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Capacitor Energy⚠ unverified

Electrical / Power Electronics · Compute the energy stored in a capacitor

Parameters

InputSymbolUnitDefaultDescription
CCF1.0Capacitance
VVV1.0Voltage across the capacitor
OutputSymbolUnitDescription
resultEJStored energy, in joules (J)

The science & history

Understanding the Parameters

Derivation (Approaching a Proof)

Charging is not free: pushing an extra charge $dq$ onto a capacitor already at voltage $v = q/C$ costs work $dW = v\,dq = \dfrac{q}{C}\,dq$. Integrating from an uncharged state to final charge $Q$:

$$E = \int_0^{Q} \frac{q}{C}\,dq = \frac{Q^2}{2C}.$$

Substituting $Q = CV$ gives the three equivalent forms

$$E = \frac{Q^2}{2C} = \frac{1}{2}QV = \frac{1}{2}CV^2.$$

The factor of $\tfrac12$ appears because the voltage rises linearly from $0$ to $V$ as charge accumulates, so the average voltage during charging is $V/2$. Physically the energy resides in the electric field, with density $u = \tfrac12 \varepsilon E^2$ integrated over the volume between the plates.

History

Energy storage in capacitors dates to the Leyden jar (1745, Musschenbroek and von Kleist), the first device to hold a substantial charge. The quantitative energy expression follows from the 19th‑century definitions of capacitance and potential, and is consistent with Maxwell's identification of energy stored in the electric field itself.

Related Concepts: Inductor Energy, Capacitive Reactance, Watt's Law

Notes: Registry calculator capacitor-energy (unverified). Assumes a linear (constant‑$C$) capacitor; ferroelectric/variable dielectrics require integrating $\int v\,dq$ directly.

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