Energy Stored by a Capacitor

The energy is the area under a Q–V graph. As charge is added, the potential difference rises in proportion, so the graph is a straight line through the origin and the area beneath it is a triangle:

W = ½ Q V

Three equivalent forms — substituting Q = CV gives

W = ½ C V² and W = Q² / (2C)

Choose whichever matches the quantities you are given.

Why the factor of ½ — this is the part that is most often lost. The first charge moves across at almost zero potential difference and takes almost no work; the last moves across the full voltage V. The average p.d. during charging is V/2, so the total work is ½QV, not QV.

Example: the missing half has a physical destination. A supply of e.m.f. V pushing charge Q round the circuit does work QV, but only ½QV ends up in the capacitor. The rest is dissipated in the resistance of the circuit — and, remarkably, that is true whatever the resistance. Halving the resistance halves the time and doubles the current, leaving the energy lost unchanged. Exactly half the energy is always wasted when charging a capacitor from a constant-voltage supply through a resistor.

Energy rises with the square of the voltage — doubling the voltage stores four times the energy, for the same capacitor. Voltage is therefore the more powerful lever than capacitance, which appears only to the first power.

Why capacitors are used for rapid discharge — a camera flash or a defibrillator needs a large power for a very short time. A capacitor stores modest energy but can release it in milliseconds, giving a very high power. A battery holds far more energy but cannot deliver it nearly as fast.

Discharging through a resistor converts all the stored energy to heat in that resistor, whatever its resistance; the resistance sets only how quickly.