Capacitor Energy Calculator
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/9/2026
The energy stored in a charged capacitor is calculated with E = ½ × C × V², where C is the capacitance in farads and V the voltage across it. A 1000 µF capacitor charged to 12 V stores 0.072 joules.
Explanation
Charging a capacitor requires doing electrical work: as charge accumulates on the plates, the voltage across them rises, and more and more energy must be supplied to add each additional bit of charge. Integrating this work over the full charge, from 0 volts up to the final voltage V, gives the formula E = ½ × C × V² — the factor of ½ comes precisely from this gradual accumulation, unlike a constant-voltage energy calculation, which would simply give Q × V. This energy, expressed in joules, depends on the capacitor's capacitance (proportional to the plate area and inversely proportional to their spacing, for a parallel-plate capacitor) and especially on the applied voltage, which enters as a square: doubling the voltage quadruples the stored energy at equal capacitance. This quadratic relationship is why capacitors used in energy-storage applications (camera flashes, defibrillators, power supply smoothing) are designed to operate at the highest voltage their insulation can handle, rather than with an oversized capacitance. Unlike a resistor, where Ohm's law directly relates voltage and current with no notion of storage, a capacitor genuinely stores this energy for the duration of the charge, before releasing it entirely on discharge (minus the real circuit's resistive losses).
Example: a 470 µF capacitor charged to 5 V
Inputs
Capacitance: 470 µF (0.00047 F). Voltage: 5 V.
Calculation
E = ½ × 0.00047 × 5² = 0.5 × 0.00047 × 25 = 0.005875 J.
Result
This capacitor stores about 5.875 millijoules (0.005875 J).
Frequently asked questions
Why does the formula use ½ × C × V² and not simply C × V?
Because the voltage across the capacitor isn't constant during charging: it starts at 0 and rises gradually to the final value V, as charge accumulates. The total energy supplied is the sum (the integral) of all these small energy contributions at rising voltage, which gives exactly half of what a charge at a constant voltage V from the start would give — hence the factor of ½.
What happens to this energy when the capacitor discharges?
It's returned to the circuit, as an electric current that powers whatever load is connected across the capacitor (a resistor, an LED, a motor…). This is the principle exploited in a camera flash: the capacitor charges slowly from the battery, then releases all its energy in a fraction of a second when triggered.
Does the stored energy depend on the type of dielectric used?
Indirectly, yes: the dielectric (the insulator between the plates) affects the capacitor's capacitance C, but once the capacitance is known, the formula E = ½ × C × V² applies the same way regardless of the material used. The same electromagnetic toolkit — see our solenoid magnetic field calculator — shows how a coil's geometry similarly shapes a magnetic rather than an electric storage effect.