The pressure a charge puts on its own metal

The charge on a conductor lives entirely on its surface, and the surface is where the interesting mechanics is.

Each element of that charge is being pushed away from every other element, and since they are all confined to the same surface the net effect is an outward pull on the surface itself. The size of it is $\sigma^2/2\varepsilon_0$, equivalently $\tfrac12\varepsilon_0 E^2$, which is the energy density of the field just outside.

The factor of a half in that expression is the whole of the physics, and getting it right is the difference between a correct answer and one that is twice too large.

Why a half #

The field at a conductor’s surface is $\sigma/\varepsilon_0$ just outside and exactly zero just inside — that discontinuity is what makes the interior field-free and is Gauss’s law applied to a pillbox straddling the surface.

The discontinuity is where the half comes from, and a pillbox is the quickest way to see it. Straddle the charged surface with a small closed box, half in the metal and half out: the flux through the outer face is $\sigma/\varepsilon_0$ times the area and the flux through the inner face is zero, because there is no field inside a conductor. So the field jumps from nothing to $\sigma/\varepsilon_0$ across a layer of charge, and a charge sitting in that layer feels neither value. It feels the average, which is half.

The natural first guess is that the force per unit area is $\sigma$ times the field it sits in. But which field — the $\sigma/\varepsilon_0$ outside, or the zero inside?

Neither. The surface charge does not exert a force on itself, so what acts on a given patch is the field produced by everything else, and that field is continuous across the layer. Splitting the field into the part made by the patch and the part made by the rest, the patch’s own contribution is $\pm\sigma/2\varepsilon_0$ on the two sides — pointing away from itself — and the rest contributes the same amount on both. Adding gives $\sigma/\varepsilon_0$ outside and zero inside, as required, and identifies the field from the rest as $\sigma/2\varepsilon_0$: the mean of the two sides.

So the pressure is

$$P = \sigma\cdot\frac{\sigma}{2\varepsilon_0} = \frac{\sigma^2}{2\varepsilon_0} = \frac{1}{2}\varepsilon_0 E^2.$$

A second derivation gets the same number with no self-force argument at all, which is the check. Imagine pushing the surface outward by $\mathrm{d}x$ at constant charge. The field in the shell swept out is destroyed — the region becomes conductor interior, where the field is zero — so the field energy falls by $\tfrac12\varepsilon_0E^2\,A\,\mathrm{d}x$. That energy has gone into work done by the surface against whatever holds it, so the outward force per unit area is $\tfrac12\varepsilon_0E^2$. Same expression, and the energy density has appeared as a pressure for the same reason it does in a gas.

The same half arrives from the other direction as an energy density. The energy per unit volume in an electrostatic field is $\tfrac12\varepsilon_0 E^2$, and the pressure on a surface bounding a field is numerically that same quantity — a stress and an energy density are the same number here, exactly, not approximately. Push the surface outward by a small distance and the field-filled volume grows by that distance times the area; the work done is the energy density times the volume, so the force per unit area is the energy density. The factor of a half is already inside it.

The half is everywhere, and it is the same half #

The factor is not a peculiarity of conductors. It appears whenever a surface distribution is asked what force it feels, and always for the same reason.

A charged plane in isolation makes a field $\sigma/2\varepsilon_0$ on each side, pointing away. Two such planes with opposite charges make $\sigma/\varepsilon_0$ between them and nothing outside, which is the capacitor. The attraction between the plates is the charge on one times the field made by the other, which is $\sigma/2\varepsilon_0$ — half the field in the gap. The same half.

The gravitational version is the surface of a self-gravitating shell, which is pulled inward by the mean of the field inside and the field outside. The magnetic version is the force on a current sheet, which feels the mean of the fields on its two sides — and that is why the pressure on the wall of a solenoid is $B^2/2\mu_0$ and not $B^2/\mu_0$, a factor that matters a great deal to anybody designing a magnet.

The general statement is that a discontinuity in a field cannot exert a force on the thing producing the discontinuity using either of its own limiting values. Only the average is well defined, and taking the average is what the self-force argument formalises.

It does not care about the sign, and it is small #

Two things follow immediately and both matter.

The pressure goes as the square of the field, so it is outward for either sign of charge. A conductor cannot be squeezed by charging it. A charged soap bubble expands, a charged drop flattens toward an oblate shape, and reversing the polarity of the charging supply changes nothing at all.

And the pressures are small. At the breakdown field of dry air, $3$ MV/m, the pull is $39.8$ pascals — four hundredths of an atmosphere, or the weight of four grams spread over a hundred square centimetres. Electrostatic forces shape soap films, dust, ink and cells, and they do not shape anything stiffer, because air stops holding off the field long before the pressure becomes structurally interesting.

A capacitor is the everyday case. Two oppositely charged plates each feel $\sigma^2/2\varepsilon_0$ pulling them toward the other — the same expression, since each plate sits in the field of the other and its own field cannot pull on it. Near breakdown that is a few tens of pascals, which sounds like nothing and is enough to bend a thin membrane measurably. It is exactly the mechanism a condenser microphone runs on, read backwards: there the membrane moves and the capacitance is measured, and the pressure computed here is the force the same arrangement exerts when it is driven instead.

Two exceptions make the number useful anyway. In a very small gap the field can be far larger than $3$ MV/m before anything breaks down, because breakdown needs an electron to gain ionising energy over a mean free path and there is not enough room; micromachined actuators run at $10^8$ V/m and pressures of tens of kilopascals. And a thin, floppy thing — a membrane, a film, a drop — responds to tens of pascals readily, which is where the rest of this essay lives.

A number for a familiar object #

A Van de Graaff sphere of $25$ cm radius charged to $200$ kV has a surface field of $800$ kV/m, well under breakdown, and a surface pressure of $2.8$ pascals — about the weight of a sheet of paper spread over the whole sphere. The charge on it is $5.6$ microcoulombs, and the outward pull on each half of it, obtained by integrating the pressure over a hemisphere, is $0.55$ newtons: enough to notice on a spring balance and nothing like enough to deform the metal.

The same sphere at breakdown would carry $21$ microcoulombs, hold a surface pressure of $40$ pascals, and pull each hemisphere with $7.8$ newtons. Since the sphere is aluminium a millimetre thick, this remains firmly in the category of things that do not matter mechanically. It is the small, floppy objects that respond.

Set against surface tension, it decides a size #

A liquid drop is held together by surface tension, which produces an inward Laplace pressure of $2\gamma/R$. Charge it and the electrostatic pressure pushes outward — the same competition of a surface against a bulk that decides which of two bubbles wins, with a new term added. The two have different dependences on the radius, so they cross.

The inward pressure that holds a drop together is $2\gamma/R$, and its dependence on radius is the whole of the competition: it grows as the drop gets smaller, without limit, which is why small drops are hard to break and why a mist is stable where a puddle is not. That is the quantity the electrostatic pull has to beat.

Put the charge $Q$ on a sphere of radius $R$: the surface density is $Q/4\pi R^2$, so the electrostatic pressure is $Q^2/32\pi^2\varepsilon_0 R^4$. Setting that equal to $2\gamma/R$ gives

$$Q^2 = 64\pi^2\varepsilon_0\gamma R^3,$$

which is the Rayleigh limit, from 1882.

For water at room temperature a one-millimetre drop is limited to $0.635$ nanocoulombs, and a ten-micrometre droplet to $0.63$ picocoulombs. The allowed charge falls with the radius, but the allowed field at the surface rises — as $R^{-1/2}$ — which is why the interesting behaviour is at the small end and why a fine capillary rather than a hosepipe is the way to reach it.

Rayleigh did not obtain the criterion by balancing pressures. He asked which small deformations of a charged drop grow, expanded the surface in spherical harmonics, and found that the $\ell = 2$ mode — the ellipsoidal one — becomes unstable exactly when the parameter now called the fissility reaches one, which is the condition above. The agreement of the two routes is worth noticing and is not automatic: a pressure balance on the undeformed sphere and a stability analysis of its deformations are different questions, and for many systems they give different numbers.

What a drop does instead of exploding #

The picture of a drop bursting into fragments is nearly right and misses the mechanism.

Past the limit, the $\ell = 2$ deformation grows: the drop elongates. Elongating concentrates charge at the two ends, because a pointed conductor holds a higher surface density, so the field there rises and the elongation accelerates. What forms is a cone with a half-angle of $49.3°$ — a shape G. I. Taylor computed in 1964 by demanding that the electrostatic and surface-tension stresses balance on a conical surface, which they do at exactly one angle — and from the tip of the cone a thin jet issues.

The jet then breaks into droplets by the ordinary instability of a liquid thread — a thread of liquid cannot stay a thread, because any bulge of long enough wavelength lowers the surface area and therefore grows. The droplets are far smaller than the parent, so each carries a charge below its own Rayleigh limit, and the system is stable at the new size.

What that costs is surface, and the arithmetic runs the other way from the usual case.

So the cascade is not the drop finding a cheaper configuration in every respect. It is the drop buying a large fall in electrostatic energy at the price of a smaller rise in surface energy, and the Rayleigh limit is precisely the size at which that trade first becomes worth making.

That cascade is an industry. Electrospray ionisation puts a protein into the gas phase, intact and charged, by exactly this route — a solution is pushed through a fine needle at a few kilovolts, forms a Taylor cone, throws a jet, and the droplets evaporate down to their own Rayleigh limits and split again until only ions are left. It made mass spectrometry of large biomolecules possible, and Fenn’s share of the 2002 chemistry prize was for it.

The droplet sizes follow from the thread rather than from the cone. A liquid cylinder is unstable to any disturbance whose wavelength exceeds its circumference, the growth rate peaks at a wavelength of about nine radii, and the drops that result are a little under twice the jet’s diameter — the same arithmetic that sets the size of the drops from a dripping tap.

The droplet sizes are worth a number. A Taylor cone at a flow rate of a microlitre a minute from a needle at a few kilovolts produces a jet of order a micrometre across, which breaks into droplets a few micrometres in diameter — each carrying a charge close to its Rayleigh limit, which for a two-micrometre drop is about $2$ femtocoulombs, or some ten thousand elementary charges. Evaporation then shrinks each one until it too crosses the line and splits again. Four or five generations later what is left is a bare ion with a few charges on it, and the whole cascade has taken a millisecond.

The same sequence happens without any apparatus. A thunderstorm’s drops carry charge, and a drop that grows by collision toward its limit disintegrates, which is one of the processes redistributing charge inside a cloud.

What the balance is really saying #

There is a way to read the Rayleigh condition that makes it less like a coincidence of two pressures and more like the general shape of a stability problem.

Both pressures come from energies. The surface energy of a drop is $4\pi R^2\gamma$ and rises as the drop is stretched, because stretching makes surface. The electrostatic energy of a charged conductor is $Q^2/8\pi\varepsilon_0 R$ and falls as the drop is stretched, because a longer conductor of the same charge has a larger capacitance and therefore a lower energy. The drop is stable while the first rises faster than the second falls.

Written that way the criterion is a competition between two energies with opposite signs and different dependences on shape, which is the standard form of every instability in this collection: a nucleus that must climb a barrier has the same shape of argument with a bulk term in place of the electrostatic one, and so does a self-gravitating cloud. What changes between them is which term is negative and which power of the size each carries.

One structural feature of such competitions is worth drawing, because it is easy to assume the instability happens where the arithmetic first favours the alternative.

A charged drop is in the same position. Well below the Rayleigh limit the elongated configurations are already energetically preferable in some respects and the drop does not adopt them, because it is sitting in a minimum with a barrier round it. What the limit marks is not where the alternative becomes cheaper but where the sphere stops being a minimum at all — which is why the criterion is stated as a stability condition rather than as a comparison of energies.

The fissility — the ratio of the two, normalised so that instability begins at one — is the quantity that makes the comparison portable. Bohr and Wheeler took it straight from Rayleigh’s drop and applied it to a nucleus in 1939, with the Coulomb energy of the protons against the nuclear surface energy, and the resulting parameter is still how one says which nuclei fission spontaneously and which do not.

Where it stops #

The drop has to be a conductor. Everything above puts the charge on the surface, which is the conducting case. A charged insulating drop holds its charge where it was put, the field penetrates the interior, and the stability analysis is different — the relevant instability may be at a different mode number, and the limit is not the same number.

And the drop has to be still. A drop in an external field, or falling, or evaporating, is being deformed by something else at the same time, and the limit shifts. Evaporating drops in an electrospray reach the limit by shrinking rather than by charging, which is the same crossing approached along a different path, and the disintegration then happens at a slightly different fissility because the drop is not spherical when it gets there.

Very high fields do something else entirely. At around $10^{10}$ V/m a metal surface emits electrons by tunnelling — field emission — and the surface loses its charge rather than tearing. That is $10^{9}$ times the pressure of the air-breakdown case, so it is reached only at the tip of something very sharp, which is precisely where the electrostatic pressure was largest anyway.

The conductor has to be able to hold the charge in the first place. A sphere of radius $R$ at potential $V$ carries $4\pi\varepsilon_0 R V$, so the capacitance is a length and small objects hold very little. Charging a ten-micrometre droplet to its Rayleigh limit puts it at about $1.1$ kilovolts, and getting it there means bringing it into contact with something at that potential or growing it from something that already is — which is why an electrospray charges the liquid before it makes the drops rather than after.

And the pressure is not a stress in a material. It is the electromagnetic field’s own stress transmitted to the charges in the surface. Where those charges are attached to something that can support a stress — a metal — the metal carries it. Where they are not, as on a free surface of liquid, the surface moves. The energy method above says nothing about which case it is in, which is why the same formula covers a drop that disintegrates and a sphere that does nothing at all.

Where the field is strong the pressure is high, and the field is strongest where the equipotentials crowd, which for a shape with a point is at the point. That is the same concentration that makes a lightning conductor work, and it is why a charged drop goes unstable at its ends rather than uniformly: the $\ell = 2$ deformation is the one that makes two points, and having made them it strengthens the field at them.

The measurement, which is harder than the theory #

Testing the limit sounds straightforward and took a century to do cleanly.

The difficulty is that a drop approaching its limit is also doing everything else: evaporating, oscillating from however it was produced, falling through air that deforms it, and sitting in the field of whatever charged it. Every one of those shifts the threshold, and the early measurements disagreed with Rayleigh by tens of per cent in both directions.

The clean version was done by levitating single droplets in an electrodynamic trap, letting them evaporate slowly, and watching for the moment of disintegration with a laser. The drop shrinks at constant charge, so it crosses the line from below, and the crossing point can be read off to a per cent. The answer is that disintegration happens at a fissility of about $1.0$, occasionally slightly below, and that the drop loses only a small fraction of its mass — around one or two per cent — while shedding a much larger fraction of its charge, typically a fifth to a third.

That asymmetry is the useful part and it is not in the criterion at all. The criterion says when the drop becomes unstable; it says nothing about what the instability does, and what it does is throw away charge preferentially because the jet comes from the region where charge has concentrated. A drop that lost mass and charge in proportion would be back at the same fissility and would disintegrate again immediately; the observed behaviour leaves the parent comfortably stable after one event.

The ladder from here #

Later rungs on this anchor: the Taylor cone derived properly, with the balance of stresses on a conical surface and the $49.3°$ that comes out of it; the stability of a charged drop in an external field, where two parameters replace one and the limit becomes a curve; the electrospray cascade quantitatively, with the droplet sizes at each generation; and the nuclear analogue, where the same fissility parameter — Coulomb energy against surface energy — decides which nuclei fission, using an expression Bohr and Wheeler took directly from Rayleigh’s drop.

The neighbouring ladders are the inside of a conductor, where the discontinuity that produces this pressure is established; the skin that is not a skin, which is the surface tension being fought; and the thread that cannot stay a thread, which is what the jet does next.