Capacitor Energy Calculator (E = ½CV²)
Calculate the energy stored in a capacitor and its charge. Enter capacitance in µF and voltage to get energy in joules and millijoules, plus charge in coulombs.
How to use this tool
- Enter capacitance (c) and voltage (v) in the fields above.
- Results update instantly as you type — or click Calculate.
- Read your energy and the full breakdown beneath it.
The energy stored in a capacitor is E = ½CV² (joules), where C is capacitance (farads) and V is voltage. Charge stored is Q = CV (coulombs). Because energy scales with V², doubling the voltage quadruples the stored energy.
Formula
Energy stored: E = ½ × C × V2
Charge stored: Q = C × V
where C is capacitance in farads and V is voltage in volts. Capacitance entered in µF is divided by 106 before use.
How it works
The energy stored in a capacitor's electric field is given by the standard formula E = ½CV², derived from integrating the work done moving charge onto the plates against the growing electric field.
Charge Q = CV follows directly from the definition of capacitance; results are accurate for ideal capacitors — real capacitors have leakage current, equivalent series resistance, and voltage-dependent capacitance that this calculator does not model.
Worked example
Worked example — 100 µF at 12 V
- Convert capacitance: C = 100 µF = 100 × 10⁻⁶ = 1 × 10⁻⁴ F.
- Energy: E = 0.5 × 1 × 10⁻⁴ × 12² = 0.5 × 1 × 10⁻⁴ × 144 = 0.0072 J.
- In millijoules: E = 0.0072 × 1000 = 7.2 mJ.
- Charge: Q = 1 × 10⁻⁴ × 12 = 0.0012 C = 1.2 mC.
Energy = 0.0072 J (7.2 mJ); charge = 0.0012 C (1.2 mC).
Key terms
- Capacitance (C)
- A component's ability to store electric charge per unit voltage, measured in farads (F); 1 F stores 1 coulomb per volt.
- Microfarad (µF)
- One millionth of a farad (10⁻⁶ F); the most common unit for practical capacitors.
- Energy stored
- The work done in charging the capacitor, held in the electric field between the plates; equals ½CV².
- Charge (Q)
- The quantity of electricity stored on the capacitor plates, measured in coulombs (C); Q = CV.
- Dielectric
- The insulating material between capacitor plates that enables charge storage; its properties determine maximum voltage and effective capacitance.
Frequently asked questions
- Can a capacitor store as much energy as a battery?
- Generally no. A typical 1 F supercapacitor at 2.7 V stores ~3.6 J, while a single AA alkaline battery stores ~15,000 J. Capacitors excel at releasing energy quickly (high peak power), not storing large amounts.
- Why does energy scale with V²?
- Each additional unit of charge added to a capacitor must be pushed against an increasing voltage, so the energy per coulomb increases linearly. Integrating Q × dV from 0 to V gives ½CV².