Electromagnetism

Electrostatics and the Laplace Equation

Guessing a solution is a legitimate method. The uniqueness theorem is what makes it one.

Updated May 2026

Electrostatics is the study of one equation and its consequences. In a charge-free region, the potential satisfies

\nabla^{2} φ = 0

and everywhere else, Poisson's equation $\nabla^{2} φ = - ρ / ε_{0}$ .

Harmonic functions have no secrets

Theorem

Mean value property

A solution of Laplace's equation equals its own average over any sphere centred on the point:

φ (r) = \frac{1}{4 π R ^{2}} \oint_{∣ r^{'} - r ∣ = R} φ d a^{'}

An immediate corollary: $φ$ can have no local maximum or minimum inside a region. Any extremum sits on the boundary.

Corollary

Earnshaw's theorem

You cannot trap a charged particle in a stable static equilibrium using electrostatic fields alone. There is no potential well to sit in, because there are no interior minima. Every practical trap therefore cheats. Paul traps oscillate the field, Penning traps add a magnetic field, optical tweezers use light.

Uniqueness, and the licence to guess

Theorem

Uniqueness

The solution of Poisson's equation in a volume $V$ is uniquely determined by $ρ$ inside $V$ together with either $φ$ on the boundary (Dirichlet) or $\partial φ / \partial n$ on it (Neumann, up to a constant).

This is the most practically useful theorem in the subject, because it converts solving into checking. If you can produce any function that satisfies the equation and matches the boundary conditions, whether by symmetry, by inspection or by luck, it is the solution. There is no other.

Images

The method of images is uniqueness weaponised. A point charge $q$ at height $d$ above a grounded conducting plane: replace the conductor with a fictitious $- q$ at $- d$ . The plane $z = 0$ is then at zero potential by symmetry, and Laplace's equation holds everywhere in $z > 0$ where the real problem lives. Uniqueness does the rest.

The force follows immediately:

F = - \frac{1}{4 π ε _{0}} \frac{q ^{2}}{( 2 d ) ^{2}}

attractive, and exactly the attraction to the image. For a grounded sphere of radius $R$ the image is a charge $q^{'} = - q R / a$ at radius $R^{2} / a$ , less obvious but the same logic.

Fig. 1

The real charge and its image. Only the upper half is physical; the image exists to enforce $φ = 0$ on the plane. Field lines meet a conductor normally, because any tangential component would drive a surface current.

Result

Induced surface charge

Differentiating the potential gives the density the conductor actually carries:

σ (r) = - \frac{q d}{2 π ( r ^{2} + d ^{2} ) ^{3/2}}

Integrating over the whole plane returns exactly $- q$ . The image charge is fictitious, but the charge it stands for is real and measurable. It is drawn up from ground.

Separation of variables

When the geometry matches a coordinate system, assume the solution factorises. In spherical symmetry with azimuthal invariance:

φ (r, θ) = ℓ = 0 \sum \infty (A_{ℓ} r^{ℓ} + \frac{B _{ℓ}}{r ^{ℓ + 1}}) P_{ℓ} (cos θ)

The Legendre polynomials are not chosen for elegance. They are what you get when you demand solutions regular on the axis, and their orthogonality is what lets boundary conditions fix the coefficients one at a time.

Multipoles

Far from a bounded charge distribution, expand in powers of $1/ r$ :

φ (r) = \frac{1}{4 π ε _{0}} [\frac{Q}{r} + \frac{p \cdot r ^}{r ^{2}} + \dots]

The leading non-vanishing term dominates, and which one that is depends on the symmetry. A neutral molecule has no monopole; a symmetric one has no dipole either and leads with a quadrupole. This is why CO₂ is infrared-active only in its bending modes, and ultimately why it is a greenhouse gas.