Energy Stored in the Electric Field

Derivation

Energy Stored in the Electric Field

D-286 Home PU-305 Threads energy · fields Depends on Electrostatic Potential and Irrotational E, Gauss's Law from Coulomb's Law

Statement

For any static charge distribution in vacuum, the total work done in assembling the charges equals the volume integral, taken over all space, of the electrostatic energy density \(u = \tfrac{1}{2}\varepsilon_0 E^2\): \[ W \;=\; \frac{\varepsilon_0}{2}\int_{\text{all space}} \left|\vec{E}(\vec{r})\right|^2 \, d^3r . \] The two equivalent forms \(W=\tfrac12\int \rho\,\phi\, d^3r\) and \(W=\tfrac{\varepsilon_0}{2}\int E^2\, d^3r\) assign the same total, but the second attributes the energy to the field rather than to the charges.

Why it matters

This result is the electrostatic ancestor of field energy in general. Once energy is written as \(\int u\, d^3r\) with a local density \(u(\vec r)\), it lives in the field itself, everywhere the field is non-zero, not "in the charges." That reinterpretation is essential for anything that carries energy without carrying charge: radiation, the Poynting flux, and the electromagnetic stress-energy tensor all descend from it.

It also supplies the practical engine behind capacitors, field-energy methods for forces (\(\vec F = -\nabla U\) applied to field energy), and the self-energy divergence that first exposed the trouble with point charges in classical electromagnetism.

Assumptions

Electrostatics (no time dependence).If \(\partial_t \neq 0\) the energy stored is not \(\tfrac12\varepsilon_0 E^2\) alone; a magnetic term \(\tfrac{1}{2\mu_0}B^2\) appears and energy flows via the Poynting vector, so "work to assemble" no longer equals the instantaneous field energy.

Linear vacuum (no dielectric).If a medium responds, \(\vec D = \varepsilon\vec E \neq \varepsilon_0\vec E\) and the density becomes \(\tfrac12\vec E\cdot\vec D\); with nonlinear or hysteretic media even that fails and one must integrate \(\int \vec E\cdot d\vec D\).

Fields fall off fast enough at infinity.For any bounded distribution \(\phi\sim 1/r\), \(E\sim1/r^2\), so the surface term in the integration by parts vanishes as the bounding sphere \(R\to\infty\). Drop this and the two forms of \(W\) differ by a surface contribution.

The charge distribution is well behaved (integrable \(\rho\), potential defined).Idealised point charges give a divergent self-energy \(\int E^2\,d^3r=\infty\); the finite continuum result quietly excludes each element's interaction with itself, whereas the field form does not.

Derivation

\[ W \;=\; \frac{1}{2}\int_{\text{all space}} \rho(\vec r)\,\phi(\vec r)\, d^3r \]

Starting point: the work to assemble a continuous charge distribution, obtained by bringing in charge in infinitesimal layers; the factor \(\tfrac12\) prevents double-counting each pair of elements. A

\[ \rho \;=\; \varepsilon_0\,\nabla\cdot\vec E \]

Gauss's law in differential form (prior result, from Coulomb's law). This trades the source \(\rho\) for the field it produces. A

\[ W \;=\; \frac{\varepsilon_0}{2}\int \left(\nabla\cdot\vec E\right)\phi \, d^3r \]

Substitute step 2 into step 1. Pure algebra; nothing yet integrated. A

\[ \left(\nabla\cdot\vec E\right)\phi \;=\; \nabla\cdot\!\left(\phi\,\vec E\right) - \vec E\cdot\nabla\phi \]

Product rule for the divergence, \(\nabla\cdot(\phi\vec E)=\phi\,\nabla\cdot\vec E+\vec E\cdot\nabla\phi\), rearranged. This is the electrostatic analogue of integration by parts. B

\[ W \;=\; \frac{\varepsilon_0}{2}\int \nabla\cdot\!\left(\phi\,\vec E\right) d^3r \;-\; \frac{\varepsilon_0}{2}\int \vec E\cdot\nabla\phi \, d^3r \]

Insert the identity from step 4 and split the integral by linearity. A

\[ \frac{\varepsilon_0}{2}\int_V \nabla\cdot\!\left(\phi\,\vec E\right) d^3r \;=\; \frac{\varepsilon_0}{2}\oint_{\partial V} \phi\,\vec E\cdot d\vec a \;\xrightarrow{\;R\to\infty\;}\;0 \]

Divergence theorem converts the first term to a surface integral over a sphere of radius \(R\). For a bounded source \(\phi\sim1/R\), \(E\sim1/R^2\), while the area grows as \(R^2\); the integrand falls as \(1/R\) and the term vanishes as \(R\to\infty\). B

\[ \vec E \;=\; -\nabla\phi \]

The electrostatic field is curl-free, so it derives from a scalar potential (prior result). This lets the remaining \(\nabla\phi\) become \(-\vec E\). A

\[ W \;=\; -\frac{\varepsilon_0}{2}\int \vec E\cdot\nabla\phi \, d^3r \;=\; +\frac{\varepsilon_0}{2}\int \vec E\cdot\vec E \, d^3r \]

Drop the vanished surface term (step 6) and substitute \(\nabla\phi=-\vec E\) (step 7). The two minus signs cancel. A

Result

\[ W \;=\; \frac{\varepsilon_0}{2}\int_{\text{all space}} E^2 \, d^3r , \qquad u(\vec r) \;=\; \frac{1}{2}\varepsilon_0 E^2 \]

Reading. The energy it took to build the charge configuration is stored, with density \(u=\tfrac12\varepsilon_0 E^2\), everywhere the electric field is non-zero. Wherever \(\vec E=0\) there is no stored energy; wherever the field is strong, energy is concentrated. The integral runs over all of space — including the vacuum between and around the charges, not just where \(\rho\neq0\).

Units check. \([\varepsilon_0]=\mathrm{C^2\,N^{-1}\,m^{-2}}\) and \([E^2]=\mathrm{(N/C)^2=N^2\,C^{-2}}\), so \([\varepsilon_0 E^2]=\mathrm{N\,m^{-2}=J\,m^{-3}}\), an energy density. Multiplying by \(d^3r\;(\mathrm{m^3})\) gives joules. Equivalently \([\varepsilon_0 E^2]=\mathrm{(F/m)(V/m)^2}=\mathrm{F\,V^2\,m^{-3}}=\mathrm{J\,m^{-3}}\). Correct.

Limiting cases

Uniform field (parallel-plate capacitor). \(E\) constant in a volume \(Ad\): \(W=\tfrac12\varepsilon_0 E^2 (Ad)\). With \(E=V/d\) and \(C=\varepsilon_0 A/d\) this reduces to \(W=\tfrac12 CV^2\).
Single point charge, energy outside radius \(a\). \(E=q/(4\pi\varepsilon_0 r^2)\) gives \(W(r>a)=q^2/(8\pi\varepsilon_0 a)\) — finite once the singular core is excluded.
Two well-separated charges. Field energy splits into two self-energies plus the interaction term \(q_1q_2/(4\pi\varepsilon_0 d)\); only the interaction part changes when you move them.
Weak-field limit. \(u\propto E^2\), so halving the field quarters the local energy density — the density is quadratic, never linear, in the field.

Breaks when

Time-varying fields. With \(\partial_t\vec B\neq0\) the electric field is no longer \(-\nabla\phi\) (it acquires a solenoidal part), step 7 fails, and the correct energy density is \(u=\tfrac12\varepsilon_0 E^2+\tfrac{1}{2\mu_0}B^2\) with energy transported by \(\vec S=\vec E\times\vec H\).
Inside linear dielectrics. Bound charge makes \(\rho\) in step 2 the total charge, not the free charge; the useful stored energy is \(u=\tfrac12\vec E\cdot\vec D\), and for nonlinear/ferroelectric media even that must be replaced by \(\int\vec E\cdot d\vec D\).
Idealised point charges. \(\int E^2 d^3r\) diverges at each point charge (self-energy \(\sim1/a\to\infty\)), so the field form assigns infinite energy where the discrete form \(\tfrac12\sum_{i\neq j}\) is finite — the two disagree by the omitted self-energies.
Unbounded distributions. If charge extends to infinity (e.g. an infinite charged plane or line), \(\phi\) does not fall off as \(1/r\), the step-6 surface term need not vanish, and \(W\) itself may be infinite or ill-defined.

Failure modes

Dropping the \(\tfrac12\). Writing \(W=\int\rho\phi\,d^3r\) double-counts every pair; the missing factor of two is the single most common error.
Integrating only where \(\rho\neq0\). The field form requires integrating \(E^2\) over all space; students often restrict to the charged region and miss most of the energy.
Equating the two forms for point charges. \(\tfrac12\int\rho\phi\) (excludes self-energy) and \(\tfrac{\varepsilon_0}{2}\int E^2\) (includes it) are equal only for smooth continuous \(\rho\); asserting equality for discrete charges is wrong.
Squaring components wrongly. \(E^2=\vec E\cdot\vec E=E_x^2+E_y^2+E_z^2\), not \((E_x+E_y+E_z)^2\); superpose fields as vectors, then square.
Keeping the surface term. Forgetting that \(\oint\phi\vec E\cdot d\vec a\to0\) only for bounded sources, and either dropping it when it survives (infinite plane) or keeping it when it vanishes.
Using \(\tfrac12\varepsilon_0 E^2\) inside a dielectric. The vacuum density understates the stored energy; the correct density there is \(\tfrac12\vec E\cdot\vec D=\tfrac12\varepsilon E^2\).

Discussion

The heart of this derivation is a reinterpretation, not a new physical law. We began with energy attributed to charges through \(\tfrac12\int\rho\phi\), and by trading \(\rho\) for \(\varepsilon_0\nabla\cdot\vec E\) and integrating by parts, we relocated it into a density spread through the field. Both bookkeeping schemes give the identical number for the total, so no experiment on total energy can distinguish them — yet the field picture is the one that survives into radiation and relativity, where energy demonstrably travels through regions containing no charge at all.

Notice how the assumptions do real work. The surface term vanishes only because bounded charges produce potentials that decay; the sign works out only because \(\vec E=-\nabla\phi\), which itself requires \(\nabla\times\vec E=0\), i.e. electrostatics. Change either premise and the clean local density \(\tfrac12\varepsilon_0 E^2\) is no longer the whole story — a foretaste of the full electromagnetic energy density and the Poynting theorem, which are the dynamical completion of this static result.

The point-charge self-energy divergence is not a computational slip but a structural feature: the field form \(\tfrac{\varepsilon_0}{2}\int E^2\) counts each element's interaction with itself, whereas \(\tfrac12\sum_{i\neq j}q_iq_j/(4\pi\varepsilon_0 r_{ij})\) does not. For a continuous \(\rho\) the two coincide because the self-energy of each infinitesimal element vanishes in the limit; for a genuine point charge it does not, giving \(\infty\). This is the classical seed of the electron self-energy problem, resolved only by regularisation and, ultimately, quantum field theory. It is also why the field energy is always non-negative (\(E^2\geq0\)) while the interaction energy \(\tfrac12\int\rho\phi\) can be negative for opposite charges — the difference is exactly the omitted, always-positive self-energies.

Common misconceptions. "The energy is in the charges" — no; the total is the same either way, but the localisation that matches dynamics puts it in the field. "The vacuum stores nothing" — the region between capacitor plates, where \(\rho=0\), is precisely where nearly all the energy resides. And \(u\propto E^2\) means energy density is not additive in the field: superposing two fields adds a cross term \(\varepsilon_0\vec E_1\cdot\vec E_2\), which is the interaction energy between the two sources.

Worked examples

Example 1 — Energy stored in a parallel-plate capacitor via the field.

\[ E \;=\; \frac{V}{d}, \qquad u \;=\; \frac{1}{2}\varepsilon_0 E^2 \]

Uniform field between the plates; energy density is the local result. A

\[ W \;=\; u\,(A\,d) \;=\; \frac{1}{2}\varepsilon_0 \frac{V^2}{d^2}\,A\,d \;=\; \frac{1}{2}\frac{\varepsilon_0 A}{d}V^2 \]

Integrate a constant density over the gap volume \(Ad\); recognise \(C=\varepsilon_0 A/d\). A

\[ A = 0.10~\mathrm{m^2},\; d = 1.0~\mathrm{mm},\; V = 100~\mathrm{V} \]\[ E = \frac{100}{1.0\times10^{-3}} = 1.0\times10^{5}~\mathrm{V/m} \]

Insert numbers only now that the symbolic form is fixed. A

\[ u = \tfrac12(8.85\times10^{-12})(1.0\times10^{5})^2 = 4.4\times10^{-2}~\mathrm{J/m^3} \]\[ W = u\,(A d) = (4.4\times10^{-2})(0.10\times1.0\times10^{-3}) \]

Evaluate density, then multiply by the gap volume \(1.0\times10^{-4}~\mathrm{m^3}\). A

\[ W \approx 4.4\times10^{-6}~\mathrm{J} = 4.4~\mu\mathrm{J} \]

Reading. A cross-check via \(W=\tfrac12 CV^2\) with \(C=\varepsilon_0A/d=8.85\times10^{-10}~\mathrm F\) gives \(\tfrac12(8.85\times10^{-10})(100)^2=4.4~\mu\mathrm J\) — identical, confirming the field integral.

Units check. \(\mathrm{(J/m^3)\cdot m^3 = J}\). Correct.

Example 2 — Self-energy of a uniformly charged sphere.

\[ E_{\text{out}}=\frac{Q}{4\pi\varepsilon_0 r^2}\;(r>R), \qquad E_{\text{in}}=\frac{Q\,r}{4\pi\varepsilon_0 R^3}\;(r<R) \]

Fields of a uniform ball of total charge \(Q\), radius \(R\), from Gauss's law. A

\[ W = \frac{\varepsilon_0}{2}\int_0^\infty E^2\,4\pi r^2\,dr = \frac{\varepsilon_0}{2}\!\left[\int_0^R E_{\text{in}}^2 4\pi r^2 dr + \int_R^\infty E_{\text{out}}^2 4\pi r^2 dr\right] \]

Spherical symmetry reduces \(d^3r\) to \(4\pi r^2 dr\); split at \(R\). B

\[ \int_R^\infty E_{\text{out}}^2\,4\pi r^2 dr = \frac{4\pi Q^2}{(4\pi\varepsilon_0)^2}\cdot\frac{1}{R}, \qquad \int_0^R E_{\text{in}}^2\,4\pi r^2 dr = \frac{4\pi Q^2}{(4\pi\varepsilon_0)^2}\cdot\frac{1}{5R} \]

Two elementary radial integrals; \(\int_R^\infty r^{-2}dr=1/R\) and \(\int_0^R r^4\,dr=R^5/5\). B

\[ W = \frac{\varepsilon_0}{2}\cdot\frac{4\pi Q^2}{(4\pi\varepsilon_0)^2}\!\left(\frac{1}{R}+\frac{1}{5R}\right) = \frac{Q^2}{8\pi\varepsilon_0}\cdot\frac{6}{5R} = \frac{3}{5}\frac{Q^2}{4\pi\varepsilon_0 R} \]

Combine the prefactors, \(\tfrac{\varepsilon_0}{2}\cdot\tfrac{4\pi}{(4\pi\varepsilon_0)^2}=\tfrac{1}{8\pi\varepsilon_0}\), and add the bracket \(1+\tfrac15=\tfrac65\). B

\[ Q = 1.0\times10^{-9}~\mathrm C,\quad R = 2.0~\mathrm{cm}=2.0\times10^{-2}~\mathrm m \]\[ W=\frac{3}{5}\,(8.99\times10^{9})\frac{(1.0\times10^{-9})^2}{2.0\times10^{-2}} \]

Numbers last, with \(1/(4\pi\varepsilon_0)=8.99\times10^9~\mathrm{N\,m^2\,C^{-2}}\). A

\[ W = \frac{3}{5}\frac{Q^2}{4\pi\varepsilon_0 R} \approx 2.7\times10^{-7}~\mathrm{J} \]

Reading. The \(3/5\) is the signature of a uniform ball; a surface-charged shell of the same \(Q,R\) stores less, \(W=\tfrac12 Q^2/(4\pi\varepsilon_0 R)\), because it lacks the interior field.

Units check. \(\mathrm{(N\,m^2\,C^{-2})(C^2/m)=N\,m=J}\). Correct.

Problems

A parallel-plate capacitor with plate area \(A=200~\mathrm{cm^2}\) and gap \(d=0.50~\mathrm{mm}\) is held at \(V=50~\mathrm V\). Find the energy density between the plates and the total stored energy.
Solution
\(E=V/d=50/(5.0\times10^{-4})=1.0\times10^{5}~\mathrm{V/m}\). \(u=\tfrac12\varepsilon_0 E^2=\tfrac12(8.85\times10^{-12})(1.0\times10^{5})^2=4.4\times10^{-2}~\mathrm{J/m^3}\). Volume \(=Ad=(2.0\times10^{-2}~\mathrm{m^2})(5.0\times10^{-4}~\mathrm m)=1.0\times10^{-5}~\mathrm{m^3}\). \(W=u\cdot Ad=4.4\times10^{-7}~\mathrm J=0.44~\mu\mathrm J\). Check: \(C=\varepsilon_0A/d=3.54\times10^{-10}~\mathrm F\), \(\tfrac12CV^2=4.4\times10^{-7}~\mathrm J\). Agree.
Show that the field energy stored outside radius \(r=a\) around a point charge \(q\) is \(W=q^2/(8\pi\varepsilon_0 a)\), and evaluate it for \(q=1.6\times10^{-19}~\mathrm C\) and the classical electron radius \(a=2.8\times10^{-15}~\mathrm m\).
Solution
\(W=\tfrac{\varepsilon_0}{2}\int_a^\infty\!\left(\tfrac{q}{4\pi\varepsilon_0 r^2}\right)^2 4\pi r^2 dr=\tfrac{q^2}{8\pi\varepsilon_0}\int_a^\infty r^{-2}dr=\tfrac{q^2}{8\pi\varepsilon_0 a}\). Numerically \(W=\tfrac12(8.99\times10^9)\tfrac{(1.6\times10^{-19})^2}{2.8\times10^{-15}}=\tfrac12(8.99\times10^9)(9.14\times10^{-24})\approx4.1\times10^{-14}~\mathrm J\approx0.26~\mathrm{MeV}\). Setting this equal to \(m_ec^2\) is how the classical radius is defined (up to a factor of order one).
A spherical shell of radius \(R\) carries total charge \(Q\) uniformly on its surface. Using the field form, find the stored energy and compare with the uniform ball's \(\tfrac35 Q^2/(4\pi\varepsilon_0R)\).
Solution
Inside the shell \(E=0\); outside \(E=Q/(4\pi\varepsilon_0 r^2)\). So \(W=\tfrac{\varepsilon_0}{2}\int_R^\infty E^2 4\pi r^2 dr=\tfrac{q^2}{8\pi\varepsilon_0}\int_R^\infty r^{-2}dr=\tfrac{Q^2}{8\pi\varepsilon_0 R}=\tfrac12\tfrac{Q^2}{4\pi\varepsilon_0R}\). The shell stores \(\tfrac12\) versus the ball's \(\tfrac35\); the ratio is \(5/6\), the deficit being the interior field energy the ball has and the shell lacks.
Two point charges \(+q\) and \(-q\) are separated by distance \(d\). Explain, using \(u=\tfrac12\varepsilon_0(\vec E_1+\vec E_2)^2\), why the change in field energy on moving them equals the interaction energy \(-q^2/(4\pi\varepsilon_0 d)\), and identify which term is responsible.
Solution
Expand \(u=\tfrac12\varepsilon_0(E_1^2+E_2^2+2\vec E_1\cdot\vec E_2)\). The \(E_1^2\) and \(E_2^2\) terms are the self-energies: each depends only on its own charge, is independent of \(d\), and is (formally) infinite for point charges. Only the cross term \(\varepsilon_0\int\vec E_1\cdot\vec E_2\,d^3r\) depends on the separation. Carrying out that integral (a standard result) gives \(q_1q_2/(4\pi\varepsilon_0 d)=-q^2/(4\pi\varepsilon_0 d)\). Thus although the self-energies diverge, the difference in total field energy between two separations is finite and equals the interaction energy — the physically meaningful, always-finite part.
A sphere of radius \(R=5.0~\mathrm{cm}\) carries charge distributed with volume density \(\rho(r)=\rho_0(1-r/R)\) for \(r\le R\), with total charge \(Q=2.0~\mathrm{nC}\). Set up (do not fully grind) the field-energy integral and estimate its order of magnitude against the uniform-ball value.
Solution
First relate \(\rho_0\) to \(Q\): \(Q=\int_0^R\rho_0(1-r/R)4\pi r^2 dr=4\pi\rho_0(R^3/3-R^3/4)=\tfrac{\pi\rho_0 R^3}{3}\), so \(\rho_0=3Q/(\pi R^3)\). The enclosed charge \(Q(r)=4\pi\rho_0(r^3/3-r^4/(4R))\) gives \(E(r)=Q(r)/(4\pi\varepsilon_0 r^2)\) inside and \(Q/(4\pi\varepsilon_0 r^2)\) outside. Then \(W=\tfrac{\varepsilon_0}{2}\int_0^\infty E^2 4\pi r^2 dr\). The exterior contributes \(\tfrac12 Q^2/(4\pi\varepsilon_0 R)\) exactly as before; the interior gives a numerical coefficient between the shell (\(0\)) and uniform ball (\(\tfrac{1}{10}\) of \(Q^2/4\pi\varepsilon_0R\)) — since charge is concentrated toward the centre, expect slightly more interior energy than uniform. Order of magnitude: \(W\sim\tfrac35\cdot\tfrac{Q^2}{4\pi\varepsilon_0 R}=\tfrac35(8.99\times10^9)\tfrac{(2.0\times10^{-9})^2}{5.0\times10^{-2}}\approx4.3\times10^{-7}~\mathrm J\), i.e. a few hundred nanojoules.