Where the energy of a field actually is

A charged capacitor holds a definite number of joules. Where those joules are sitting sounds like a question of preference — on the plates, in the charges, in the gap between them — and it stays a question of preference right up to the point where something is done that forces an answer.

Everything below is where those two numbers come from, the three arrangements in which one of the accounts collapses and the other does not, and what the surviving account costs.

Two ways of counting the same joules #

The first account never mentions space. Charging a capacitor means moving charge from one plate to the other against a potential difference that grows as the transfer proceeds, so the work is the integral of $V\,\mathrm{d}q$ from zero to $Q$. With $V = q/C$ that gives $Q^2/2C$, or equivalently $\tfrac12 QV$, and every quantity in it is a property of the charges and of the single number attached to each point that the potential is. For the figure’s capacitor — 200 cm² of plate, 15 nC on it, 2.00 mm apart — the capacitance is $\varepsilon_0 A/d = 88.8$ pF, the voltage is 169 V, and the stored energy is 1.27 µJ.

The second account never mentions charges. It says that the space between the plates holds energy at a density of $\tfrac12\varepsilon_0E^2$ joules per cubic metre, and asks how many cubic metres there are. The field in the gap is $Q/\varepsilon_0 A = 84.7$ kV/m, which puts 31.8 millijoules into every cubic metre; the gap contains $3.99\times10^{-5}$ of a cubic metre; and the product is 1.27 µJ.

$$U = \tfrac12 QV = \int \tfrac12\varepsilon_0E^2\,\mathrm{d}V.$$

The agreement is exact rather than close, and it is not evidence for anything. The two expressions are the same theorem in different variables, related by an integration by parts that turns a volume integral over the field into a surface integral over the charges.

Capacitance is geometry, which is what makes the first account available at all. For plates of 100 cm² the capacitance reads 177.1 pF at 0.5 mm, 88.5 pF at 1 mm, 44.3 pF at 2 mm and 22.1 pF at 4 mm — nothing but the area, the separation and $\varepsilon_0$ enters, and the reciprocal dependence on the gap is exact for as long as the plates are close compared with their size. The essay’s own capacitor has twice that area at twice the 1 mm separation, which is why it lands on the same 88.8 pF.

So the question is not which account is right, but which still has anything to say when its own ingredients are taken away. Three arrangements answer that sharply.

The field the second account integrates over is not quite the ruled set of parallel lines the formula assumes. Traced from discrete charges on each plate it comes out straight in the middle — because the physics makes it straight, not because it was drawn that way — and bows outward near the ends, with the field nine-tenths of the way to the edge at 72 per cent of the field at the centre. The 1.27 µJ computed above assumes uniformity across the whole area, so the fringing is exactly where that idealisation sits and exactly how much it is worth.

The force, computed twice, from curves that go opposite ways #

The two plates attract each other, and the attraction is $-\mathrm{d}U/\mathrm{d}x$: the rate at which the stored energy changes as the gap widens. Doing that twice, at fixed charge and at fixed voltage, is the cleanest test the bookkeeping has.

At fixed charge the field in the gap does not change when the plates move apart, because it is set by the surface charge density alone. Widening the gap therefore adds field to the world at unchanged strength, and the energy rises in proportion to the separation. Differentiating gives an attraction of $Q^2/2\varepsilon_0A$, which for 15 nC on 200 cm² is 635 µN — about the weight of 65 milligrams.

At fixed voltage the same operation gives the wrong sign. Now $U = \tfrac12 CV^2$ with $C \propto 1/d$, so the stored energy falls as the plates separate, and $-\mathrm{d}U/\mathrm{d}x$ comes out positive: a repulsion. That answer is wrong, and it is wrong for a reason that is entirely visible in the accounting. Holding the voltage fixed means a battery is attached, and as the capacitance drops the battery takes charge back. At the crossing point the field energy falls at 635 microjoules for every metre of separation, while the battery recovers 1.27 millijoules per metre — twice as much. The difference, 635 microjoules per metre, is what an external hand has to supply, and it is an attraction of 635 µN.

The same number, from two calculations whose intermediate quantities have opposite signs. That is the first thing the plate-bound account cannot produce, because $Q^2/2\varepsilon_0A$ is not an expression about plates. Divided by the area it is $\tfrac12\varepsilon_0E^2$: the attraction per square metre is numerically the energy per cubic metre, 31.8 millipascals against 31.8 millijoules per cubic metre. The force is the field’s own stress, and the plates are where the stress happens to be terminated.

Contours of constant potential round a charge, with field lines crossing them at right angles, make the first account’s bookkeeping visible: no work is needed along a contour and none depends on the route between two of them, which is what makes $\int V\,\mathrm{d}q$ well defined and $\tfrac12 QV$ an honest total. The potential is a genuine economy and it is not a location. The contours say what work a charge would cost; they say nothing at all about which cubic metres the joules are in.

The energy arrives through the sides #

The second arrangement is a resistor, and it is the centre of the argument because the answer sounds wrong before it is computed.

A wire carrying a steady current has an electric field along it, of magnitude $V/L$; that is what drives the current. Inside a conductor in equilibrium the field is exactly nothing, but a current-carrying conductor is not in equilibrium. It also has a magnetic field around it, circling the wire at $\mu_0 I/2\pi r$ — the field that wraps a current. The two are perpendicular everywhere on the surface, one axial and one circumferential, so their cross product is radial, and it points the same way at every point of the surface: inward.

The flux of $\mathbf{S} = \mathbf{E}\times\mathbf{B}/\mu_0$ inward through the wire’s own surface is the dissipation, to every figure the arithmetic can carry. Nothing crosses the ends. The energy that heats the wire enters through its curved sides, out of the field in the space surrounding it, having travelled from the battery through that space rather than through the copper.

Two features of the drawing make the claim harder to dismiss than a single surface would. The flux density falls as $1/r$ while the area it crosses grows as $r$, so every coaxial cylinder carries the identical 1.67 W: the energy is not manufactured at the wire’s surface, it is passing through a region and being absorbed at the end of it. And reversing the current reverses both E and B, leaving the cross product unchanged — which is why a resistor heats up whichever way the current runs.

This is not an interpretation laid over the equations. Taking $\partial/\partial t$ of the energy density and substituting the field equations gives

$$\frac{\partial}{\partial t}\left(\tfrac12\varepsilon_0E^2 + \frac{B^2}{2\mu_0}\right) + \nabla\cdot\mathbf{S} = -\mathbf{E}\cdot\mathbf{J},$$

which is a continuity equation: energy density changes only by flowing in or by being handed to charges. The right-hand side integrated over the wire is $I^2R$. There is no term in it for energy carried along a conductor by its current, and no way to add one without breaking the identity. The wires guide; the field delivers.

A wave that is nothing but field #

The third arrangement removes the charges altogether. A light wave has no charge anywhere in it, and it carries energy — which leaves $\tfrac12 QV$ with nothing to be an account of.

At an amplitude of 1 V/m the time-averaged density is 4.43 picojoules per cubic metre, and multiplying by $c$ gives an intensity of 1.33 mW/m². Sunlight above the atmosphere runs at 1361 W/m² — 4.54 microjoules per cubic metre, an amplitude of 1.01 kV/m paired with 3.38 µT. Worth holding next to the capacitor: the gap at 84.7 kV/m holds 31.8 millijoules per cubic metre, some seven thousand times the energy density of full sunlight, while sitting still.

The two constants in the density were measured with no light involved at all, which is the part of the story worth keeping. Weber and Kohlrausch obtained $3.107\times10^8$ m/s in 1856 from one quantity of charge measured in two unit systems; Fizeau had measured light at $3.153\times10^8$ m/s in 1849 with a toothed wheel. $\varepsilon_0$ and $\mu_0$ are the constants of proportionality in the two halves of the energy density, so the same pair of laboratory measurements fixes how many joules a cubic metre of field holds and how fast it travels.

What it costs #

An air gap stores 39.8 joules per cubic metre and no more. Dry air breaks down at about 3 MV/m, and $\tfrac12\varepsilon_0E^2$ at that field is 39.8 J/m³. Petrol carries about $3.4\times10^{10}$ J/m³. The ratio is nearly $10^9$, and it is the reason no vehicle has ever been powered by an air-gap capacitor.

A dielectric buys four orders of magnitude and not five. A polymer film with a relative permittivity of 3 and a breakdown field near 300 MV/m holds $\tfrac12\varepsilon_r\varepsilon_0E^2 \approx 1.2$ MJ/m³. That is a thirty-thousandth of petrol, and it is close to the practical ceiling for any electrostatic store, because the density goes as the square of a field that materials refuse to survive.

The magnetic term is where the joules actually are. $B^2/2\mu_0$ at 1 T is 398 kJ/m³, ten thousand times the air-gap electric figure at its own breakdown limit, and at 10 T it is 39.8 MJ/m³. That asymmetry is why large-scale field storage is magnetic, and it comes with a structural bill: the same expression read as a pressure is 398 kPa at 1 T — four atmospheres — and 39.8 MPa at 10 T, which is what a superconducting magnet’s former has to hold in.

An electrostatic actuator pays the square twice over. The stress available is the energy density, and 31.8 millipascals is not a useful pressure. A useful one means a large field at a low voltage, which means a small gap: at 1 µm and 10 V the field is 10 MV/m and the stress is 443 Pa. Every micromechanical comb drive follows from that arithmetic, and so does the fact that such devices are microscopic.

Intensity is a field quantity, and that is what an instrument reports. A power meter, a solar cell rating, an antenna specification and an exposure limit are all statements about $|\mathbf{S}|$ in watts per square metre at a place where no charge need be present. The plate-and-charge account has no way to express any of them.

Where the model stops #

Everything above is vacuum, linear response and a steady state, and each of those is doing real work.

In matter the density is $\tfrac12\mathbf{D}\cdot\mathbf{E}$, not $\tfrac12\varepsilon_0E^2$. The difference, $\tfrac12\mathbf{P}\cdot\mathbf{E}$, is the energy that went into pulling the material’s own charges apart rather than into the field, and it is not a correction: water at low frequency has a relative permittivity near 80, so the density at a given field is eighty times the vacuum figure and seventy-nine parts in eighty of it is polarisation. Which parts belong to the field and which to the matter has no unique answer, and the rival conventions disagree about the momentum inside a dielectric.

The resistor calculation assumes a steady current in a uniform wire. The 1.67 W is exact for a direct current with the return conductor far away. At high frequency the current crowds into a skin at the surface, the axial field is no longer uniform, and the flux pattern changes shape — though not its total, which the conservation law fixes regardless.

The field of a point charge holds infinite energy, and no radius rescues it.

The two marked radii differ by exactly two, and the factor is a definition rather than a physical distinction. At 1.409 fm the field energy outside equals $m_ec^2$. The classical electron radius, $q^2/4\pi\varepsilon_0m_ec^2 = 2.818$ fm, is defined without the half, and outside that radius the field holds precisely half the rest energy, 255 keV. Neither number is the size of an electron: scattering finds no structure in it down to $10^{-18}$ m, so the classical radius is at least 2,800 times too large. What it is is the length at which this bookkeeping stops making sense. The divergence is a genuine failure of classical electromagnetism, and no classical model of an extended electron repaired it — it was absorbed into quantum electrodynamics by renormalisation, where the infinite piece is folded into the measured mass.

Why the density diverges so violently is a matter of two competing powers. The field falls as the inverse square because a fixed number of lines is spread over an area growing as $r^2$; the energy density is the square of the field, so it goes as $1/r^4$; and the volume available goes only as $r^3$. The shells nearest the charge win by one power, and one power is exactly what makes the integral diverge.

Radiation pressure, momentum, and one honest objection #

The same expression that carries energy carries momentum, at a density of $\mathbf{S}/c^2$. That is why light exerts a pressure: sunlight’s 1361 W/m² divided by $c$ is 4.54 micropascals on a black surface and twice that on a mirror. The quantum version of the same statement is a photon with a momentum of $E/c$, and the classical and quantum accounts agree on the pressure because the momentum-to-energy ratio is the same in both.

Field energy also settles arguments with no light in them. When a charge accelerates, the near field and the far field disagree about where it is, and the mismatch propagates outward carrying energy away — which is why a turning charge must radiate. What leaves was field energy already there, cut loose.

The standard objection is worth stating in full, because it is correct as far as it goes. A charged capacitor sitting in a static magnetic field has crossed E and B with no light, no current and nothing happening — and a Poynting vector. For the gap above at 84.7 kV/m in a field of 1 T, $EB/\mu_0$ is 67.4 GW/m², fifty million times the flux density of sunlight, circulating round the apparatus for as long as it is left alone and delivering nothing anywhere. Its divergence is zero everywhere, so no joule ever accumulates or departs.

Two things follow and only one is a concession. The concession is that $\mathbf{S}$ is not uniquely determined: adding the curl of any vector field leaves $\nabla\cdot\mathbf{S}$ unaltered, so the theorem fixes the flux through a closed surface and not the route. The resistor’s 1.67 W is not negotiable; the picture of it arriving radially is the simplest field meeting that constraint rather than a measured trajectory. What is not a concession is the momentum. The circulating density $\mathbf{S}/c^2$ above is $7.5\times10^{-7}$ kg m⁻²s⁻¹, and it is required: without it, angular momentum is not conserved when the charge is allowed to leak away, and a freely suspended arrangement begins to rotate with no external torque. The circulation nobody can detect while nothing changes is what balances the books when something does.

Poynting, 1884, and what the picture had to survive #

Maxwell had put the energy in the medium from the start. His 1865 paper and the Treatise of 1873 both treat the field as the seat of the energy, following Faraday’s conviction that space near a charge is under a strain, and the $\tfrac12\varepsilon_0E^2$ expression is his. What was missing was the flow.

John Henry Poynting supplied it in 1884, in a paper for the Philosophical Transactions called On the Transfer of Energy in the Electromagnetic Field. He had been Maxwell’s student at the Cavendish before taking the chair at Mason College in Birmingham, and the argument he gave is the continuity equation above, derived rather than guessed. Oliver Heaviside reached the same result independently within about a year, and insisted on the consequence Poynting had also seen: the energy delivered to a conductor travels in the space outside it, which is a statement about geometry rather than about metal.

The resistance the picture met was not to the mathematics, which nobody disputed, but to the sideways flow. That the heat in a wire should arrive through its surface read as an absurdity to physicists entirely comfortable with the equations requiring it, and the objection about the non-uniqueness of $\mathbf{S}$ was raised early and correctly. Poynting computed the resistor case in the original paper, then spent much of the rest of his career on gravitation, measuring the gravitational constant with a balance in 1891.

One further observation. The field concept was accepted in the 1880s because it predicted a disturbance that leaves its source and travels at a finite speed — and the energy accounting is what makes that prediction mean anything, since a disturbance carrying no energy would be a change of notation. Every argument that the field is a thing rather than a bookkeeping device runs through the joules.

What the picture cannot show #

Energy density has no direction, and every figure here is made of arrows. The 31.8 mJ/m³ in the capacitor gap is a scalar attached to each cubic metre. The plate figure draws field lines, which carry direction and not amount, and the two things are being read off the same picture by different rules.

The resistor figure is drawn at the wrong aspect ratio, deliberately. The wire is 1.6 mm across and 2 m long, a ratio of 1,250 to 1, and the drawing shows it perhaps twenty times longer than thick so that the coaxial surfaces can be labelled. At the true proportions the inward arrows would be indistinguishable from a line, which is why the sideways flow is invisible to intuition.

The divergence cannot be drawn. The lower panel of the density figure is a straight line on logarithmic axes, and the whole content of the argument is that the line never stops. The drawing stops at 0.4 fm because a canvas has an edge. Nothing in the figure distinguishes “this rises without limit” from “this rises steeply and the plot ends”.

No figure can show that the flow is non-unique. The radial arrows around the wire are one field meeting the closed-surface constraint; the alternatives differing from it by the curl of something would look completely different while predicting the identical 1.67 W. A picture of the ambiguity would have to draw several incompatible flows at once.

The capacitor curves hide where in the gap the energy is. The 1.27 µJ assumes the uniform field everywhere, and the traced figure measures the failure: 72 per cent of the central field at nine-tenths of the way to the edge. The fringing field outside the gap holds joules that no curve on the plot accounts for.

The ladder from here #

Later rungs on this anchor: the Poynting theorem in full, with the $-\mathbf{E}\cdot\mathbf{J}$ term as the only channel between field and matter; the energy density in a polarisable medium, where $\tfrac12\mathbf{D}\cdot\mathbf{E}$ replaces $\tfrac12\varepsilon_0E^2$; the Maxwell stress tensor, of which the 31.8 millipascals above is one diagonal component; inductance read as $B^2/2\mu_0$ over a coil, the magnetic mirror of this whole essay; and the self-energy problem, from Abraham and Lorentz to what renormalisation does with it.

The neighbouring ladders are Maxwell’s four equations, which are where the energy density and the flow are both derived from; magnetism and the force that does no work, which supply the $B^2/2\mu_0$ half and explain why a magnetic field can hold energy while never changing a particle’s; conductors, where the geometry that fixes a capacitance also fixes how many joules a given voltage buys; Gauss’s law, which is the flux argument this one is built on; and the attraction that needs no charge, where a field with no net charge anywhere still stores energy and still pulls.