Energy Stored in a Capacitor
Worked example: 100 uF at 12 V → 7.2 mJ — press Try an example to run it live, then adjust anything.
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Grade 12Grade 12 Physics
Capacitance →
UniversityCircuits & Electrical Power
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Energy Stored in a Capacitor explained
Why the half? Charging a capacitor is like stretching a spring: the first coulomb slides on easily, but every later coulomb must be pushed against the voltage the earlier ones built up. The voltage ramps linearly from 0 to V as charge accumulates, so the average push is V/2, and the total work is Q×V/2 = ½CV² — the triangular area under the Q–V line. Equivalent forms E = Q²/(2C) and E = QV/2 follow directly from C = Q/V.
Worked example: a camera-flash capacitor of 1000 µF charged to 300 V stores E = ½ × 0.001 × 300² = 45 J, dumped through the xenon tube in about a millisecond — a burst of tens of kilowatts from a pocket battery. The same math sizes defibrillators and grid-scale supercapacitor banks. Solving for V takes the positive square root, since the formula gives the voltage magnitude.
Energy Stored in a Capacitor formula
- = Stored energy (J)
- = Capacitance (μF)
- = Voltage (V)
Missing one of these? Work it out first, then come back
- Stored energy — Energy Stored in an Inductor, Electrical Energy (E = Pt)
- Capacitance — RC Time Constant, Capacitance (C = Q/V)
- Voltage — Ohm's Law, Electrical Power (P = VI)