The charge that has to be somewhere else

A point charge sits a couple of centimetres above a large earthed metal sheet. The problem is to find the field above the sheet, and it looks hard: the charge attracts electrons in the metal, which move, which changes the field, which changes where they move. The distribution has to be solved for self-consistently, and the equation to solve is a partial differential one in three dimensions with a boundary condition on an infinite plane.

The answer takes one line. Delete the sheet, put an equal and opposite charge at the mirror position, and write down the potential of the two.

Why a guess can be a proof #

Electrostatics away from charge obeys Laplace’s equation, $\nabla^2 V = 0$. The theorem that makes the trick work is about that equation and nothing else: in a region bounded by surfaces on which the potential is specified, Laplace’s equation has exactly one solution.

The proof is three lines and worth having, because the whole method rests on it. Suppose two solutions $V_1$ and $V_2$ both satisfy the equation inside and match the same values on every boundary. Their difference $U = V_1 - V_2$ satisfies Laplace’s equation inside and is zero on every boundary. Green’s identity then gives

$$\int_{\text{volume}} |\nabla U|^2 \, dV = \oint_{\text{surface}} U \,\nabla U \cdot d\mathbf{A} - \int_{\text{volume}} U \nabla^2 U \, dV = 0,$$

because $U$ vanishes on the surface and $\nabla^2 U$ vanishes inside. An integral of a square is zero only if the square is zero everywhere, so $\nabla U = 0$, so $U$ is constant, and being zero on the boundary it is zero throughout. The two solutions are the same.

The consequence is a licence to guess. Produce, by any means at all, a function that satisfies Laplace’s equation in the region of interest and takes the right values on the boundary, and it is the answer — there is nothing else it could be. How the function was arrived at is irrelevant, and no error estimate is required, because there is no error.

For the charge above a plane, the boundary condition is that an earthed conductor sits at $V = 0$. Two charges $+q$ at height $d$ and $-q$ at depth $d$ give, at a point on the plane a distance $r$ from the foot of the perpendicular,

$$V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{\sqrt{r^2 + d^2}} - \frac{q}{\sqrt{r^2 + d^2}}\right) = 0,$$

at every point on it, because both distances are the same. Above the plane there is no charge except the real one, so Laplace’s equation holds. That is the whole verification.

A guess can be a proof because of the uniqueness theorem, and the geometry makes it obvious. The plane halfway between two opposite charges is an equipotential — the surface where the two contributions cancel exactly — and it is flat. So a configuration that reproduces a flat equipotential at the right place satisfies the same boundary condition as the conductor, and the theorem says there is only one field that does. Having guessed it, one has found it.

That last observation is the general form of the trick and is worth more than the special case. Any equipotential surface of any charge arrangement may be replaced by a thin conductor held at that potential, and the field elsewhere does not change. The charge-above-a-plane problem is the simplest instance because the equipotential in question happens to be flat.

What the metal is really doing #

The image charge is not there, and something real is. The field just above a conductor’s surface is perpendicular to it — tangential field would drive a current and a static situation would not be static — and its magnitude is fixed by the surface charge density through $E = \sigma/\varepsilon_0$.

The condition being satisfied is visible in what a conductor does to a field: the lines meet its surface square on, and the interior is empty. Square-on is the same statement as constant potential across the surface, since any component along the surface would drive the charges until it was gone. The image charge is chosen for no other reason than to produce that.

Differentiating the two-charge potential and evaluating on the plane gives the induced density directly:

$$\sigma(r) = -\varepsilon_0 \frac{\partial V}{\partial z}\bigg|_{z=0} = -\frac{q\,d}{2\pi\,(r^2 + d^2)^{3/2}}.$$

It is largest directly under the charge and falls as $1/r^3$ far away — a cube, not a square, because the near cancellation of the two contributions costs a power of the distance in the same way a dipole’s field falls one power faster than a monopole’s.

The integral is worth doing by hand once. With $dA = 2\pi r\,dr$,

$$Q_{\text{induced}} = \int_0^\infty -\frac{q d}{2\pi (r^2 + d^2)^{3/2}}\, 2\pi r\, dr = -qd\left[-\frac{1}{\sqrt{r^2+d^2}}\right]_0^\infty = -q.$$

Every $d$ cancels. The plane collects exactly the image’s worth of charge no matter how far above it the real charge is held — which it must, since every field line leaving the charge has to end somewhere, and above an infinite earthed plane the only place available is the plane.

The total induced charge is forced by bookkeeping rather than computed. Wrap a closed surface around the real charge and close it below the metal: it encloses the real charge and all the induced charge together, and the flux through it is zero because the field below the metal is zero and the field far above has fallen off fast enough. So the induced charge is exactly minus the real one — a number obtained without integrating anything.

The force, and the energy that is not twice it #

The force on the real charge is the field of the image evaluated at the charge’s position, which is the Coulomb force between two charges $2d$ apart:

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

attractive, and correct. For a nanocoulomb 20 mm up it is 5.6 microlitres of newton — about the weight of a large grain of sand — and it goes as the inverse square of the height in the usual way.

The energy is where the image has to be handled carefully, and it is the standard place the method is misused. The work done bringing the charge in from infinity is the integral of the force:

$$W = -\int_\infty^d \frac{q^2}{16\pi\varepsilon_0 z^2}\,dz = -\frac{q^2}{16\pi\varepsilon_0 d}.$$

Whereas two real charges $q$ and $-q$ a distance $2d$ apart have interaction energy $-q^2/8\pi\varepsilon_0 d$ — twice as much. The factor of two is not a slip. A real partner would stay put while the charge came in; the induced distribution does not, because it is dragged along, spreading and concentrating as the charge descends. Half the work goes into that rearrangement, and the image picture, which draws the partner as a fixed point charge, has no term for it.

The rule that comes out of it: the image gives the field and every quantity computed pointwise from the field, and it does not give an energy by inspection. Energies have to be integrated from the force, or computed from $\varepsilon_0 E^2/2$ over the real region only — where “only” means above the plane, since below it there is no field.

The energy is where the field is, and integrating the energy density over the region above the plane gives the same halved answer as integrating the force along the approach. The halving is not a correction: it is the statement that the field exists only on one side, so only half of what a genuine pair of charges would store is actually there. Two routes, one number, and the agreement is the check.

The pull on the metal, counted a second way #

The force was got by evaluating the image’s field at the charge. It can be got again from the other end — from the metal — and the two answers have to agree, which makes the second calculation a check on the first rather than a repetition of it.

A charged surface feels an outward pressure. The field is $\sigma/\varepsilon_0$ just outside and zero just inside, so the charge on the surface sits in the average of the two, $\sigma/2\varepsilon_0$, and the force per unit area is

$$P = \frac{\sigma^2}{2\varepsilon_0},$$

directed away from the conductor whatever the sign of $\sigma$ — an attraction toward the charge above, in this case, since that is the only thing to be attracted to. Integrating that pressure over the whole plane, with the density from before,

$$\int_0^\infty \frac{1}{2\varepsilon_0}\left(\frac{qd}{2\pi(r^2+d^2)^{3/2}}\right)^2 2\pi r\,dr = \frac{q^2 d^2}{4\pi\varepsilon_0}\int_0^\infty \frac{r\,dr}{(r^2+d^2)^3} = \frac{q^2}{16\pi\varepsilon_0 d^2},$$

which is the image force exactly. The plane is pulled up with the same force the charge is pulled down with, which is Newton’s third law arriving in a problem where one of the two bodies was replaced by a fiction partway through.

The pressure is also why the plane’s own weight never enters. It is a stress on a surface, and it does not care what is behind the surface as long as the surface is a conductor.

The induced density falls as the inverse cube, and the reason is the same one that governs any dipole. Two nearly cancelling contributions leave only their difference, and a difference falls one power faster than either term — so a distribution whose net effect at a distance is a dipole’s declines as $1/r^3$ where a point charge’s would decline as $1/r^2$.

Curved mirrors: the sphere #

The plane is the easy case, and the method is not confined to it. For an earthed sphere of radius $a$ with a charge $q$ at distance $s$ from its centre, the boundary condition can be met by a single image of magnitude $-qa/s$ placed at distance $a^2/s$ from the centre, on the line joining them. The verification is the same as before — the potential of the pair is zero everywhere on the sphere — and the checking is a page of algebra rather than a line, because the two distances are no longer equal and their ratio has to be shown constant.

Two consequences follow that the plane conceals. The image is smaller than the real charge, so the sphere collects less than $|q|$ of induced charge; the remainder has gone to earth. And if the sphere is insulated and neutral instead of earthed, a second image has to be added at the centre to restore the total to zero, which makes the field outside that of three charges rather than two.

With two conducting boundaries instead of one the problem changes character. A single image suffices for a plane; two facing planes need an infinite series of images, each one the reflection of the last, and the capacitance that comes out is the ordinary parallel-plate answer with a correction from the tail of that series. The method survives, and it stops being a single guess.

Why any neutral conductor attracts #

The insulated sphere case has a consequence general enough to be worth stating on its own, because it explains a fact everyone has met and few people can derive.

Take the sphere, leave it insulated and neutral, and work out the force. Two images are needed: the one that makes the surface an equipotential, of strength $-qa/s$, and a second at the centre of strength $+qa/s$ to restore the total to zero. The first is nearer the charge than the second, so the attraction wins, and the net force is attractive at every separation. Far away the two images are a dipole of moment proportional to $a^3/s^2$, and the force between a charge and the dipole it induces falls as $a^3/s^5$ — very steep, and never zero, and never repulsive.

That is the general statement. A neutral, uncharged, ungrounded conductor is attracted to any charge brought near it, whatever the sign of the charge, because the charge polarises it and pulls harder on the near end than it pushes on the far one. A scrap of aluminium foil jumps to a charged rod for this reason, and so does a stream of water, and so does a dust particle to a screen.

The steepness is what makes it feel like a threshold. Halving the distance to a sphere multiplies the attraction by thirty-two, so a body that felt nothing at ten radii is snatched at three — which is why the effect reads as a sudden grab rather than as a gradually increasing pull.

Two planes, and when the images run out #

The parallel-plate remark above is the general question in disguise: when does the method terminate?

Take two earthed planes meeting along a line, with the charge in the wedge between them. The charge must be imaged in one plane; that image must be imaged in the other; the result imaged again; and so on. Whether the process closes depends on the angle. If the planes meet at $\pi/n$ for an integer $n$, the chain of reflections returns to its starting point after finitely many steps and there are exactly $2n-1$ images. A right-angled corner, the case $n = 2$, needs three: one under each plane and one diagonally opposite, which is the arrangement anybody would guess and which the theorem justifies.

At any other angle the chain never closes. Reflections in two planes at an incommensurate angle generate an infinite set of points, and the method delivers a series rather than a formula. Parallel planes — the limit $n \to \infty$ — are that case: an infinite ladder of images stretching away in both directions, which converges and is a perfectly good solution, but not a closed one.

There is a satisfying reason the condition is what it is. Reflecting in two planes at angle $\theta$ composes to a rotation by $2\theta$, so the images close into a finite set exactly when that rotation has finite order — when $2\theta$ divides a full turn. The question of whether an electrostatics problem has a finite image solution is a question about a group of rotations, and it has nothing to do with charge at all.

Where the model stops #

Three assumptions are doing work and each fails somewhere.

The plane is infinite. A real sheet has edges, and near an edge the boundary condition is no longer “an infinite equipotential plane”. The correction is small while the charge is much closer to the sheet than to its nearest edge — the induced charge is concentrated within a few times $d$, which is exactly the calculation in the figure above — and it is not small for a charge held a metre above a hand-sized plate.

The conductor is perfect and the situation is static. A real metal responds in a time of order its dielectric relaxation time, around $10^{-18}$ s for copper, so for anything short of optical frequencies the assumption is excellent. It fails outright for a poor conductor, where the induced charge lags and the force acquires a component along the surface — which is how a charged object dragged over a semiconductor experiences drag, and it is the same lag that makes a conductor exclude a changing field only to a depth rather than absolutely.

The charge is a point. Bring a real charged sphere close and the image is no longer a point either; it is imaged again in the sphere, and that image imaged in the plane, and so on. The series matters only when the separation is comparable with the sphere’s size, and it is the reason two conducting spheres attract even when they carry the same sign of charge, once they are close enough.

The electron near the metal, where the picture ends #

The most consequential use of the image charge is the smallest one: a single electron sitting just outside a metal surface. The same calculation gives it a potential energy

$$U(z) = -\frac{e^2}{16\pi\varepsilon_0 z},$$

with the same factor of sixteen, for the same reason — the halving that the energy section above insists on. At a nanometre from the surface that is 0.36 electronvolts, which is not a small energy by the standards of anything happening at a surface.

Two things depend on it directly. The barrier an electron has to climb to leave a metal is not a step but a step with this attraction subtracted from it, and applying an external field lowers the top of that barrier — the field pulls the electron out while the image pulls it back, and the two cross at a height below the flat-barrier value. That lowering is why the current from a hot filament rises with the field applied to collect it rather than saturating, and it is a measurable effect with this expression’s square root in it.

And the attraction has bound states. An electron trapped between the image potential and the metal’s own reflecting barrier sits in a Coulomb-like well, so it has a hydrogen-like series of levels converging on the vacuum level, the lowest bound by about 0.85 electronvolts. Those levels are real, they are measured by two-photon photoemission, and their spacing is the $1/n^2$ of a Rydberg series with a factor of sixteen in the constant.

What is striking is where the whole picture stops. The expression diverges at $z = 0$, and it must be wrong before that: the metal’s electrons cannot respond to a disturbance on a scale finer than their own screening length, which is an ångström or so, and at that separation the electron is not outside the metal in any meaningful sense. So the classical image force is excellent from a few ångströms out to as far as anyone can measure, and undefined inside — a domain of validity that the uniqueness theorem cannot supply, because the theorem is about a boundary condition and the failure is in whether the boundary exists.

What the pictures cannot show #

Every figure here draws the field above the plane, and the region below is drawn as flat shading because there is nothing in it. That is a true statement and an unhelpful picture: the interesting thing about the metal is the surface charge, which lives on a plane of zero thickness and cannot be drawn as a field. The density figure shows it as a graph instead, which is honest but loses the geometry.

Nothing here shows the induced charge arriving. The whole account is static, and the transient — electrons flowing in from earth as the charge is lowered — is where the halved energy actually goes.

And the field-line figure draws lines, which are a choice rather than a thing. The perpendicularity at the surface is a real property of the field; the number of lines and where they start is not.

Where the ladder goes next #

The step from a conductor’s interior being empty to how much charge a shape will hold to this one is a step in generality: from a statement about the inside, to a number for the whole body, to a method for the field in the space around it.

What follows is the method applied to boundaries that are not equipotentials of anything simple — where images fail and the equation has to be solved by separation of variables, or by relaxation on a grid. The rung after that is the dielectric version, where the boundary condition is not $V = 0$ but a jump in the normal component, and the image charge acquires a fractional strength $(\varepsilon_1 - \varepsilon_2)/(\varepsilon_1 + \varepsilon_2)$ — the same coefficient that turns up as the reflection coefficient at an optical boundary, for the same reason.

The habit is the one the uniqueness theorem licenses. Where a problem is hard to solve and easy to check, stop solving it. Guess something that satisfies the equation, test it against the boundary, and let the theorem convert the test into a proof.