Electric Field Energy Density
On this page
Direct answer
Wherever an electric field exists, energy is stored at a density u = (1/2)ε0E² joules per cubic metre, and the total energy of any charge configuration is the integral of this density over all space. For a charged conducting sphere of radius R the entire energy lives outside, giving the celebrated self-energy U = Q²/(8πε0R); a uniformly charged non-conducting sphere of the same Q and R stores more, U = 3Q²/(20πε0R), because field also fills the interior (one-sixth of the total sits inside). The same density idea gives the force per unit area on a capacitor plate, and merging n identical charged drops into one big drop multiplies both the potential and the electrostatic energy by n^(2/3) — the JEE Advanced favourite that fuses surface tension thinking with field energy.
What you must remember
- Density formula: u = (1/2)ε0E² = σ²/(2ε0) at a conductor surface with charge density σ; equivalently u = (1/2) × (charge × potential) spread over volume.
- Conducting sphere (shell): U = Q²/(8πε0R) = (1/2)kQ²/R, all of it in the field outside since the interior field is zero; the potential at the surface is kQ/R and U = (1/2)QV.
- Uniformly charged solid sphere: U = 3Q²/(20πε0R); the outside portion alone equals the shell value, and the extra Q²/(40πε0R) — one-sixth of the total — hides inside.
- Parallel-plate capacitor: u = (1/2)ε0E² between plates; total energy (1/2)CV²; force attracting each plate F = Q²/(2ε0A), the "one-half factor" force that students perpetually drop.
- Battery-connected versus isolated capacitor: on disconnecting the battery, Q is frozen and halving d (or inserting a dielectric) changes energy as Q²/2C; with the battery connected, V is fixed and energy (1/2)CV² moves oppositely — the standard two-case drill.
- Drop arithmetic: n identical drops (charge q, radius r each) merging into one gives Q = nq and R = n^(1/3)r, so the big drop's potential and self-energy are n^(2/3) times the sum of the parts; conversely, splitting one drop into n raises the total field energy by the factor n^(1/3).
- Energy density is always positive: fields carry energy even in charge-free regions, which is the physical reason field lines "repel" each other and why the midplane between like charges stores real energy.
One sphere, three accountings
Charge a conducting sphere to Q = 1 microcoulomb on R = 10 cm. Its self-energy is U = kQ²/(2R) = 9 × 10⁹ × 10⁻¹²/(0.2) = 0.045 J. Now spread the same charge uniformly through a solid insulating sphere of the same radius: U = 3Q²/(20πε0R) = (3/5) × kQ²/R = 0.108 J — larger, because the interior now holds field E(r) = kQr/R³ growing linearly outward, contributing exactly one-sixth of the total. The comparison itself is a JEE multiple-correct staple: the conductor sits at lower energy because charge migrates to the surface, and the difference (0.063 J here) is the energy released by letting the charge rearrange.
The drop problem closes the loop. Coalesce n identical mercury drops, each carrying charge q on radius r, into a single large drop: volume conservation forces R = n^(1/3)r while charge conservation gives Q = nq. The big drop's potential is k nq/(n^(1/3)r) = n^(2/3) × kq/r — twenty-seven drops at 1 volt each merge into one drop at 27^(2/3) = 9 volts. Its self-energy k n²q²/(2 n^(1/3)r) = n^(5/3) times a single small drop's energy, which against the pre-merger total of n small drops is again a factor n^(2/3) higher: merging costs field energy, paid for by the surface energy the drops shed as their combined surface shrinks. Run the film backwards — one drop bursting into n — and the total field energy still rises, but only by n^(1/3), because each daughter starts from the big drop's already-concentrated charge.
Where students slip
The half-factor in u = (1/2)ε0E² and in the plate force F = Q²/(2ε0A) is the most-dropped mark in electrostatics — the ½ survives because a plate does work against a field that builds continuously as charge arrives, not against the final value. The second slip is quoting Q²/(8πε0R) for a uniformly charged solid sphere; the conductor result assumes zero interior field. In energy-change questions, students mix the two capacitor regimes: battery connected fixes V (energy ∝ C), battery removed fixes Q (energy ∝ 1/C) — state which quantity is conserved before writing any energy expression.
Frequently asked questions
What is the energy density of an electric field in vacuum?
u = (1/2)ε0E², in joules per cubic metre, obtained by dividing the energy of a parallel-plate capacitor by the volume between its plates.
Why do conducting and uniformly charged spheres have different self-energies?
A conductor confines all field outside, giving Q²/(8πε0R); a uniform solid sphere also has field inside, adding a sixth more for a total of 3Q²/(20πε0R).
What force do the plates of a charged capacitor exert on each other?
Each plate feels F = Q²/(2ε0A) = (1/2)QE, attraction — the factor one-half appears because a plate sits in the field of the other plate only, not its own.
How does capacitor energy change when the battery stays connected versus disconnected?
Connected, V is fixed and energy (1/2)CV² tracks C; disconnected, Q is fixed and energy Q²/2C moves inversely with C — the two regimes change in opposite directions.
Where is the energy of a charge configuration actually stored?
In the field itself, at density (1/2)ε0E² over all space — including regions far from any charge, which is why field energy is the cleaner physical bookkeeping than "potential of charges".