A charged capacitor stores energy in its electric field, and the amount is E = ½CV² — half the capacitance times the voltage squared. It also holds a charge Q = CV. The two behave very differently with voltage: charge rises in step with it, but energy climbs with its square, so a small jump in voltage stores far more energy. That square law is why capacitors are prized for fast energy delivery — camera flashes, defibrillators, snubbers — and why high-voltage ones demand respect even after the power is cut.
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the capacitor energy relation E = ½CV² and Q = CV.
The capacitor equations
Convert capacitance to farads first — microfarads are 10⁻⁶ F. Energy is the area under the charge-versus-voltage line, a triangle giving ½CV²; charge is simply the product CV. To find the capacitance needed for a target energy at a known voltage, use C = 2E/V²; to find the voltage a known capacitor must reach, V = √(2E/C). The charge always follows directly as Q = CV.
Worked example — a smoothing capacitor
Scenario: A 100 µF capacitor is charged to 12 V. How much energy and charge does it hold?
The capacitor stores 7.2 mJ and 1.2 mC. Raise the voltage to 24 V and the charge merely doubles to 2.4 mC, but the energy quadruples to 28.8 mJ — the V² term at work. Drop to 5 V and it falls to 1.25 mJ. This is exactly why energy-storage capacitor banks run at the highest safe voltage: every extra volt buys disproportionately more stored energy.
Frequently Asked Questions
E = ½CV². 100 µF at 12 V = ½ × 100×10⁻⁶ × 144 = 7.2 mJ. Convert µF to F first.
Q = CV (coulombs). 100 µF × 12 V = 1.2 mC. Charge scales linearly with voltage.
Energy is the area under the Q–V line (½CV²). Double V → 2× charge but 4× energy.
7.2 mJ, 1.2 mC. At 24 V it's 28.8 mJ; at 5 V, 1.25 mJ.
Small ones no; large high-voltage caps can store joules and shock even when off. Discharge before handling.