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Sumivo · Electrical

Capacitor Energy Calculator

Energy stored in a capacitor is E = ½·C·V²: enter capacitance in farads (F) and voltage in volts (V) to get joules (J). At fixed capacitance, doubling voltage stores four times the energy. The tool also shows electric charge in coulombs.

01 / INPUTS

F
V

02 / RESULT

STORED ENERGY

0.05 J

Stored energy
0.05 J
Charge
0.01 C

Calculation trace

  1. 0.5 × 0.001 F × 10 V × 10 V = 0.05 J
    0.05 J
    Stored energy
  2. 0.001 F × 10 V = 0.01 C
    0.01 C
    Charge

WORKED EXAMPLES

Each example uses E = 0.5 × C × V × V, exactly as the calculator does. Voltages are illustrative operating voltages, not advice to exceed a component rating. Wh values below are rounded conversions; the tool itself displays J and C.

Electrolytic capacitor: 1000 µF at 25 V

  1. Convert: 1000 µF ÷ 1,000,000 = 0.001 F. Enter 0.001 F and 25 V.
  2. Energy: E = 0.5 × 0.001 × 25 × 25 = 0.3125 J.
  3. Convert to Wh: 1 Wh = 3600 J, so 0.3125 ÷ 3600 ≈ 0.00008680556 Wh.
  4. Charge shown alongside energy: Q = 0.001 × 25 = 0.025 C.

Supercapacitor-scale example: 1 F at 5 V

  1. Capacitance is already in farads: 1 F = 1,000,000 µF; dividing by 1,000,000 returns 1 F. Enter 1 F and 5 V.
  2. Energy: E = 0.5 × 1 × 5 × 5 = 12.5 J.
  3. Convert to Wh: 1 Wh = 3600 J, so 12.5 ÷ 3600 ≈ 0.00347222222 Wh.
  4. Charge: Q = 1 × 5 = 5 C. This is ideal stored energy, not a usable-runtime estimate.

Double the voltage: 1000 µF at 10 V and 20 V

  1. Keep capacitance fixed: 1000 µF ÷ 1,000,000 = 0.001 F.
  2. At 10 V: E = 0.5 × 0.001 × 10 × 10 = 0.05 J; Q = 0.001 × 10 = 0.01 C.
  3. At 20 V: E = 0.5 × 0.001 × 20 × 20 = 0.2 J; Q = 0.001 × 20 = 0.02 C.
  4. Energy ratio: 0.2 ÷ 0.05 = 4. Charge only doubles.
  5. Using 1 Wh = 3600 J: 0.05 ÷ 3600 ≈ 0.00001388889 Wh and 0.2 ÷ 3600 ≈ 0.00005555556 Wh.

UNIT REFERENCE

ConversionRule
µF → F for inputDivide by 1,000,000; 1 µF = 0.000001 F.
J → mJMultiply by 1000; 0.3125 J = 312.5 mJ.
J → WhDivide by 3600; 1 Wh = 3600 J.
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How it works

Energy stored in a capacitor is E = ½·C·V², with capacitance C in farads (F), voltage V in volts (V), and energy E in joules (J). Enter farads, not microfarads: divide µF by 1,000,000 first. At fixed capacitance, doubling the voltage stores four times the energy because voltage is squared. As charge is added, the capacitor voltage rises; integrating that increasing voltage over charge gives the factor ½. The calculator also reports Q = C·V in coulombs (C). It displays J and C, not mJ or Wh; the conversions in the examples are separate arithmetic.

Assumptions & limits

  • Models an ideal capacitor and does not account for leakage, equivalent series resistance, dielectric loss or voltage-dependent capacitance.
  • Capacitance and voltage must both be positive finite numbers, entered in F and V. Although the ideal formula gives zero energy at zero voltage, this tool rejects zero and negative inputs.
  • Use the actual voltage across the capacitor. A printed voltage rating is a component limit, not necessarily its operating voltage; the calculator does not check that rating.
  • Stored energy is not a prediction of usable output energy, discharge time, power or circuit efficiency. The supercapacitor example uses the same constant-capacitance approximation.
  • Calculations use JavaScript number arithmetic. Extremely large or small inputs can overflow or underflow during multiplication; the engine checks input finiteness but does not guard the computed energy or charge.
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FAQ

What is the formula for energy stored in a capacitor?
Use E = ½·C·V². Capacitance must be in farads and voltage in volts to obtain energy in joules. For example, 0.001 F at 25 V stores 0.5 × 0.001 × 25² = 0.3125 J in the ideal model.
How do I calculate joules in a capacitor from µF and V?
Convert microfarads to farads before entering capacitance: 1000 µF ÷ 1,000,000 = 0.001 F. At 25 V, E = 0.5 × 0.001 × 25² = 0.3125 J, or 312.5 mJ. Enter 0.001 in the F field, not 1000.
Why does capacitor energy increase with voltage squared?
At fixed capacitance, more voltage means more stored charge and more work per added charge. Integrating the rising voltage gives E = ½·C·V². For 0.001 F, raising voltage from 10 V to 20 V changes stored energy from 0.05 J to 0.2 J: four times as much, while charge only doubles.
How do I convert capacitor energy from joules to watt-hours?
Divide joules by 3600 because 1 Wh = 3600 J. A 1 F capacitor at 5 V stores 12.5 J, or approximately 0.00347222222 Wh. This calculator reports joules; it does not display Wh or estimate battery runtime.
How is energy stored in a capacitor different from energy in a battery?
An ideal capacitor stores energy in an electric field; a battery stores chemical energy made available through electrochemical reactions. Conventional capacitors generally store much less energy per mass than batteries but can suit brief pulses. Supercapacitors increase capacitance, yet energy still depends on voltage. Compare both in J or Wh, and do not confuse charge in coulombs with energy.
Can a capacitor release only part of its stored energy?
If discharge stops at a nonzero final voltage, energy remains stored. For constant capacitance, the decrease is ΔE = ½·C·(Vinitial² − Vfinal²). A circuit may stop working before voltage reaches zero, and losses reduce the energy delivered usefully to a load. This tool calculates energy at one positive voltage, not discharge behavior; compare two positive-voltage results to find the ideal energy decrease.
What is the difference between capacitor energy and capacitor charge?
Energy E = ½·C·V² is measured in joules; charge Q = C·V is measured in coulombs. They are different quantities, and this page reports both with energy first. The Capacitor Charge Calculator linked below reports the same ideal quantities with charge first; neither page calculates charging time.
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