Capacitor Energy Calculator

Calculate energy stored in a capacitor from capacitance and voltage.

1000 μF = 0.001 F
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Why Capacitor Energy Depends on Voltage Squared

A charged capacitor stores energy in the electric field between its conductors. The key relationship is Q = CV: for a fixed capacitance, adding more charge raises the voltage. Charging is therefore not performed against a constant voltage. Early charge is added when the voltage is small, while later charge must be pushed onto the capacitor against a larger voltage.

That changing voltage explains the factor of one-half in the energy formula. Integrating the work needed to move charge from zero to the final charge gives E = ½QV. Replacing Q with CV gives E = ½CV2. The squared voltage is especially important: doubling voltage stores four times as much energy in the same capacitor, provided the capacitor's voltage rating is not exceeded. The equivalent form E = Q2/(2C) is useful when charge rather than voltage is known.

E = ½CV2 = ½QV = Q2/(2C)    and    Q = CV
SymbolMeaningUnits / role
EStored electrical energyJoules (J)
CCapacitanceFarads (F); larger C stores more energy at the same V
VVoltage across the capacitorVolts (V); energy scales with V2
QStored chargeCoulombs (C); Q = CV

Capacitance prefixes matter: 1 µF = 10−6 F and 1 mF = 10−3 F. The calculated energy is the ideal field energy at the stated voltage. Real capacitors also have leakage, equivalent series resistance, dielectric losses, and maximum voltage ratings that affect practical storage and discharge.

Worked Examples

Example 1: Camera flash: C=1000μF, V=300V
E = 0.5×0.001×300²
Result: 45 J
Enough to trigger a bright flash
Example 2: Supercapacitor: C=500F, V=2.7V
E = 0.5×500×2.7²
Result: 1822.5 J = 0.506 Wh
Small but fast charge/discharge
Example 3: Small decoupling capacitor
C = 100 µF = 1 × 10−4 F, V = 5 V → E = 0.5 × 10−4 × 52
Result: E = 0.00125 J = 1.25 mJ
The energy is small, but the capacitor can deliver it quickly to help stabilize a local supply during a brief load change.
Example 4: Same capacitance at twice the voltage
C = 100 µF, V = 10 V → E = 0.5 × 10−4 × 102
Result: E = 0.005 J = 5 mJ
Doubling voltage from 5 V to 10 V increases energy from 1.25 mJ to 5 mJ, exactly fourfold because energy depends on V2.

Common Mistakes

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Entering microfarads as farads

A 1000 µF capacitor is 0.001 F, not 1000 F. Because energy is proportional to capacitance, a prefix error transfers directly into an enormous energy error.

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Forgetting to square the voltage

The formula is E = ½CV2. Using V instead of V2 understates stored energy increasingly as voltage rises.

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Dropping the factor of one-half

The capacitor voltage rises from zero to its final value during charging. The average voltage during an ideal quasistatic charge is one-half the final voltage, which is why the stored energy is ½QV rather than QV.

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Using calculated energy to justify exceeding the voltage rating

The V2 relationship does not mean voltage can be increased without limit. Exceeding a capacitor's rated voltage can damage the dielectric and create a safety hazard.

Frequently Asked Questions

How does voltage affect energy storage?
Energy scales with V² — doubling voltage quadruples energy stored. This is why high-voltage capacitors store so much energy in compact packages.
Capacitor vs battery energy density?
Batteries: ~100–250 Wh/kg. Capacitors: ~0.01–10 Wh/kg. Capacitors lose on energy density but win on power density and cycle life (millions of cycles vs thousands for batteries).
Why is capacitor energy one-half CV squared?
The voltage is not constant while the capacitor charges. It rises from zero to the final voltage as charge accumulates. Integrating the incremental work V dQ over that changing voltage gives E = ½CV2.
Can capacitor energy be calculated from charge instead of voltage?
Yes. Since Q = CV, the energy can also be written as E = ½QV or E = Q2/(2C). Choose the form that matches the quantities given in the problem.
Where is the energy physically stored in a capacitor?
In the electromagnetic description, the energy is stored in the electric field in the dielectric region between and around the conductors. The capacitor plates provide the separated charge that creates this field.
Does a capacitor release all of its ideal stored energy instantly?
No. Discharge rate depends on the connected circuit, especially resistance, inductance, and the capacitor's equivalent series resistance. The energy formula gives the amount available at the starting voltage, not the time required to deliver it.

Formula Explorer connections

Interpretation: This relationship connects motion, force, momentum, work or energy in a mechanical system. Assumption: Choose a consistent reference direction and unit system. The model may assume constant acceleration, rigid bodies, negligible losses or an isolated system.

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