Case 1
- Capacitance C (F)
- 0.001F
- Voltage V (V)
- 10V
0.050000J
Open with these valuesCapacitor Energy Calculator
0.050000J
Result: 0.050000 JHalf the capacitance times the voltage squared gives joules: 0.001 F at 10 V holds 0.05 J. Voltage counts double because it is squared — double it and the energy quadruples, while double the capacitance only doubles it. Enter farads, not microfarads: 1000 µF is 0.001 F.
The numbers at a glance
Held fixed: Capacitance C (F) 0.001000000000 F.
| Voltage V (V) (V) | Result (J) |
|---|---|
| 0.000 | 0.000000 |
| 2.500 | 0.003125 |
| 5.000 | 0.012500 |
| 7.500 | 0.028125 |
| 10.000Your value | 0.050000 |
| 12.500 | 0.078125 |
| 15.000 | 0.112500 |
| 17.500 | 0.153125 |
| 20.000 | 0.200000 |
Worked examples
Case 2
- Capacitance C (F)
- 0.0047F
- Voltage V (V)
- 9V
0.190350J
Open with these valuesCase 3
- Capacitance C (F)
- 1F
- Voltage V (V)
- 2V
2.000000J
Open with these valuesHow it's calculated
E = ½ × C × V²
- StepEnter the capacitance in farads — 1000 µF is 0.001 F.
- StepEnter the voltage across the capacitor in volts.
- ResultRead the stored energy in joules; one joule is one watt-second.
What this number means
Voltage is squared, capacitance is not
Doubling the capacitance doubles the energy; doubling the voltage quadruples it and tripling it multiplies by nine. 0.001 F holds 0.05 J at 10 V and 0.2 J at 20 V.
Farads, not microfarads
Component values are printed in µF or nF, but this field takes farads: 1000 µF is 0.001 F and 1 nF is 0.000000001 F. Typing 1000 for a 1000 µF part at 10 V returns 50000 J instead of 0.05 J.
A charged capacitor stays charged
The energy sits in the electric field between the plates until the capacitor is discharged, and it can come back out very quickly. That is why even a modest voltage on a large capacitor can deliver a strong jolt.
Commonly misread
E = C × V², so 0.001 F at 10 V holds 0.1 J.
The half belongs in the formula: E = ½ × C × V² gives 0.05 J. Dropping it doubles every answer.
Twice the voltage means twice the energy.
Voltage enters squared, so twice the voltage is four times the energy: 0.001 F goes from 0.05 J at 10 V to 0.2 J at 20 V.
My capacitor is 1000 µF, so I enter 1000.
1000 µF is 0.001 F. Entering 1000 at 10 V returns 50000 J, a million times the real 0.05 J.
The same formula gives the energy in a coil.
A coil stores ½ × L × I², with the current squared, not the voltage. The two formulas look alike and describe different quantities.
Reference table
| C (F), V (V) | Typical part | Energy (J) |
|---|---|---|
| 0.000001, 12 | 1 µF ceramic | 0.000072 |
| 0.001, 10 | 1000 µF electrolytic | 0.05 |
| 0.01, 5 | 10 mF bank | 0.125 |
| 0.0047, 9 | 4700 µF electrolytic | 0.19035 |
| 1, 2 | 1 F supercapacitor | 2 |
Questions
How do I calculate the energy stored in a capacitor?
Multiply half the capacitance by the voltage squared: E = ½ × C × V². Use farads and volts and the answer comes out in joules. A 1000 µF capacitor (0.001 F) charged to 10 V stores 0.05 J.
Why does doubling the voltage quadruple the energy?
Because the voltage is squared in the formula. Doubling it multiplies the stored energy by four and tripling it by nine. That is why even a modest voltage on a large capacitor can deliver a strong jolt.
Does capacitance or voltage matter more?
Both raise the energy, but not equally. Capacitance enters in direct proportion, so doubling it doubles the energy. Voltage is squared, so doubling it quadruples the energy.
Which units should I enter?
Farads for capacitance and volts for voltage, which gives joules. Printed component values are usually microfarads or nanofarads, so convert first: 1000 µF is 0.001 F and 1 nF is 0.000000001 F. One joule equals one watt-second.
Where is the energy actually stored?
In the electric field between the plates. It was supplied while the capacitor charged and comes back out when it discharges, sometimes very quickly. It is measured in joules, like every other form of energy.
Sources and last check
Information, not professional advice.