Charge and Energy Stored in a Capacitor

Finds the charge on a capacitor as Q = CV and the energy it stores as E = ½CV². The energy grows with the square of the voltage.

Explanation

Apply a voltage to a capacitor and charge accumulates on its plates, storing energy in the process.

Q = CV \qquad E = \dfrac{1}{2} C V^2

Here $Q$ is the charge in coulombs, $C$ the capacitance in farads, $V$ the voltage and $E$ the energy in joules.

Example

Put 12 V across a 100 μF capacitor.

Q = 100 \times 10^{-6} \times 12 = 1.2 \times 10^{-3}\ \text{C}

E = \dfrac{1}{2} \times 100 \times 10^{-6} \times 12^2 = 7.2 \times 10^{-3}\ \text{J}

That is 1.2 mC of charge holding 7.2 mJ of energy.

Where the ½ comes from

Why not simply $E = QV$ ?

Because during charging the voltage across the capacitor climbs from zero up to $V$ . The first charges slip in against almost no voltage at all and cost nothing to place. Only the last ones have to be forced in against the full $V$ .

Averaged over the whole process, the charge went in against $V/2$ , so $E = Q \times \dfrac{V}{2} = \dfrac{1}{2} CV^2$ .

The square matters

The energy goes as the square of the voltage. Double the voltage and you store four times as much.

Camera flashes and defibrillators live on this. They trickle-charge a capacitor to a high voltage over seconds, then dump the stored energy in an instant. A defibrillator releases 100 to 200 joules in a few milliseconds.

Watch out

A large capacitor stays charged after the power is switched off. Unplugging a device does not make it safe. This is exactly why service manuals insist on discharging capacitors before anyone reaches inside.