Energy Stored in Capacitors

Start with the capacitor empty. To move another dq from one plate to the other when charge q is already in place, you must carry it across the potential difference the plates already have, q/C, so the work is dW = (q/C) dq. Charging creates no charge; it moves charge from one plate to the other, and the device stays neutral overall. Integrate from empty to the final charge: U = ∫₀^Q (q/C) dq = Q²/(2C). The factor of one half is the only thing here worth committing to memory, and it is not mysterious. Plot ΔV = q/C against q and you get a straight line through the origin, so the work — the area beneath it — is the area of a triangle, ½ × Q × (Q/C). The first coulomb crossed nothing; the last crossed the full final voltage. For a 10 μF capacitor taken to 12 V the final charge is Q = CΔV = 1.2×10⁻⁴ C, and U = (1.2×10⁻⁴ C)²/(2 × 1.0×10⁻⁵ F) = 7.2×10⁻⁴ J, or 0.72 mJ.

Substituting Q = CΔV turns one result into three: U = Q²/(2C) = ½QΔV = ½C(ΔV)². They are algebraically identical, so any of them returns the right number for a capacitor sitting still. They stop being interchangeable the moment something changes, because a real circuit holds one variable fixed and lets the other move. A capacitor still wired to a battery has ΔV pinned, so ½C(ΔV)² is the form whose supposedly fixed quantity really is fixed. A capacitor that has been disconnected has Q pinned instead, and Q²/(2C) is the honest form there. The middle form, ½QΔV, is the one that keeps the accounting visible, since QΔV is exactly what a constant-voltage source pays out. Note the powers while you are here: at fixed capacitance U goes as (ΔV)², so running a capacitor at twice the voltage stores four times the energy. That is why the voltage rating, rather than the capacitance, usually decides how much energy a bank can hold.

An ideal source held at ΔV does not care how much charge has already moved: every coulomb it pushes round the circuit is lifted through the full ΔV. Deliver Q in total and it has done W = QΔV = C(ΔV)², while the capacitor keeps ½C(ΔV)². Half of what the source spent is missing, and better wiring does not recover it. Put a resistance R in series and check. The current is I = (ΔV/R)e^(−t/RC), so the resistor dissipates ∫₀^∞ I²R dt = (ΔV²/R) × (RC/2) = ½C(ΔV)². R has cancelled: halve it and the peak current doubles while the transient lasts half as long, leaving the total dissipation untouched. Drive R towards zero and the energy leaves as radiation or heats the source's own internal resistance instead. In the 10 μF example the 12 V battery spends QΔV = 1.44 mJ, the capacitor keeps 0.72 mJ, and 0.72 mJ is gone. The only escape is to stop dropping the charge through the whole ΔV at once — charge in voltage steps, or through an inductor.

Where does the stored energy sit? Write ½C(ΔV)² for a parallel-plate capacitor with a vacuum gap, using C = ε₀A/d and ΔV = Ed for the uniform field between the plates: U = ½(ε₀A/d)(Ed)² = ½ε₀E² × (Ad). Both plate quantities have cancelled from the coefficient, leaving the field strength squared multiplied by Ad — precisely the volume the field occupies. So define an energy density u = ½ε₀E², in J m⁻³. Take A = 0.20 m², d = 0.50 mm and E = 1.0×10⁶ V m⁻¹, giving ΔV = Ed = 500 V and C = ε₀A/d = 3.54 nF. Then ½C(ΔV)² = 4.43×10⁻⁴ J, and u = ½(8.854×10⁻¹²)(1.0×10⁶)² = 4.43 J m⁻³ spread over 1.0×10⁻⁴ m³ gives the same 4.43×10⁻⁴ J. The local statement is the more general one, because it still applies where E varies from point to point: U = ∫ ½ε₀E² dV. Electrostatics alone cannot tell the two pictures apart by measurement, but the field version is the one that survives into radiation, where energy travels with no charge to sit on.

Before choosing a formula, ask which quantity the circuit is holding. Take the 10 μF capacitor at 12 V — Q = 120 μC, U = 0.72 mJ — and double the plate separation, halving C to 5.0 μF. Disconnect it first and Q stays at 120 μC: ΔV climbs to Q/C = 24 V and U = Q²/(2C) = 1.44 mJ. The energy doubled, and it came from your hand, since the plates attract and pulling them apart takes work. Leave the battery connected and ΔV is pinned at 12 V instead: the charge falls to 60 μC and U = ½C(ΔV)² = 0.36 mJ, so the capacitor's energy halves. It has not vanished — 60 μC flowed back through the battery, returning ΔQ ΔV = 0.72 mJ to the source, and the balance Wₑₓₜ = ΔU − W_source = (−0.36 mJ) − (−0.72 mJ) = +0.36 mJ says you still did positive work pulling. Same mechanical action, opposite sign for ΔU, because in the second case the source is inside the energy account.

The field picture earns its keep when you want a force. At fixed charge the field between the plates is E = σ/ε₀ = Q/(ε₀A), which does not depend on the separation, so with a gap x the stored energy U = ½ε₀E²(Ax) is linear in x. The plates therefore attract with a force of magnitude dU/dx = ½ε₀E²A = Q²/(2ε₀A) = ½QE. That one half is physics, not carelessness: each plate sits in the field of the other plate alone, which is E/2, not the full gap field. Watch the units of the energy density too — J m⁻³ is N m⁻², so ½ε₀E² is at once an energy per unit volume and the electrostatic pressure pulling the plates together. With A = 0.20 m² and E = 1.0×10⁶ V m⁻¹ that pressure is 4.43 Pa and the attraction is 4.43 × 0.20 = 0.89 N: negligible on a bench, dominant across a micrometre gap, which is how condenser microphones and MEMS actuators work. It is also the force you worked against when the fixed-charge energy went up.