About Capacitor Charge & Energy Calculator — Q = CV, E = ½CV²
A charged capacitor holds two numbers worth knowing: the charge Q = C·V, which sets how much current it can deliver for how long, and the energy E = ½·C·V², which sets how big a bang — or how many seconds of backup — it stores. This calculator computes both from the capacitance and voltage, with a unit select that spans the real capacitor world from picofarad ceramics to multi-thousand-farad supercapacitors, and prints results with engineering prefixes (2.5 mC, 31.25 mJ) rather than raw exponents.
The ½ in the energy formula is the part intuition misses: it is not half the charge, it is the price of charging. Each increment of charge is pushed onto the plates against the voltage already there, which grows from 0 to V — so the average push is V/2, and the stored energy is Q·V/2 = ½CV², exactly half of what Q·V would suggest. The same square law explains why doubling the voltage quadruples the stored energy, and why a camera-flash capacitor at 330 V is dangerous while the same capacitance at 5 V is a toy.
How It Works
- Enter the capacitance and pick its unit. The select covers pF (RF and timing ceramics), nF (decoupling, filters), µF (electrolytics, motor-run), mF (bulk storage), and full farads (supercapacitors). Switching units converts your entry in place — 100 µF becomes 0.1 mF, not an error.
- Enter the capacitor voltage. For energy questions use the actual charged voltage, not the rating printed on the can — a 450 V capacitor charged to 325 V (rectified 230 V mains) stores ½CV² at 325 V.
- Read Q = C·V: the charge in coulombs, with the prefix chosen automatically. Charge is the quantity that pays for constant-current loads: a 1 F supercap delivering 10 mA drops 10 mV every second (I = C·dV/dt).
- Read E = ½CV²: the stored energy in joules and its watt-hour equivalent (E ÷ 3600). Watt-hours make supercap-vs-battery comparisons honest: 12.5 J at 5 V is just 3.5 mWh — a coin cell holds about seventy times more.
- Remember the square law when derating: charging the same capacitor to half the voltage stores a quarter of the energy. That is also the discharge-safety rule in reverse — most of the energy lives in the top half of the voltage.
Worked Example
A 100 µF electrolytic charged to 25 V: C = 100 × 10⁻⁶ = 10⁻⁴ F, so the charge is Q = 10⁻⁴ × 25 = 2.5 × 10⁻³ C = 2.5 mC, and the energy is E = ½ × 10⁻⁴ × 25² = ½ × 10⁻⁴ × 625 = 31.25 mJ — enough to blink an LED, not enough to feel. Scale up to a 1 F supercapacitor at 5 V and the numbers change character: Q = 5 C (two thousand times more charge) but E = ½ × 1 × 25 = 12.5 J, which is still only 3.47 mWh — the supercap can dump its charge far faster than any battery, but it stores far less total energy than its physical size suggests.
Formulas
- Stored charge
Q = C × V- Stored energy
E = ½ × C × V² = Q²/(2C) = Q·V/2- Watt-hour equivalent
E_Wh = E / 3600
Standards & References
- Q = CV and E = ½CV² are the SI defining relations of capacitance (farad = coulomb/volt); the energy integral ∫q/C dq = Q²/2C appears in any physics text (e.g. Halliday, Resnick & Walker, Fundamentals of Physics)
- Unit factors pF/nF/µF/mF are exact powers of ten by SI prefix definition
- Safety note: stored energy above a few joules at more than ~50 V can deliver a hazardous shock or arc — discharge through a resistor, never a screwdriver
Frequently Asked Questions
Why is the energy ½CV² and not just Q·V = CV²?
Because the capacitor is not charged against a constant voltage. The first coulomb arrives when the plates are near 0 V and costs almost nothing; the last arrives against the full V. The work is the integral of q/C dq from 0 to Q, which is Q²/2C = ½CV² — the average charging voltage is exactly V/2. The "missing" half is real: charge a capacitor from a fixed supply through any resistor and exactly ½CV² is dissipated in the resistor, no matter its value.
How much energy does a typical capacitor actually store?
Far less than intuition says. A 100 µF/25 V electrolytic: 31 mJ. A hefty 470 µF at 325 V (rectified mains): about 25 J — genuinely dangerous, but still only 7 mWh. A 3,000 F supercapacitor at 2.7 V: about 10.9 kJ = 3 Wh, roughly one percent of a phone battery. Capacitors win on power (how fast) and cycle life, not on energy (how much) — that is why they buffer and batteries store.
Is a charged capacitor dangerous?
It depends on voltage and joules together. Below ~50 V the shock hazard is minimal regardless of size, though a big supercap can still weld a wrench across its terminals — 5 C of charge is a lot of current for a short time. Mains-side capacitors (hundreds of volts, tens of joules) can deliver a painful to hazardous shock long after power-off, which is why bleeder resistors exist. Discharge through a resistor sized for the energy (a 1 kΩ / 5 W resistor handles typical PSU caps), and verify with a meter.
What is the difference between coulombs and joules here?
Coulombs count charge — how many electrons' worth the capacitor moved; joules count energy — the work that charge can do. They answer different questions: charge governs voltage droop under constant current (ΔV = I·Δt/C, straight from Q = CV), while energy governs heating, flash brightness, and backup runtime. Two capacitors can hold the same charge and very different energy: 1 F at 5 V and 0.2 F at 25 V both hold 5 C, but the second stores five times the joules.
Can I use this for capacitors in series or parallel?
Yes — combine first, then enter the equivalent. Parallel capacitances add (C = C₁ + C₂ + …) and every unit sees the same voltage. Series combine as the reciprocal sum (1/C = 1/C₁ + 1/C₂), the same charge Q lands on every capacitor, and the voltage divides inversely with capacitance — which is why series supercap stacks need balancing: the smallest capacitance ends up with the biggest share of the voltage.