Result

Stored energy

(E4.4.1) E4.4 · Energy Stored in a Charged Capacitor →

U = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q Δ V = \frac{1}{2} C (Δ V)^{2}

Statement

The work done in charging the capacitor, recoverable when it discharges.

Symbols and units

$U$	energy stored	J
$Q$	charge held	C
$Δ V$	voltage across the capacitor	V
$C$	capacitance	F

When it applies

Any capacitor, charged from zero.

Don't use when

Assumes the capacitor is charged from zero through a resistive or otherwise gradual path — it does not describe energy delivered instantaneously by an ideal, lossless connection to a fixed-voltage source.

Derivation

Suppose charge q has already been moved. The voltage across the plates right now is

Δ V = \frac{q}{C}

Moving the next small amount dq costs:

d W = Δ V d q = \frac{q}{C} d q

Integrate from empty to full charge Q:

U = \int_{0}^{Q} \frac{q}{C} d q = \frac{Q ^{2}}{2 C}

Substituting Q = CΔV gives the other two forms:

U = \frac{1}{2} Q Δ V = \frac{1}{2} C (Δ V)^{2}

Worked examples

Photographic flash

100 µF charged to 300 V:

U = \frac{1}{2} C (Δ V)^{2} = 4.5 J

Discharged in 1 ms:

P = \frac{4.5}{1 0 ^{- 3}} = 4.5 kW

Modest energy delivered as large power.

Plates separated after disconnection

Charge fixed, gap doubled:

C \to \frac{1}{2} C, U = \frac{Q ^{2}}{2 C} \to 2 U

The additional energy is the work done against the attraction between the plates.

Watch out for

The factor ½ is the whole point of the derivation: the voltage rises from zero to its final value, so on average you pay only half the final voltage per unit charge.

Connects to

Energy density Capacitance Charging