Energy Density of the E Field

Chapter 1 — Electrostatics

Where does the energy of a charge configuration "live"? You can think of it as bookkeeping among the charges, or — more profoundly — as energy stored in the electric field itself. Both views give the same total, but the field-density picture is the one that survives into the dynamic theory where the field propagates as light.

The formula

The electrostatic energy of any charge configuration equals

\[U = \int u\, dV,\qquad u = \frac{\epsilon_0}{2}|\vec E|^2,\]

integrated over all of space. Units of \(u\): J/m\(^3\). This is the energy density of the field.

Parallel-plate motivator

Two large parallel plates have charge \(\pm Q\), area \(A\), surface density \(\sigma = Q/A\). Field between them: \(E = \sigma/\epsilon_0\). Separating them by \(\Delta x\) against their mutual attraction takes work

\[\Delta W = F\,\Delta x = (QE_{\text{on plate}})\,\Delta x = \frac{Q^2}{2\epsilon_0 A}\Delta x = \frac{\epsilon_0}{2}E^2\cdot A\Delta x.\]

But \(A\Delta x\) is the new volume of \(\vec E\) field you created. Energy per unit of new field volume: \(u = \tfrac12 \epsilon_0 E^2\). The factor of \(1/2\) (vs. naive thinking) comes from the fact that the force on one plate is due only to the field from the other plate, which is half the total field between them — see the conductor surface field discussion.

Equivalence to the "charge" formula

For a smooth distribution, you can rewrite

\[U = \tfrac12 \int \rho V\, dV = \tfrac12 \int \epsilon_0(\nabla\cdot\vec E)V\,dV,\]

and integrate by parts (using \(\nabla\cdot(V\vec E) = V\nabla\cdot\vec E + \vec E\cdot\nabla V = V\nabla\cdot\vec E - |\vec E|^2\)) to get the all-space integral of \(\tfrac12 \epsilon_0|\vec E|^2\). The boundary term vanishes if fields fall off fast enough at infinity. You'll do this cleanly once you learn the divergence and divergence theorem in Chapter 2.

Point-charge caveat

Apply \(u = \tfrac12\epsilon_0 E^2\) to a single point charge: \(E = kq/r^2\), integrate over all space using \(dV = 4\pi r^2\,dr\):

\[U = \int_0^\infty \tfrac12 \epsilon_0\left(\frac{kq}{r^2}\right)^2 4\pi r^2\,dr \propto \int_0^\infty \frac{dr}{r^2},\]

which diverges at \(r \to 0\). A point charge has infinite self-energy in this picture. This isn't a bug — it's absorbed into the particle's rest mass and doesn't participate in any interaction energy differences, which are what we actually observe.

Sanity check: uniform sphere

Compute the field energy of a uniform ball of radius \(R\), charge \(Q\):

\[U = \int_0^R \tfrac12 \epsilon_0\left(\frac{kQr}{R^3}\right)^2 4\pi r^2\,dr + \int_R^\infty \tfrac12\epsilon_0\left(\frac{kQ}{r^2}\right)^2 4\pi r^2\,dr.\]

Inside: \(\tfrac{kQ^2}{10R}\). Outside: \(\tfrac{kQ^2}{2R}\). Total: \(\tfrac{3}{5}\tfrac{kQ^2}{R}\) — the same self-energy we got from the shell-by-shell approach.

Practice Problems