Boundary-Value Problems in Electrostatics

Here is a problem that sounds impossible. A single point charge q floats a few centimetres above a large sheet of grounded metal. The charge's own field is simple enough — but the metal responds: electrons in the conductor rearrange, pile up beneath the charge, and set up a field of their own. To know the real electric field anywhere you would seem to need to know exactly how all those countless electrons have arranged themselves, which depends on the field, which depends on the electrons… a snake swallowing its own tail.

Almost every electrostatics problem worth solving has this flavour: you are told the charges in some places and the potential on some surfaces (a grounded plate is held at \varphi = 0; a battery holds a plate at a fixed voltage), and you must find the potential \varphi everywhere in between. This is a boundary-value problem. This page is about the master equation these problems obey, a theorem that makes them tractable, and a beautiful trick — the method of images — that solves the grounded-plane puzzle above in a single line, with no snakes.

The master equation: Poisson and Laplace

Everything starts from one of Maxwell's equations, Gauss's law in the form \nabla\cdot\mathbf{E} = \rho/\varepsilon_0, together with the fact that the electrostatic field is the (downhill) gradient of a potential, \mathbf{E} = -\nabla\varphi. Substitute the second into the first and the field drops out, leaving a single equation for the potential alone:

So the task "find the field" becomes "solve a single second-order partial differential equation for one scalar function \varphi." That is a huge simplification — one function instead of three field components — but a differential equation alone does not pin down a unique answer. Laplace's equation has infinitely many solutions (any constant, any linear ramp ax + by + cz, and endless combinations). What selects the physical one is the extra information at the edges: the boundary conditions.

Two kinds appear again and again. A Dirichlet condition fixes the value of \varphi itself on each boundary surface (a grounded conductor gives \varphi = 0; a biased plate gives \varphi = V). A Neumann condition fixes instead the normal derivative \partial\varphi/\partial n on the surface — equivalently the surface charge or the normal field. Give a well-posed boundary condition all the way round a region and the potential inside is completely determined. Why "completely"? That is the next, load-bearing idea.

The uniqueness theorem: why any trick that works, works

Here is the theorem that turns clever guessing into rigorous physics.

The proof is short and worth seeing. Suppose two solutions \varphi_1 and \varphi_2 both satisfy Poisson's equation in V and agree on the boundary. Their difference u = \varphi_1 - \varphi_2 then satisfies \nabla^2 u = 0 inside (the \rho terms cancel) and u = 0 on the boundary. Now integrate \nabla\cdot(u\nabla u) over the volume and use the divergence theorem:

\int_V |\nabla u|^2 \, dV = \oint_S u\,(\nabla u)\cdot d\mathbf{a} - \int_V u\,\nabla^2 u\, dV = 0,

because the surface term vanishes (u = 0 on S) and the last term vanishes (\nabla^2 u = 0). But |\nabla u|^2 \ge 0 everywhere, and an integral of something non-negative can only be zero if the thing is zero everywhere. So \nabla u = 0, meaning u is constant; and since it is zero on the boundary, that constant is zero. Therefore \varphi_1 = \varphi_2. The solution is unique.

This licence — find a solution any way you like and it is guaranteed correct — is exactly what makes the next trick legitimate.

The method of images: a mirror charge does the whole job

Back to the opening puzzle: a charge +q at height d above an infinite grounded plane (take the plane as z = 0, the charge at (0,0,d)). We need a potential in the region z > 0 that (i) solves Poisson's equation with just that one point charge, and (ii) equals zero all over the plane. The genius move: throw the conductor away entirely and pretend there is a second, image charge -q sitting at the mirror point (0,0,-d). Reveal the figure step by step to watch the plane turn into a mirror.

Write down the potential of this pair, valid for z > 0:

\varphi(x,y,z) = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{\sqrt{x^2 + y^2 + (z-d)^2}} - \frac{q}{\sqrt{x^2 + y^2 + (z+d)^2}}\right].

Now check it against the two requirements. On the plane (z = 0) the two distances become equal, \sqrt{x^2+y^2+d^2}, so the bracket is q - q = 0: the potential vanishes on the whole plane, exactly the grounded boundary condition. In the region z > 0 the only real charge is the +q at (0,0,d) — the image sits at z = -d, outside our region — so \varphi solves Poisson's equation with precisely the right source. Both boxes ticked. By the uniqueness theorem, this is the potential above a grounded plane. We solved a hard conductor problem by writing down the field of two point charges.

What the mirror charge tells us: induced charge, force, energy

With \varphi in hand, everything else follows by differentiation.

Induced surface charge. The charge that gathers on the conductor's surface is \sigma = -\varepsilon_0\,\partial\varphi/\partial z evaluated at z = 0. Carrying out the derivative and writing r = \sqrt{x^2+y^2} for the distance from the foot of the charge,

\sigma(r) = -\frac{q\,d}{2\pi\,(r^2 + d^2)^{3/2}}.

It is negative (opposite charge is drawn up under the +q), sharply peaked right beneath the charge at \sigma(0) = -q/(2\pi d^2), and falls off with distance. Integrate it over the whole plane and the total induced charge comes to exactly -q — the plane carries precisely the image's worth of charge, spread out.

The force on the charge. The real charge is attracted to the induced charge on the plane. But by the images construction, the field acting on the +q is just the field of the image -q sitting a distance 2d away. So the force is a plain Coulomb attraction across 2d, directed toward the plane:

F = -\frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{(2d)^2} = -\frac{q^2}{16\pi\varepsilon_0\,d^2}.

Worked example — the force. Take q = 1\ \text{nC} = 1\times 10^{-9}\ \text{C} at d = 1\ \text{mm} = 1\times 10^{-3}\ \text{m}, with 1/(4\pi\varepsilon_0) \approx 8.99\times 10^9\ \text{N·m}^2/\text{C}^2. Then

|F| = \frac{(8.99\times 10^9)(1\times 10^{-9})^2}{(2\times 10^{-3})^2} = \frac{8.99\times 10^{-9}}{4\times 10^{-6}} \approx 2.2\times 10^{-3}\ \text{N}.

A couple of millinewtons — tiny, but enough to hold a speck of dust against a charged surface, which is exactly how electrostatic dust precipitators and photocopier toner work.

The energy. The energy of the configuration — the work needed to carry the charge in from infinity — is

W = -\frac{q^2}{16\pi\varepsilon_0\,d}.

Notice the 16\pi, not 8\pi. This is half of what two real charges +q and -q a distance 2d apart would have, -q^2/(8\pi\varepsilon_0 d). The factor of two is the subtle heart of the images method — see the "Watch out!" box.

It feels like cheating that a single fictitious -q stands in for the whole messy rearrangement of conduction electrons. The uniqueness theorem is what makes it honest. The real conductor does just one thing that matters to the region above it: it holds its surface at \varphi = 0. Anything — real charges, imaginary charges, a wish — that produces a potential which solves Laplace's equation above the plane and is zero on the plane must, by uniqueness, give the identical field there. The image charge is simply the cheapest gadget that reproduces that one boundary value. It doesn't matter that the electrons and the image bear no physical resemblance; they are indistinguishable from outside.

The same idea scales to prettier geometries. A charge outside a grounded sphere of radius R at distance a from the centre is imaged by a charge q' = -qR/a placed a distance R^2/a from the centre — a smaller, off-centre mirror charge that again zeroes the potential on the sphere. Two grounded planes meeting at a right angle need three image charges; parallel planes need an infinite train of them. The mirror trick is a whole toolbox.

Two traps snare almost everyone the first time they meet image charges.

The image is fictitious, and only lives in the real region. The two-charge potential is the correct answer only for z > 0, the region where the real charge lives. Below the plane there is no field at all — a grounded conductor screens it completely, so \varphi = 0 throughout z < 0. If you blindly evaluate the two-charge formula at z < 0 you get a perfectly definite number that is physically wrong. The image is a computational scaffold for the real half-space, never a description of what is behind the mirror.

The energy is not the energy of two real charges. You might expect W = -q^2/(8\pi\varepsilon_0 d), the interaction energy of +q and -q a distance 2d apart. The true answer is half that, -q^2/(16\pi\varepsilon_0 d). The reason: as you push the real charge in, the image obediently moves too, but no work is done on the image — it isn't real, there is no field energy stored in the fictitious lower half-space (where the true field is zero). Only the upper half-space holds energy, so you collect only half the naive value. Compute image energies from the work or the field integral over the real region, never by copying the two-real-charge formula.

The other great technique: separation of variables

Images are magic when a charge sits near a simple conductor, but many boundary-value problems have no convenient mirror — a charge inside a rectangular box, the potential in a slot between biased plates, a sphere with a patterned surface potential. For these the workhorse is separation of variables. You look for solutions of Laplace's equation of the product form

\varphi(x,y,z) = X(x)\,Y(y)\,Z(z),

which turns the one partial differential equation into three ordinary ones (each factor is a sine, cosine, or exponential in Cartesian coordinates; Legendre polynomials and powers of r in spherical coordinates). Because Laplace's equation is linear, you may add such product solutions freely, and you tune the coefficients of the resulting Fourier (or Legendre) series to match the boundary conditions term by term. Uniqueness again guarantees that once the series fits the boundary, it is the answer. Images and separation of variables are the two pillars of analytic electrostatics; between them they crack a remarkable share of the classic problems.