Capacitor Potential Energy Calculator

Formula used

Capacitor stored energy can be written in three equivalent forms:

• U = ½ C V² (when capacitance and voltage are known)
• U = Q² / (2C) (when charge and capacitance are known)
• U = ½ Q V (when charge and voltage are known)

For a parallel-plate capacitor estimate: C = ε₀ εᵣ A / d, where ε₀ is vacuum permittivity, εᵣ is relative permittivity, A is plate area, and d is separation.

How to use this calculator

1. Select a calculation mode that matches your known values.
2. Enter values and choose their units (capacitance, voltage, charge, or geometry).
3. Select an output unit for energy (J, mJ, or µJ).
4. Press Calculate to lock the results.
5. Use Download CSV or Download PDF to save a summary.

Capacitor potential energy guide (360 words)

1) Meaning of stored energy

Capacitor potential energy is the electrical energy stored in an electric field between conductors. It represents how much work was done to separate charge and raise the voltage. This calculator converts common input sets into the same stored energy result, so you can compare designs quickly and consistently in practice.

2) Inputs and unit choices

The most direct method uses capacitance and voltage. Enter C in F, mF, µF, nF, or pF, then enter V in mV, V, or kV. The tool standardizes everything to joules in the background, and can optionally show charge in coulombs using Q = C·V. Use the unit dropdowns to avoid scientific notation and reduce entry mistakes.

3) Voltage-squared scaling

Energy grows with the square of voltage: doubling voltage increases stored energy by four times. For example, a 10 µF capacitor at 5 V stores 0.000125 J, while at 10 V it stores 0.0005 J. This strong scaling is why voltage rating matters so much.

4) Typical stored-energy values

Realistic numbers help with intuition. A 470 µF capacitor at 12 V stores 0.03384 J. A 1000 µF capacitor at 50 V stores 1.25 J. A 1 F supercapacitor at 2.7 V stores 3.645 J. These values fit typical small electronics and energy buffering.

5) Geometry-based estimate

If you only know geometry, the parallel‑plate estimate is useful: C ≈ ε0·εr·A/d. With A = 10 cm² (0.001 m²), d = 1 mm (0.001 m), and εr = 3, capacitance is about 26.6 pF. At 1 kV, energy is roughly 0.0133 J. Changing spacing from 1 mm to 0.5 mm doubles capacitance and doubles energy at the same voltage.

6) Charge-based calculation modes

Charge‑based inputs are common in lab work. If you enter Q and V, the calculator uses U = ½ QV. With Q = 1 mC and V = 100 V, energy is 0.05 J. If you enter C and Q, it uses U = Q²/(2C), highlighting how smaller C increases energy for the same charge.

7) Limits and safe handling

Always consider limits. Exceeding rated voltage risks dielectric breakdown, heating, and venting. The stored energy may look small in joules, but discharge currents can be high due to low ESR. Use bleeder resistors for high‑voltage setups and verify polarity for electrolytics.

FAQs

1. Which formula should I choose?

Pick the input mode that matches what you know. C & V uses U = ½CV², Q & V uses U = ½QV, and C & Q uses U = Q²/(2C). All represent the same stored energy when values are consistent.

2. Why does energy jump when voltage increases?

Stored energy scales with voltage squared. If you double the voltage, energy becomes four times larger. This is why increasing voltage rating or operating voltage has a much bigger impact than small capacitance changes.

3. Can I calculate energy for supercapacitors?

Yes. Enter capacitance in farads and the working voltage. Supercapacitors often store several joules or more, so discharge can be intense. Stay within the specified voltage and consider series balancing if using multiple cells.

4. How does the geometry method estimate capacitance?

It uses the parallel‑plate approximation: C ≈ ε0·εr·A/d. Convert area to square meters and spacing to meters for best accuracy. This estimate works best when plates are large compared with the gap and edge effects are small.

5. What does one joule mean here?

One joule equals one watt‑second of energy. For capacitors, it’s the field energy available to be released into a circuit. Actual delivered energy depends on voltage sag, resistance, and the minimum usable voltage of your load.

6. Is capacitor energy dangerous at low joules?

Sometimes. Even modest joules can be hazardous at high voltage because shock risk depends on voltage and current path. Low ESR capacitors can dump large peak currents. Discharge safely, use insulation, and add a bleeder resistor for high‑voltage setups.