Free Capacitor Energy Calculator
Find the energy stored in a capacitor from its capacitance and voltage using E = ½·C·V². Enter capacitance in microfarads (µF) and voltage in volts to get the stored energy in joules and millijoules, along with the stored charge in coulombs and microcoulombs.
Enter capacitance and voltage to find the stored energy with E = ½·C·V².
E = ½·C·V² · Q = C·V. Capacitance is converted from µF to farads (µF × 10⁻⁶). A charged capacitor can store a dangerous amount of energy — discharge before handling.
Quick answer
The energy stored in a capacitor is E = ½·C·V², where C is capacitance in farads and V is voltage in volts. For a 1000 µF capacitor charged to 12 V, E = ½ × (1000 × 10⁻⁶ F) × 12² = 0.072 J (72 mJ). The capacitor also holds a charge of Q = C·V = 0.012 C (12000 µC).
Formula & method
E = ½ · C · V²
- E — Energy stored, in joules (J)
- C — Capacitance, in farads (F); µF × 10⁻⁶
- V — Voltage across the plates, in volts (V)
Energy in joules. C must be in farads (multiply µF by 1e-6) and V in volts. Equivalent forms: E = ½·Q·V = Q²/(2C).
Q = C · V
- Q — Charge, in coulombs (C)
- C — Capacitance, in farads (F)
- V — Voltage, in volts (V)
Charge stored on the plates, in coulombs (C). Multiply by 1e6 to get microcoulombs (µC).
Examples
- Input
- C = 1000 µF, V = 12 V
- Result
- E = 0.072 J (72 mJ), Q = 0.012 C (12000 µC)
- Why
- Convert capacitance: 1000 µF = 1000 × 10⁻⁶ = 0.001 F. Energy E = ½ × 0.001 × 12² = ½ × 0.001 × 144 = 0.072 J = 72 mJ. Charge Q = 0.001 × 12 = 0.012 C = 12000 µC.
- Input
- C = 470 µF, V = 25 V
- Result
- E = 0.146875 J (≈146.88 mJ), Q = 0.01175 C (11750 µC)
- Why
- 470 µF = 0.00047 F. Energy E = ½ × 0.00047 × 25² = ½ × 0.00047 × 625 = 0.146875 J ≈ 146.88 mJ. Charge Q = 0.00047 × 25 = 0.01175 C = 11750 µC.
- Input
- C = 2200 µF, V = 50 V
- Result
- E = 2.75 J (2750 mJ), Q = 0.11 C (110000 µC)
- Why
- 2200 µF = 0.0022 F. Energy E = ½ × 0.0022 × 50² = ½ × 0.0022 × 2500 = 2.75 J = 2750 mJ. Charge Q = 0.0022 × 50 = 0.11 C = 110000 µC. This is enough stored energy to be a real discharge hazard.
- Input
- C = 10 µF, V = 400 V
- Result
- E = 0.8 J (800 mJ), Q = 0.004 C (4000 µC)
- Why
- 10 µF = 0.00001 F. Energy E = ½ × 0.00001 × 400² = ½ × 0.00001 × 160000 = 0.8 J = 800 mJ. Charge Q = 0.00001 × 400 = 0.004 C = 4000 µC. Note how high voltage dominates energy because of the V² term.
When to use this tool
- Sizing or comparing filter, bulk, and bypass capacitors in a power-supply design.
- Estimating the discharge hazard of a charged capacitor before servicing equipment such as power supplies, amplifiers, or camera flashes.
- Designing energy-storage circuits like camera flashes, defibrillators, or pulse-discharge systems where joules of stored energy matter.
- Checking homework or lab results in a physics or electronics course involving E = ½CV².
Common mistakes
- Forgetting to convert microfarads to farads. You must multiply µF by 10⁻⁶ before applying E = ½CV²; using the raw µF number inflates the answer by a million.
- Dropping the ½ factor. The energy formula is E = ½CV², not CV². Leaving out the one-half doubles your result.
- Treating voltage linearly. Because of the V² term, doubling the voltage quadruples the stored energy — not just doubles it.
- Confusing energy (E, joules) with charge (Q, coulombs). Q = C·V grows linearly with voltage, while E grows with the square of voltage; they are different quantities.
Frequently asked questions
What is the formula for energy stored in a capacitor?
The energy stored in a capacitor is E = ½·C·V², where C is the capacitance in farads and V is the voltage across the plates in volts. The result is in joules. Equivalent forms are E = ½·Q·V and E = Q²/(2C), where Q is the stored charge.
How do I convert microfarads (µF) to farads for this calculation?
Multiply the value in microfarads by 10⁻⁶ (one millionth). For example, 1000 µF = 1000 × 10⁻⁶ = 0.001 F, and 100 µF = 0.0001 F. This calculator does the conversion automatically when you enter capacitance in µF.
How much energy does a 1000 µF capacitor at 12 V store?
It stores E = ½ × 0.001 F × 12² = 0.072 joules, or 72 millijoules. The same capacitor holds a charge of Q = C·V = 0.012 coulombs (12000 µC).
Why does doubling the voltage more than double the stored energy?
Because energy depends on the square of the voltage (E = ½CV²). Doubling V multiplies V² by four, so the stored energy quadruples even though the charge Q = C·V only doubles. This is why high-voltage capacitors can store dangerous amounts of energy.
Is the energy in a charged capacitor dangerous?
It can be. A few joules is enough to deliver a painful or harmful shock, and larger banks (high capacitance and high voltage) store much more. Always discharge capacitors through an appropriate resistor and verify with a meter before touching the terminals — this tool helps you estimate how much energy is present.
What is the difference between charge (Q) and energy (E) in a capacitor?
Charge Q = C·V (in coulombs) is the amount of separated electric charge on the plates and grows linearly with voltage. Energy E = ½CV² (in joules) is the work stored in the electric field and grows with the square of voltage. They measure different physical quantities, so report whichever your application needs.
Sources & references
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Provided “as is” for general information only — results may be inaccurate, so verify before you rely on them. No warranty; use at your own risk.
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