Capacitor charge and energy: Q = CV and E = ½CV²
A capacitor stores charge on two plates and energy in the field between them. How much of each depends on its capacitance and the voltage across it: the charge grows with the voltage, the energy with its square. This calculator gives both for any capacitor and voltage, shows the charge as electrons, draws the plates, and tells you what delivering that charge means in current and time, or how long a steady load can run on it, with a warning when the stored energy is enough to hurt.
How to use the capacitor charge calculator
- Type the capacitance as it is printed: 100n,
4.7u, 1000uF, or 10 for a 10 F supercapacitor. The disc beside it shows the three-digit code. - Type the voltage it is charged to. Keep it under the capacitor's rating; the energy depends on the square of it.
- Read the charge in coulombs (and as a number of electrons) and the energy in joules. The plates in the drawing gain symbols with the charge; the gauge turns red when the energy could injure.
- Optional: type a time to see the average current and power needed to deliver the charge that fast (a flash, a pulse), or a steady current to see how long the charge lasts (a backup capacitor).
- Other voltages tabulates the same capacitor at common supply voltages; Calculation writes the maths out; click the capacitor in the drawing for more.
Charge: Q = C × V
Q = C × V
Capacitance is charge per volt: a 1 F capacitor holds 1 coulomb for every volt across it. Connect a capacitor to a supply and current flows until the plates carry enough charge for their voltage to equal the supply's; then it stops. The charge sits as a surplus of electrons on one plate and a shortage on the other, held in place by their attraction across the insulating gap. One coulomb is about 6.24 × 10¹⁸ electrons, so even a nanocoulomb is billions of them.
Your capacitor Q = 1000 µF × 12 V = 12 mC, about 7.49 × 10¹⁶ electrons.
Energy: E = ½ C V²
E = ½ C V² = ½ Q V = Q² / 2C
Pushing charge onto a capacitor takes work, and the work per coulomb is the voltage at that moment. The first coulomb arrives when the voltage is still near zero and costs almost nothing; the last arrives at the full voltage. On average the charge was moved against V/2, so the energy is Q × V / 2 — the famous half. Because Q is itself C × V, the energy grows with the square of the voltage: twice the voltage, four times the energy. That is why a 400 V capacitor of modest size stores far more than a large low-voltage one.
| Capacitor | Charge | Energy |
|---|---|---|
| 100 nF decoupling at 3.3 V | 330 nC | 544 nJ |
| 1000 µF at 12 V | 12 mC | 72 mJ |
| 330 µF flash at 330 V | 109 mC | 18 J |
| 10 F supercapacitor at 2.7 V | 27 C | 36.5 J |
Current and time
I = Q / t t = Q / I
Current is charge per second. Deliver the whole charge in a millisecond and the average current is a thousand times the charge in coulombs; that is how a flash capacitor gives a short, enormous pulse. The other way round, a capacitor backing up a circuit that draws a steady current lasts Q / I seconds, but its voltage falls in a straight line the whole time (dV/dt = I / C), so what matters is how long it stays above the voltage the circuit needs. The RC time constant calculator covers discharge through a resistor, where the current falls as the voltage does.
Safety: when a capacitor can hurt
A capacitor keeps its charge after the power is off, sometimes for days. The commonly quoted threshold is about 10 J stored, or more than 50 V, for a capacitor to be dangerous: the discharge can burn, weld a tool to the terminals, or disturb the heart. Flash units, switch-mode power supplies and motor drives all have such capacitors. Discharge them through a resistor (a few hundred ohms, rated for the energy) rather than a screwdriver, and check with a meter afterwards. The calculator's gauge turns red at 10 J.
Your capacitor, step by step
- Charge: 1000 µF × 12 V = 12 mC (7.49 × 10¹⁶ electrons).
- Energy: ½ × 1000 µF × (12 V)² = 72 mJ.
Worked example: 1000 µF at 12 V
Q = 1000 µF × 12 V = 12 mC, about 7.49 × 10¹⁶ electrons. E = ½ × 1000 µF × 12² = 72 mJ: safe to handle, though enough to make a small spark. Drawing 20 mA, the charge would last 600 ms, with the voltage dropping 20 V every second — in practice a circuit that needs at least 6 V gets 300 ms. A camera flash is the opposite case: 330 µF at 330 V holds 18 J, fired in a millisecond that is an average of 109 A.
Questions
Is the charge on a capacitor really zero overall?
Yes. One plate has +Q, the other −Q; the capacitor as a whole is neutral. "The charge on a capacitor" means the amount moved from one plate to the other, Q.
Why can't a capacitor replace a battery?
A battery holds its voltage as it empties; a capacitor's voltage falls in proportion to the charge taken out, and its energy with the square. Supercapacitors (farads at a few volts) bridge short gaps and smooth pulses well, but for the same energy they are much bigger than a battery and deliver it over a sliding voltage.
Does the energy depend on the dielectric or the size?
Only through C. The dielectric and plate area set the capacitance; once C is known, Q and E follow from the voltage alone. The dielectric does set the voltage rating, which caps the energy you can actually store.
What happens to the energy when I short the terminals?
It goes into the spark and the wire as heat and light, all at once. Through a resistor it goes into the resistor over about five time constants, and the resistor must be rated for the energy as well as the peak current V / R.