Energy Density of the Electric Field

The previous article priced the assembly of a charge configuration and left the bill pinned to the charges: $W=\tfrac12\int\rho V\,\mathrm{d}\tau$ , an integral that lives only where charge lives. This article moves the money. One integration by parts — Gauss's law in, boundary term out — converts the same total into $W=\tfrac{\varepsilon_0}{2}\int E^2\,\mathrm{d}\tau$ , an integral over all of space, charge or no charge. Read literally, it says something startling: the energy is not attached to the charges at all. It is stored in the field, at density $u=\tfrac{\varepsilon_0}{2}E^2$ , wherever the field reaches. In statics you cannot tell the two bookkeepings apart — they always agree on the total, and no experiment probes "where" a static joule sits. But the field version is the one that survives. When Maxwell's equations let fields detach from their sources and travel as light, the energy demonstrably travels with the fields, through vacuum, years after the source stopped caring. The sunlight warming your face left its charges eight minutes ago; the only place its energy could have been in the meantime is the field. The formula you derive here is the static limit of that conservation law, and adopting it is the moment electrostatics stops being about action-at-a-distance and starts being about a medium with local properties. The price of the new ontology is one honest debt: for a true point charge, $\tfrac{\varepsilon_0}{2}\int E^2$ diverges. Chasing that infinity leads to the classical electron radius, and eventually to renormalization — a crack in classical physics visible from the second chapter of Griffiths.

Key takeaways

The assembly energy of a static charge configuration can be rewritten as a field integral: $W=\frac{\varepsilon_0}{2}\int_{\text{all space}}E^2\,\mathrm{d}\tau$ — no reference to where the charges sit.
The conversion from $\tfrac12\int\rho V\,\mathrm{d}\tau$ is one integration by parts; the surface term dies as $1/r$ for any localized distribution, which is why the field integral must run over all space.
The integrand $u=\tfrac{\varepsilon_0}{2}E^2$ is an energy density: always positive, concentrated where fields are strong (it quadruples when the field doubles), and numerically equal to the electrostatic pressure on a conductor surface.
Field energy does not superpose: for $\mathbf{E}=\mathbf{E}_1+\mathbf{E}_2$ the cross term $\varepsilon_0\int\mathbf{E}_1\cdot\mathbf{E}_2\,\mathrm{d}\tau$ is exactly the interaction energy of the pair sums.
For a point charge the field integral diverges as the inner cutoff shrinks — the classical self-energy problem; setting $W=m_ec^2$ defines the classical electron radius $r_e\approx2.82\,\mathrm{fm}$ , the scale where classical electrodynamics indicts itself.
Key idea: electrostatic energy is not bound to the charges but distributed through space as $u=\tfrac{\varepsilon_0}{2}E^2$ ; integrating this density over all space reproduces the total assembly energy, and the two pictures — energy-in-charges vs energy-in-field — agree because an integration by parts trades one for the other.

Where this lives today

The formula $u=\tfrac{\varepsilon_0}{2}E^2$ (and its dielectric extension $\tfrac12\varepsilon E^2$ ) is the design equation of electrical energy storage. A supercapacitor engineer maximizing joules per litre is maximizing $\varepsilon E^2$ subject to breakdown; the reason your phone runs on lithium chemistry rather than capacitors is precisely the $\sim40\,\mathrm{J/m^3}$ air-gap ceiling of Worked Example 1, clawed up by a factor of $\sim10^6$ in modern double-layer devices — nanometre gaps and high- $\varepsilon$ interfaces are the only levers the formula offers. The same density, read as stress, sets the pull on capacitor plates, MEMS actuators, and the outward pressure on charged conductor surfaces — the next article derives that pressure and finds it is numerically $u$ itself. The self-energy debt, meanwhile, became one of the most productive embarrassments in physics. The classical electron radius $r_e=2.82\,\mathrm{fm}$ from Derivation C sets the Thomson cross-section $\sigma_T=\tfrac{8\pi}{3}r_e^2=6.65\times10^{-29}\,\mathrm{m^2}$ — the number through which cosmic-microwave-background photons scatter off electrons, hard-wired into every CMB analysis code fitting the universe's parameters in 2026. And the divergence itself, exported to quantum field theory, forced the invention of renormalization: the modern electron "self-energy" is still infinite before subtraction, and the machinery for handling it — born of the integral $\int_a^{\infty}E^2\,\mathrm{d}\tau$ refusing to converge — now underwrites the most precise predictions in science.

The physical picture

Charge a capacitor and ask the childlike question: while it sits there charged, where is the energy? The charge picture shrugs — "in the configuration." The field picture answers with coordinates: it is in the gap, packed at $\tfrac{\varepsilon_0}{2}E^2$ joules per cubic metre, and nowhere else — outside the gap the field (nearly) vanishes and so does the energy. Discharge the capacitor through a lamp and you drain exactly that gap-volume times that density. The field picture pays out like a warehouse with an address.

Figure

Where the energy lives — a capacitor gap at uniform density, a point charge in a fading halo

Because $u\propto E^2$ , the energy crowds into high-field real estate: against a charged surface, in a capacitor gap, at a conductor's sharp tip. Doubling the field quadruples the local density — which is why the regions that store the most energy per litre are exactly the regions where dielectrics break down and field emission ignites. Air gives up near $3\times10^6\,\mathrm{V/m}$ , capping any air-gap device at a modest $u\approx40\,\mathrm{J/m^3}$ ; the whole engineering discipline of dielectric materials is a hunt for media that let $E$ — and therefore $E^2$ — climb higher before failing. The two bookkeepings disagree, sometimes comically, about location. For two separated charges, the pair sum from the previous article posts the interaction energy "at the charges"; the field ledger spreads it through the region where the two fields overlap. For a bound pair the interaction entry is negative — yet the field density is positive at every point in space. Both statements are true, because the field ledger also carries each charge's own always-positive self-energy, which the pair sum deliberately omitted. In statics the disagreement has no experimental consequence. The moment fields radiate, only the field ledger balances — and physics votes for the warehouse.

From charge bookkeeping to field bookkeeping

Work and Energy of a Charge Configuration ended with the continuous assembly formula, and it is our only input:

W=\frac{1}{2}\int\rho\,V\,\mathrm{d}\tau \qquad (1)

where: $\rho$ is the (smooth) charge density, $V$ the potential of the full final configuration, and the integral needs only the charged region — elsewhere $\rho V=0$ . The factor $\tfrac12$ was earned there by the double-counting argument; we do not re-derive it. Everything in this article flows from massaging equation (1) with two facts about electrostatics: Gauss's law in differential form, $\rho=\varepsilon_0\nabla\cdot\mathbf{E}$ , and $\mathbf{E}=-\nabla V$ . The result of that massage — Derivation A below — is the headline pair. Locally,

u=\frac{\varepsilon_0}{2}E^2 \qquad (2)

where: $u$ is energy per unit volume ( $\mathrm{J/m^3}$ ), $\varepsilon_0=8.854\times10^{-12}\,\mathrm{F/m}$ , and $E$ is the local field magnitude. Note what is absent: any reference to $\rho$ or $V$ . A cubic millimetre of field at $10^5\,\mathrm{V/m}$ stores the same $4.4\times10^{-11}\,\mathrm{J}$ whether the nearest charge is a millimetre or a mile away. Globally,

W=\frac{\varepsilon_0}{2}\int_{\text{all space}} E^2\,\mathrm{d}\tau \qquad (3)

where: the domain is genuinely all of space — the field form has no license to stop at the edge of the charge. Equations (1) and (3) are equal totals with incompatible opinions about location; the exchange rate between them is one boundary term that dies at infinity.

Which energy formula answers which question

you have	use	watch out for
discrete point charges	$W=\tfrac12\sum_i q_iV_i$ with $V_i$ from the others	excludes self-energy; can be negative
smooth $\rho$ , potential known	$W=\tfrac12\int\rho V\,\mathrm{d}\tau$ (eq. 1)	integrate over the charge only
field known everywhere	$W=\tfrac{\varepsilon_0}{2}\int E^2\,\mathrm{d}\tau$ (eq. 3)	ALL space; includes self-energy; never negative
local density / stress questions	$u=\tfrac{\varepsilon_0}{2}E^2$ (eq. 2)	equals the outward pressure on a conductor surface
two overlapping systems	$W=W_1+W_2+\varepsilon_0\int\mathbf{E}_1\cdot\mathbf{E}_2\,\mathrm{d}\tau$	energy does not superpose — the cross term is the interaction

Derivation ATrading the charge integral for the field integral

The goal is to convert equation (1) into equation (3) and to watch, step by step, where the energy bookkeeping moves. The strategy is the standard one for shifting derivatives between factors: substitute Gauss's law to make the integrand field-flavoured, integrate by parts via the product rule and the divergence theorem, and then argue the boundary term to death by taking the surface to infinity. The one subtlety worth guarding is the domain: equation (1) needed only the charged region, but the by-parts manoeuvre only kills the surface term if the volume is all of space — so the field form inherits an all-space domain as a condition, not a convenience.

Step 1. Substitute $\rho=\varepsilon_0\nabla\cdot\mathbf{E}$ into (1):

W=\frac{\varepsilon_0}{2}\int (\nabla\cdot\mathbf{E})\,V\,\mathrm{d}\tau

Why this is legal everywhere: Gauss's law holds pointwise, and extending the integral beyond the charge costs nothing since $\nabla\cdot\mathbf{E}=0$ there.

Step 2. Move the derivative off $\mathbf{E}$ with the product rule $\nabla\cdot(V\mathbf{E})=V(\nabla\cdot\mathbf{E})+(\nabla V)\cdot\mathbf{E}$ , so $(\nabla\cdot\mathbf{E})V=\nabla\cdot(V\mathbf{E})-(\nabla V)\cdot\mathbf{E}$ , and use $\mathbf{E}=-\nabla V$ to turn the second piece into $+E^2$ :

W=\frac{\varepsilon_0}{2}\left[\int \nabla\cdot(V\mathbf{E})\,\mathrm{d}\tau+\int E^2\,\mathrm{d}\tau\right]

Step 3. Convert the total-divergence piece with the divergence theorem, over a large sphere of radius $r$ :

W=\frac{\varepsilon_0}{2}\left[\oint_{\mathcal S} V\mathbf{E}\cdot\mathrm{d}\mathbf{a}+\int_{\mathcal V} E^2\,\mathrm{d}\tau\right]

Why both terms depend on $r$ while their sum does not: the equality holds for any volume enclosing all the charge, so as $\mathcal V$ grows, whatever the (positive) volume integral gains, the surface term must lose.

Step 4. Send $r\to\infty$ . For a localized distribution, far away $V\sim1/r$ and $E\sim1/r^2$ , so the integrand $V\mathbf{E}$ falls as $1/r^3$ while the surface area grows only as $r^2$ : the surface term scales as $1/r$ and dies. What remains is equation (3), with the volume integral now over all space.

Step 5. Read off the local statement. The surviving integrand, $\tfrac{\varepsilon_0}{2}E^2$ , is nonnegative and defined at every point, and we declare it the energy density, equation (2). In statics this attribution is a choice (only the total is measured); Poynting's theorem later promotes it to a local conservation law with a flux, and the choice becomes physics. Verified: reproduces Griffiths eq. 2.45 and Jackson §1.11; the scaling audit $V E\,\mathrm{d}a\sim r^{-1}\cdot r^{-2}\cdot r^{2}=r^{-1}$ confirms Step 4, and the monopole term dominates any localized multipole at large $r$ , so the argument is general.

The cross-term identity in the table deserves its one line of algebra, because it reconnects this article to the pair sums of the last one. Write $\mathbf{E}=\mathbf{E}_1+\mathbf{E}_2$ and expand $E^2=E_1^2+E_2^2+2\,\mathbf{E}_1\cdot\mathbf{E}_2$ :

W=W_1+W_2+\varepsilon_0\int \mathbf{E}_1\cdot\mathbf{E}_2\,\mathrm{d}\tau \qquad (4)

where: $W_1$ and $W_2$ are the self-energies each system would have alone, and the overlap integral is the interaction energy — for two point charges it evaluates to the familiar $q_1q_2/4\pi\varepsilon_0 d$ , positive for like charges, negative for opposite ones. Energy does not superpose even though fields do, because squaring is not linear: bring two systems together and the books change by exactly the cross term, which is the only part the pair sum of the previous article ever tracked.

Figure

Two ledgers, one total — the charge picture posts energy at the charges, the field picture spreads it through the overlap

The classic checkpoint: the uniformly charged ball

Work and Energy of a Charge Configuration assembled a uniform ball of charge shell by shell and found $W=\tfrac{3}{5}\,kQ^2/R$ . That number was earned by the charge route; the honest test of equation (3) is to recompute it from the field route and watch the two ontologies agree to the digit.

Go deeper · Derivation B · 1 minField energy of a uniformly charged ball

Derivation BField energy of a uniformly charged ball

The goal is the total field energy of charge $Q$ spread uniformly through a ball of radius $R$ , computed purely from equation (3), as an independent cross-check of the shell-assembly result $W=\tfrac{3}{5}Q^2/(4\pi\varepsilon_0R)$ derived in the previous article. The strategy: quote the interior and exterior fields from Applications of Gauss's Law and Symmetry (they are that article's first trophy — we do not re-derive them), square, and integrate in spherical shells $\mathrm{d}\tau=4\pi r^2\,\mathrm{d}r$ . Watch how the two regions contribute in the fixed ratio 1 : 5 — a detail invisible to the assembly route, which never asks where the energy sits.

Step 1. The fields, by Gauss's law: inside, $E_{\text{in}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qr}{R^3}$ for $r<R$ ; outside, $E_{\text{out}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ for $r>R$ .

Step 2. Exterior contribution:

W_{\text{out}}=\frac{\varepsilon_0}{2}\int_R^{\infty}\!\left(\frac{Q}{4\pi\varepsilon_0 r^2}\right)^{2}4\pi r^2\,\mathrm{d}r=\frac{Q^2}{8\pi\varepsilon_0}\int_R^{\infty}\frac{\mathrm{d}r}{r^2}=\frac{Q^2}{8\pi\varepsilon_0 R}

Why this piece is universal: outside $R$ the ball is indistinguishable from a point charge, so this is also the energy stored beyond radius $R$ of any point charge — a fact Derivation C will weaponize.

Step 3. Interior contribution:

W_{\text{in}}=\frac{\varepsilon_0}{2}\int_0^{R}\!\left(\frac{Qr}{4\pi\varepsilon_0 R^3}\right)^{2}4\pi r^2\,\mathrm{d}r=\frac{Q^2}{8\pi\varepsilon_0 R^6}\int_0^R r^4\,\mathrm{d}r=\frac{Q^2}{8\pi\varepsilon_0 R}\cdot\frac{1}{5}

Why so little lives inside: the interior field grows only linearly from zero, and $u\propto E^2$ punishes weak fields quadratically — the ball's core is energetically almost empty.

Step 4. Add:

W=\frac{Q^2}{8\pi\varepsilon_0 R}\left(1+\frac{1}{5}\right)=\frac{3}{5}\,\frac{Q^2}{4\pi\varepsilon_0 R}

The field route lands on exactly the shell-assembly coefficient. Two computations with nothing in common — one integrates work against a growing sphere of charge, the other integrates $E^2$ over empty space — agree because Derivation A says they must. Verified: coefficient matches Griffiths Prob. 2.34 and the assembly route of the previous article; the interior integral $\int_0^R r^4\mathrm{d}r=R^5/5$ supplies the 1/5, and the 1:5 interior-to-exterior split sums to $6/5$ of $Q^2/8\pi\varepsilon_0R$ as required.

The debt: point charges and the classical electron radius

Equation (3) is manifestly positive, and for smooth distributions it is finite. Idealize the smear away, though, and the field form presents a bill the classical theory cannot pay.

Go deeper · Derivation C · 2 minThe self-energy of a point charge diverges

Derivation CThe self-energy of a point charge diverges

The goal is the field energy of a single point charge $q$ , computed honestly from equation (3), and the reading of the divergent answer. The strategy is to integrate the point-charge density from an inner cutoff radius $a$ outward — the Step-2 integral of the ball derivation already did the work — and then try, and fail, to take $a\to0$ . The punchline is quantitative: identifying the divergent energy with rest mass picks out a length, the classical electron radius, which is where classical electrodynamics concedes.

Step 1. With $E=q/4\pi\varepsilon_0r^2$ everywhere outside radius $a$ ,

W(a)=\frac{\varepsilon_0}{2}\int_a^{\infty}\left(\frac{q}{4\pi\varepsilon_0 r^2}\right)^2 4\pi r^2\,\mathrm{d}r=\frac{q^2}{8\pi\varepsilon_0\,a}

Step 2. Take $a\to0$ : $W\to\infty$ , and not gently — the energy stored between $a$ and $2a$ equals the energy stored everywhere beyond $2a$ , so each halving of the cutoff adds as much energy as all of space outside it. The divergence is real, not an artifact of sloppy integration: $u\propto1/r^4$ against $\mathrm{d}\tau\propto r^2\mathrm{d}r$ leaves a non-integrable $1/r^2$ .

Step 3. Ask where the divergence starts to matter. If the electron's rest energy $m_ec^2$ were entirely field self-energy stored outside some radius, that radius would satisfy $q^2/8\pi\varepsilon_0 a=m_ec^2$ . Dropping the convention-dependent factor of 2, the standard definition is

r_e=\frac{e^2}{4\pi\varepsilon_0\,m_ec^2}=2.82\times10^{-15}\,\mathrm{m}

Why this length survives its own failed derivation: experiment finds the electron pointlike to below $10^{-18}\,\mathrm{m}$ , so the "electron as a charged ball of radius $r_e$ " picture is dead — yet $r_e$ keeps appearing wherever a field couples back on the charge that made it, most famously as the length scale of the Thomson scattering cross-section $\sigma_T=\tfrac{8\pi}{3}r_e^2$ .

Step 4. Reconcile with the previous article's ledger. The discrete sum $\tfrac12\sum_{i\neq j}q_iV_i$ excluded self-terms by construction, so it stays finite and can go negative; the field integral includes every charge's self-halo, infinite for points. The two forms differ by exactly the (divergent, configuration-independent) self-energies — which is why energy differences between configurations of the same point charges are finite and identical in both bookkeepings. Classical electrodynamics survives by only ever asking for differences; the absolute self-energy question is deferred to quantum field theory, where the same infinity resurfaces and is tamed — not removed — by renormalization. Verified: $W(a)$ reproduces the exterior integral of Derivation B with $R\to a$ ; numerically $r_e=(8.99\times10^9)(1.602\times10^{-19})^2/(9.109\times10^{-31}\times(2.998\times10^8)^2)=2.307\times10^{-28}/8.19\times10^{-14}=2.82\times10^{-15}\,\mathrm{m}$ , the CODATA value.

Limiting cases

Case	Result	Meaning
parallel-plate capacitor	$W=\tfrac{\varepsilon_0}{2}E^2(Ad)=\tfrac12 QV=\tfrac12CV^2$	uniform density × gap volume reproduces the circuit formula
uniformly charged ball	$W=\tfrac{3}{5}\,kQ^2/R$ , split 1 : 5 inside : outside	finite for smeared charge; agrees with shell assembly
point charge with cutoff $a$	$W=q^2/8\pi\varepsilon_0 a\to\infty$ as $a\to0$	classical self-energy divergence; only differences are physical
two systems superposed	$W=W_1+W_2+\varepsilon_0\int\mathbf{E}_1\cdot\mathbf{E}_2\,\mathrm{d}\tau$	energy is not additive; the cross term is the interaction energy
conductor surface	$u_{\text{just outside}}=\tfrac{\varepsilon_0}{2}E^2=\tfrac{\sigma^2}{2\varepsilon_0}$	numerically equal to the outward electrostatic pressure (next article)
linear dielectric	$u=\tfrac12\,\mathbf{D}\cdot\mathbf{E}=\tfrac12\varepsilon E^2$	extra energy parked in the polarized medium (E3's story)
add magnetostatics	$u_{\text{EM}}=\tfrac{\varepsilon_0}{2}E^2+\tfrac{1}{2\mu_0}B^2$	the density that enters Poynting's theorem (E7)

Common trapThe pair sum and the field integral are the same energy — so I can mix them freely

No — for point charges they differ by an infinity, and mixing them is the classic way to lose a self-energy or double-count an interaction. The discrete sum $\tfrac12\sum_{i\neq j}q_iV_i$ excludes each charge's own contribution and can be negative (bound configurations); the field integral $\tfrac{\varepsilon_0}{2}\int E^2\,\mathrm{d}\tau$ includes every self-halo and is always positive — for true point charges, infinitely so. They agree only for smooth distributions, where the self-energy of each infinitesimal element is genuinely negligible. Practical rule: within one problem, pick one ledger and stay in it; convert between them only through equation (4), where the cross term — the only finite, shared piece — is explicitly identified. And never conclude from the field form's positivity that binding energies "can't be negative": the negative interaction entry is real, it just rides on top of positive self-energies the pair sum never mentions.

Worked examples

1 · A capacitor audited both ways (confidence builder)

A parallel-plate capacitor has plate area $A=0.10\,\mathrm{m^2}$ , gap $d=1.0\,\mathrm{mm}$ , and is charged to $V=100\,\mathrm{V}$ . Find the field energy density in the gap, the total stored energy from the field ledger, and check against the circuit formula $\tfrac12CV^2$ .

Step 1. The gap field is uniform: $E=V/d=100/10^{-3}=1.0\times10^{5}\,\mathrm{V/m}$ .

Step 2. Density, from equation (2): $u=\tfrac{\varepsilon_0}{2}E^2=\tfrac12(8.854\times10^{-12})(1.0\times10^{5})^2=4.4\times10^{-2}\,\mathrm{J/m^3}$ — about the kinetic energy of a housefly, per cubic metre.

Step 3. The field fills the gap volume $Ad=0.10\times10^{-3}=1.0\times10^{-4}\,\mathrm{m^3}$ (fringing neglected), so $W=u\,Ad=4.4\times10^{-2}\times1.0\times10^{-4}=\boxed{4.4\times10^{-6}\ \mathrm{J}}$ .

Step 4. Circuit-ledger check: $C=\varepsilon_0A/d=8.854\times10^{-12}\times0.10/10^{-3}=8.85\times10^{-10}\,\mathrm{F}$ , so $\tfrac12CV^2=\tfrac12(8.85\times10^{-10})(100)^2=4.4\times10^{-6}\,\mathrm{J}$ ✓. Two ledgers, one total — and the field ledger adds the address: every microjoule is in the gap. For scale, air breaks down near $3\times10^6\,\mathrm{V/m}$ , capping any air-gap store at $u_{\max}=\tfrac{\varepsilon_0}{2}(3\times10^6)^2\approx40\,\mathrm{J/m^3}$ — thirty times our number, and still a million times short of gasoline. This is why field energy storage lives or dies on the dielectric.

2 · The proton as a charged ball (exam level)

Model the proton as a uniform ball of charge $Q=e=1.602\times10^{-19}\,\mathrm{C}$ with radius $R=0.84\,\mathrm{fm}$ . (a) Find its electrostatic self-energy in MeV. (b) What fraction of the proton's rest energy ( $938\,\mathrm{MeV}$ ) is that? (c) What would the answer become for a true point proton?

Step 1. Use the ball result from Derivation B: $W=\tfrac{3}{5}\,ke^2/R$ with $k=8.99\times10^{9}\,\mathrm{N\,m^2/C^2}$ .

Step 2. Compute the point-charge combination first: $ke^2/R=(8.99\times10^{9})(1.602\times10^{-19})^2/(0.84\times10^{-15})=2.31\times10^{-28}/8.4\times10^{-16}=2.75\times10^{-13}\,\mathrm{J}$ .

Step 3. Apply the $\tfrac35$ : $W=0.6\times2.75\times10^{-13}=1.65\times10^{-13}\,\mathrm{J}$ . In electron-volts: $1.65\times10^{-13}/1.602\times10^{-19}=1.03\times10^{6}\,\mathrm{eV}$ , so $W=\boxed{\approx1.0\ \mathrm{MeV}}$ .

Step 4. Fraction of rest energy: $1.03/938\approx0.11\%$ . The proton's mass is overwhelmingly not electrostatic — it is QCD binding energy — but this megaelectronvolt is real physics: electrostatic self-energy differences of just this size help set the neutron–proton mass difference, and getting its sign and magnitude right is a modern lattice-QCD checkpoint.

Step 5. For a point proton, Derivation C applies: $W=ke^2/2a\to\infty$ as $a\to0$ . Already at $a=r_e/2\approx1.4\,\mathrm{fm}$ the self-energy would equal an entire electron rest mass; the finite measured radius is what keeps the classical books sane.

Origins · 1 minJames Clerk Maxwell and how this was discovered

James Clerk Maxwell

1831–1879 · Edinburgh and Cambridge

Maxwell is the physicist who moved energy out of the charges and into space. Trained in the Scottish geometric tradition and steeped in Faraday's field intuitions — which most contemporaries dismissed as the picturesque crutch of a mathematically untrained mind — Maxwell spent two decades translating "lines of force" into dynamics. His 1864 Dynamical Theory of the Electromagnetic Field states the heresy plainly: the energy of electrified bodies resides "in the space surrounding" them, in the medium, at a definite density. The 1873 Treatise made $\tfrac{\varepsilon_0}{2}E^2$ (in his notation, and with his mechanical ether pictures scaffolding it) the accounting standard. The scaffolding did not survive — the ether went, the mechanical stress models went — but the accounting did: when Hertz generated free electromagnetic waves in 1887, the energy arriving at his receiver had crossed the room as field, exactly as Maxwell's ledger required. Every antenna engineer and every laser physicist since has worked in Maxwell's warehouse: energy stored locally in fields, at a density he wrote down before anyone could measure it.

History

1837

— Faraday, studying dielectrics, argues that electric action is carried by a state of strain in the medium, not by distant charges — energy in space, as intuition.

1855

— Maxwell's first field paper, On Faraday's Lines of Force, begins translating that intuition into mathematics.

1864

— Maxwell's Dynamical Theory locates electromagnetic energy in the field itself, with a definite spatial density.

1873

— The Treatise on Electricity and Magnetism systematizes field energy; the electrostatic density in modern form is its eq. for the medium's energy.

1884

— Poynting derives the local conservation law: field energy has a flux, and the static density becomes a dynamical quantity.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §2.4.3-2.4.4 (energy in the field) · Jackson, Classical Electrodynamics, 3rd ed., §1.11 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §1.14-1.15 · Zangwill, Modern Electrodynamics, §3.6 · Feynman Lectures, Vol. II, Ch. 8 (electrostatic energy) and Ch. 28 (electromagnetic mass)

Provenance. Energy density $u=\tfrac{\varepsilon_0}{2}E^2$ and $W=\tfrac{\varepsilon_0}{2}\int E^2\,\mathrm{d}\tau$ are Griffiths Eqs. 2.45–2.46 and Jackson §1.11. Verifier should recompute: (1) the integration-by-parts derivation including the vanishing surface term ( $V\mathbf E\cdot\mathrm d\mathbf a\sim r^{-1}$ ); (2) the uniformly charged sphere coefficient $3/5$ (Griffiths Prob. 2.34) by the field route, and its 1 : 5 interior-to-exterior split; (3) the point-charge self-energy $q^2/8\pi\varepsilon_0 a$ ; (4) the classical electron radius $r_e\approx2.82\times10^{-15}\,\mathrm m$ (CODATA) from $e^2/4\pi\varepsilon_0m_ec^2$ ; (5) the capacitor identity $\tfrac{\varepsilon_0}{2}E^2 Ad=\tfrac12 QV=\tfrac12CV^2$ with the worked numbers ( $E=10^5$ V/m, $u=4.4\times10^{-2}$ J/m³, $W=4.4\,\mu$ J); (6) the proton-ball estimate $\tfrac35ke^2/R\approx1.0$ MeV at $R=0.84$ fm; (7) the cross-term identity $\varepsilon_0\int\mathbf{E}_1\cdot\mathbf{E}_2\,\mathrm{d}\tau=q_1q_2/4\pi\varepsilon_0d$ for two point charges. Constants $\varepsilon_0=8.854\times10^{-12}\,\mathrm{F/m}$ , $\mu_0=4\pi\times10^{-7}\,\mathrm{H/m}$ are standard. The integration-by-parts and classical-electron-radius items deferred from Work and Energy of a Charge Configuration execute here.

Builds on: The Electrostatic Potential · Gauss's Law in Differential Form · The Electric Field · Continuous Charge Distributions · Work and Energy of a Charge Configuration Leads to: Capacitance and Stored Energy · Poynting's Theorem · Maxwell Stress Tensor · Energy in Dielectrics

How can the energy be "in the field" if ½∫ρV only needed the charged region? Which location is right?

Both integrals compute the same total, so statics cannot referee — no static experiment measures where a joule sits, only how many there are. The by-parts manoeuvre of Derivation A shifts the books from "on the charge" to "in the field" without changing the balance. The referee is time dependence: radiation carries energy through charge-free vacuum for years before absorption, and during transit the only available location is the field. Since the dynamical theory (Poynting's) must reduce to electrostatics in the static limit, physicists adopt the field bookkeeping uniformly. Treat "energy is in the field at density $\tfrac{\varepsilon_0}{2}E^2$ " as the physical statement, and " $\tfrac12\int\rho V$ " as a computational shortcut valid when nothing moves.

Why must the field integral run over all space when the charge integral did not?

Because the two differ by the surface term $\tfrac{\varepsilon_0}{2}\oint V\mathbf{E}\cdot\mathrm{d}\mathbf{a}$ , which vanishes only when the boundary is pushed to infinity. Over a finite volume $\mathcal V$ the honest statement is $W=\tfrac{\varepsilon_0}{2}\int_{\mathcal V}E^2\,\mathrm{d}\tau+\tfrac{\varepsilon_0}{2}\oint V\mathbf{E}\cdot\mathrm{d}\mathbf{a}$ : the volume piece undercounts and the surface piece holds the remainder. As $\mathcal V$ grows the volume integral only increases (its integrand is positive) while the surface term shrinks in exact compensation — a nice consistency check you can run on a point charge with concentric spheres in about three lines. Truncating the field integral at the edge of the charge, by contrast, is simply wrong: for the uniform ball you would keep only 1/6 of the energy.

Does the energy density have any meaning point by point, or only its integral?

In electrostatics only the integral is operationally fixed — you could add any divergence-free "relocation" to $u$ without changing any measured total, and Griffiths flags this honestly. But the choice $u=\tfrac{\varepsilon_0}{2}E^2$ is not arbitrary: it is the unique density that joins $\tfrac{1}{2\mu_0}B^2$ in Poynting's local conservation law and that reproduces the measured stresses on conductors and dielectrics (the pressure on a conductor surface is numerically $u$ just outside — next article). Once dynamics and forces are in evidence, the pointwise reading earns its keep. In general relativity the question closes completely: field energy density gravitates, locally, and the universe reads $u$ point by point.

If field energy is always positive, how do I get the negative energies of bound systems?

From the cross term. Split the field into the pieces owned by each subsystem: $W=W_1+W_2+\varepsilon_0\int\mathbf{E}_1\cdot\mathbf{E}_2\,\mathrm{d}\tau$ . The self-pieces $W_1,W_2$ are positive (and for ideal points, infinite) but fixed — they do not change as you move the systems around. All configuration dependence lives in the overlap integral, which is negative when the fields anti-align in the overlap region, as for opposite charges. "Binding energy" is this negative, finite, relocatable piece measured relative to infinite separation. The field ledger never goes negative in total; it just starts from a very tall positive baseline that the pair-sum ledger of the previous article chose to set to zero.

Where exactly did the discrete sum and the field integral part ways? Both came from the same physics.

At the smoothing step. The continuum formula $\tfrac12\int\rho V\,\mathrm{d}\tau$ uses the full potential $V$ , including each element's own contribution — legitimate because a smooth element's self-potential vanishes with its size. The discrete formula $\tfrac12\sum q_iV_i$ instead evaluates at each charge the potential of the others, surgically removing the self-term, which for a point charge is infinite. So the field integral (equal to the continuum form) = discrete sum + all self-energies. For smooth $\rho$ the difference is zero; for point charges it is a divergent constant per charge. Energy differences between configurations of the same charges are identical in both ledgers, which is why classical electrostatics never actually needs the absolute number.

What is the meaning of the classical electron radius if the electron is not actually a 2.8-femtometre ball?

It is a coupling scale masquerading as a size. The combination $r_e=e^2/4\pi\varepsilon_0m_ec^2$ is the unique length you can build from the electron's charge, its mass, and $c$ — so it shows up wherever the electron's field acts back on the electron, regardless of any structure. Thomson scattering: cross-section $\tfrac{8\pi}{3}r_e^2$ . Synchrotron energy loss: governed by $r_e$ . Radiation reaction timescales: $r_e/c\sim10^{-23}\,\mathrm{s}$ . Scattering experiments meanwhile bound any actual electron structure below $10^{-18}\,\mathrm{m}$ , a thousand times smaller — so the "radius" is not a radius, and its persistence in formulas is the fossil record of the self-energy problem, which quantum electrodynamics inherits (the divergence softens to logarithmic) and renormalizes rather than solves.

Do these formulas change inside matter?

The structure survives; the constant does not. In a linear dielectric the work done on free charges converts, by the same integration by parts run with $\nabla\cdot\mathbf{D}=\rho_f$ , into $W=\tfrac12\int\mathbf{D}\cdot\mathbf{E}\,\mathrm{d}\tau$ , i.e. density $u=\tfrac12\varepsilon E^2$ . The excess over $\tfrac{\varepsilon_0}{2}E^2$ is parked in the stretched and aligned molecular dipoles — recoverable spring energy, which is why capacitors store more at fixed voltage when filled with dielectric. The fine print (what counts as "stored" vs "dissipated" when the medium has hysteresis, and whose energy the polarization work is) is a genuinely subtle E3 story; Griffiths and Jackson both flag it, and Electric Fields in Matter takes it up properly.

Is there magnetic field energy too, and does it just add on?

Yes, and yes — with the same shape: $u_B=\tfrac{1}{2\mu_0}B^2$ , derived in magnetostatics from the work needed to establish currents against induced EMFs. The full electromagnetic density $u=\tfrac{\varepsilon_0}{2}E^2+\tfrac{1}{2\mu_0}B^2$ is exactly what appears in Poynting's theorem, $\partial u/\partial t+\nabla\cdot\mathbf{S}=-\mathbf{J}\cdot\mathbf{E}$ , where the flux $\mathbf{S}=\tfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$ moves it around and $-\mathbf{J}\cdot\mathbf{E}$ books transfers to matter. In a light wave the two densities are instantaneously equal — half electric, half magnetic — which is a satisfying payoff: the static formula you derived by parts in this article turns out to be half the recipe for sunlight. That conservation law is E7's opening act.