi }_0/{\phi }_1$ of the potential at the center to that at a corner of a uniformly charged cube.

Solution Sketch

Think of the cube as $8$ smaller cubes. By symmetry, each sub-cube contributes equally to ${\phi }_0$. The potential at the corner of a sub-cube scales as $\phi \sim \rho s^2$ (dimensional analysis with $s$ the side length), and under $s\to 2s$, the integrals give ${\phi }_0/{\phi }_1=2$ (confirmed by explicit substitution $x\to 2x$ and same for $y$ and $z$, which we can do as we can write down explicit integrals).

3.14 — Variation on Coulomb’s Law

If $\phi (r)=(1/4\pi {ϵ}_0)r^{-(1+ϵ)}$, find $\phi (r)$ inside a uniform spherical shell of radius $R>r$. To first order in $ϵ$:

$$\frac{V(R)-V(r)}{V(R)}=\frac{ϵ}{2}\left[\frac{R}{r}\ln \frac{R+r}{R-r}-\ln \frac{4R^2}{R^2-r^2}\right]$$

Solution Sketch

Integrate the modified potential over the shell surface. The angular integral gives $|r+R{|}^{1-\eta }-|r-R{|}^{1-\eta }$ (same structure as Problem 2.14). Expand to first order in $ϵ$ using $r^{-ϵ}\approx 1-ϵ\ln r$.

Since Cavendish, formulae of this type have been used to test the exactness of the inverse-square law.

3.15 — Energy Between Concentric Spheres

The space between spheres of radii $a$ and $R\ge a$ is uniformly filled with charge $Q$. Find $U_E(x)$ with $x=a/R$.

Solution Sketch

$$U_E=\frac{1}{4\pi {ϵ}_0}\frac{Q^2}{2R}\left[1+\frac{1-5x^3+9x^5-5x^6}{5(1-x^3{)}^2}\right]$$

Limits: $x=0$ gives the uniform ball; $x\to 1$ (L’Hôpital) gives the hollow shell $Q^2/(8\pi {ϵ}_{0}R)$.

$U_E$ is monotonically decreasing in $x$ — the minimum is at $x=1$ (all charge on the outer surface), consistent with Thomson’s theorem.

Source: L. Brito and M. Fiolhais, Eur. J. Phys. 23, 427 (2002).

3.19 — Interaction Energy of Two Spherical Distributions

Two non-overlapping spherical charge distributions ${\rho }_1,{\rho }_2$ with total charges $Q_1,Q_2$ separated by $R$. Use the stress tensor to show $V_E=Q_1{Q}_2/(4\pi {ϵ}_{0}R)$.

Solution Sketch

Integrate $T_{ij}$ over any surface enclosing one distribution but not the other. By Gauss’s law, $\mathbf{E}$ on such a surface is identical to that of two point charges $Q_1,Q_2$ at the distribution centers. Therefore the force and energy are the same as for point charges.

This illustrates the power of the stress tensor: when sources are well-separated, the surface integral “sees” only the multipole structure.

3.20 — Two Electric Field Formulae

(a) Show that for uniform $\rho$ in volume $V$ with surface $S$:

$$\mathbf{E}(\mathbf{r})=\frac{\rho }{4\pi {ϵ}_0}{\oint }_S\frac{\mathrm{d}{\mathbf{S}}^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

(b) Show that for arbitrary localized $\rho$:

$$\mathbf{E}(\mathbf{r})=-\frac{1}{4\pi {ϵ}_0}\int {\mathrm{d}}^3{r}^{\prime }\frac{{\nabla }^{\prime }\rho ({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

Solution Sketch

(a): Start from $\mathbf{E}=-\nabla \phi$, move $\nabla$ inside (acts on $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|$), convert to $-{\nabla }^{\prime }$, then for constant $\rho$ pull it outside and use the divergence theorem on ${\nabla }^{\prime }(1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)$ to get the surface integral.

(b): Same start, but integrate by parts: ${\nabla }^{\prime }[\rho /|\mathbf{r}-{\mathbf{r}}^{\prime }|]=({\nabla }^{\prime }\rho )/|\mathbf{r}-{\mathbf{r}}^{\prime }|+\rho {\nabla }^{\prime }(1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)$. The total derivative integrates to a surface term at infinity which vanishes for localized $\rho$.

Source: O.D. Jefimenko, Electricity and Magnetism (1966).

3.21 — Potential of a Charged Line Segment (Coordinate-Free)

Uniform $\lambda$ on segment from $P$ to $P^{\prime }$. Vectors: $\mathbf{a}$ along the segment, $\mathbf{b}$ and $\mathbf{c}$ from observation point to the endpoints.

Solution Sketch

Parameterize by $s^{\prime }$ measuring distance along $\mathbf{a}$ from the closest approach point. The perpendicular distance is $h=|\mathbf{b}\times \mathbf{a}|/a$. The integration limits are $s^{\prime }\in [(\mathbf{c}\cdot \hat{\mathbf{a}}),(\mathbf{b}\cdot \hat{\mathbf{a}})]$:

$$\phi (\mathbf{r})=\frac{\lambda }{4\pi {ϵ}_0}\ln \left[\frac{\frac{\mathbf{b}\cdot \mathbf{a}}{a}+\sqrt{{\left(\frac{\mathbf{b}\cdot \mathbf{a}}{a}\right)}^2+\frac{|\mathbf{b}\times \mathbf{a}{|}^2}{a^2}}}{\frac{\mathbf{c}\cdot \mathbf{a}}{a}+\sqrt{{\left(\frac{\mathbf{c}\cdot \mathbf{a}}{a}\right)}^2+\frac{|\mathbf{b}\times \mathbf{a}{|}^2}{a^2}}}\right]$$

The trick is decomposing everything into projections parallel and perpendicular to $\mathbf{a}$ — a useful technique for any line-source problem.

Source: H.A. Haus and J.R. Melcher, Electromagnetic Fields and Energy (1989).

3.24 — Energy Outside a Charged Volume

$\phi ={\phi }_0$ on closed surface $S$ bounding $V$, total charge $Q$ inside, no charge outside. Show $U_E(\text{out})=\frac{1}{2}Q{\phi }_0$.

Solution Sketch

Write $U_E=\frac{1}{2}\int {\mathrm{d}}^{3}r\rho \phi =\frac{{ϵ}_0}{2}\int {\mathrm{d}}^{3}r\phi \nabla \cdot \mathbf{E}$. Integrate by parts and use the fact that $\rho$ vanishes outside $V$:

$$U_E=\frac{{ϵ}_0}{2}{\oint }_S\mathrm{d}\mathbf{S}\cdot \mathbf{E}\phi +\frac{{ϵ}_0}{2}{\int }_V{\mathrm{d}}^{3}r|\mathbf{E}{|}^2$$

The surface term, with $\phi ={\phi }_0$ constant on $S$, gives $\frac{1}{2}{\phi }_{0}Q$ by Gauss’s law. The volume term is $U_E(\text{in})$. Therefore $U_E(\text{out})=U_E-U_E(\text{in})=\frac{1}{2}Q{\phi }_0$.

Source: A.M. Portis, Electromagnetic Fields (Wiley, 1978).

3.25 — Overcharging of Macro-ions

Large macro-ions ($Q<0$) in a solution of micro-ions ($q>0$). $N$ micro-ions adsorb onto the macro-ion surface (radius $R$). The minimum energy of $N\gg 1$ charges on a sphere is:

$$E(N)=\frac{q^2}{4\pi {ϵ}_{0}R}\frac{1}{2}\left(N^2-N^{3/2}\right)$$

Find the equilibrium $N$ and show the macro-ion is overcharged.

Solution Sketch

(a): Smear the micro-ion charge uniformly over the sphere: total charge $Nq$, self-energy $\sim (Nq{)}^2/(8\pi {ϵ}_{0}R)\sim N^2$. This gives the $N^2$ term.

(b): The $N^{3/2}$ correction: smearing ignores that discrete charges avoid each other. Each micro-ion has $\sim \sqrt{N}$ nearest neighbors at average spacing $\sim R/\sqrt{N}$. Subtracting the smeared charge within each micro-ion’s “territory” (a cap of area $\sim 4\pi R^2/N$) gives a correlation energy $\sim N\times q^2/(4\pi {ϵ}_0\cdot R/\sqrt{N})\sim N^{3/2}$.

(c): Total energy = $E(N)$ + macro-ion/micro-ion interaction $=E(N)-NqQ/(4\pi {ϵ}_{0}R)$. Minimize over $N$:

$$\frac{\partial }{\partial N}\left[\frac{1}{2}{N}^2-\frac{1}{2}{N}^{3/2}-N\frac{|Q|}{q}\right]=0⟹N-\frac{3}{4}\sqrt{N}-\frac{|Q|}{q}=0$$

Solving: we find that $N=\frac{|Q|}{q}+\frac{9}{32}+\frac{9}{32}\sqrt{1+\frac{64}{9}\frac{|Q|}{q}}>\frac{|Q|}{q}$, meaning the micro-ions overcharge the macro-ion. The correlation energy ($N^{3/2}$ term) is what drives overcharging beyond simple neutralization.