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Energy Stored in a Capacitor

E=12CV2E = \tfrac{1}{2} C V^{2}

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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
E=12CV2E = \tfrac{1}{2} C V^{2}
Where
  • EE= Stored energy
  • CC= Capacitance
  • VV= Voltage
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