Capacitor Energy Calculator
Enter any two of capacitance, voltage and stored energy — the calculator returns the third using E = ½ C V², and also reports the charge Q = C·V on the plates.
Energy
0.0072 J
- In millijoules
- 7.2000 mJ
- In microjoules
- 7200.0000 µJ
- In watt-hours
- 2.000000e-6 Wh
- Stored charge Q = C·V
- 0.0012 C
- Capacitance used
- 1.0000e-4 F
- Voltage used
- 12.0000 V
E = ½ C V². The factor of one half comes from integrating the work done as the capacitor charges from 0 to V against its own rising voltage.
How to use this calculator
Pick what you want to solve for from the dropdown — energy (E), capacitance (C), or voltage (V). Then fill in the other two quantities; the unused field is ignored, so you can leave its default in place. Capacitance is entered in farads (F): a 100 µF electrolytic is 0.0001 F or 1×10⁻⁴; a 22 nF ceramic is 2.2×10⁻⁸; a 33 pF trimmer is 3.3×10⁻¹¹. Voltage is in volts. Energy is in joules (1 J = 1 W·s). The breakdown converts the answer to the SI prefixes that make practical sense — microfarads down to picofarads, millijoules down to microjoules — and also shows the stored charge Q = C·V in coulombs so you can sanity-check against current pulses (Q = I·t) when designing flash circuits, defibrillators or motor-start banks.
How the calculation works
A capacitor stores energy in the electric field between its plates. As you push charge onto one plate it raises that plate’s potential by V = q/C, and pushing the next increment dq across that growing potential costs work dW = V·dq = (q/C)·dq. Integrating from 0 to the final charge Q gives U = Q²/(2C); substituting Q = CV produces the three equivalent forms U = ½ C V² = ½ Q V = Q²/(2C). The factor of one half is not an approximation — it comes directly from the linear voltage rise during charging, just as the kinetic energy of a mass accelerated from rest has the same factor for the same reason. All three forms hold exactly for an ideal linear capacitor (constant C, no dielectric loss). Real electrolytics and ceramics have effective series resistance (ESR), dielectric absorption and voltage-dependent capacitance, but for design-level energy budgeting E = ½CV² is the correct first-order figure. The energy is stored in the electric field, with energy density u = ½ ε E² where ε is the dielectric permittivity and E here is the field strength (not the total stored energy).
Worked example
A 100 µF electrolytic capacitor charged to 12 V — a common smoothing cap on a 12 V rail — stores E = 0.5 × 100×10⁻⁶ × 12² = 7.2 mJ, with charge Q = 100×10⁻⁶ × 12 = 1.2 mC on each plate. That is enough energy to deliver a 1.2 A current pulse for 1 ms at the rated voltage. Scaling up: a defibrillator’s 32 µF cap charged to 5 000 V stores E = 0.5 × 32×10⁻⁶ × 5000² = 400 J — roughly the same as a 1 kg object dropped 40 m, delivered through the chest in milliseconds. Scaling down: a 33 pF trimmer at 5 V stores just E = 0.5 × 33×10⁻¹² × 25 = 4.1×10⁻¹⁰ J = 0.41 nJ — negligible by itself, but the tank-circuit oscillations a trimmer enables can carry far more cycled energy.
Frequently asked questions
Why is there a factor of ½ in E = ½ C V²?
Because the voltage across the capacitor rises linearly with the charge already on it. The first electrons cross at almost zero potential difference and cost almost no work; the last electrons have to push against the full final voltage. Integrating V dq from 0 to Q against V = q/C gives ½ Q V — the average of zero and V times the total charge. The factor of one half is the same reason that the kinetic energy of a mass accelerated from rest is ½ m v² rather than m v² — both come from integrating a linearly-rising quantity.
What is the difference between energy and charge stored on a capacitor?
Charge Q = C·V is what physically sits on the plates — measured in coulombs. Energy E = ½ Q V = ½ C V² is the work you had to do to put it there — measured in joules. They are related but distinct: doubling the voltage doubles the charge but quadruples the energy, which is why high-voltage capacitor banks deliver so much more punch than low-voltage ones of similar capacitance. When you discharge through a load, charge flows as current and energy flows as power; their integrals match charge and energy respectively.
How big is a capacitor in farads, microfarads, nanofarads, picofarads?
One farad is enormous — a typical 1 F supercapacitor is a small puck the size of a watch battery, and a 1 F electrolytic would fill a soda can. Standard electrolytic capacitors range from about 1 µF (one millionth of a farad) to 10 000 µF. Film and ceramic capacitors typically run 1 nF (10⁻⁹ F) to 1 µF (10⁻⁶ F). RF tuning capacitors and integrated-circuit decoupling drop into picofarads (10⁻¹² F). The earth’s ionosphere acts as a capacitor against the surface with capacitance around 0.05 F. Supercapacitors used in regenerative braking and short-term backup commonly reach 1–3000 F.
Can I exceed a capacitor’s rated voltage to store more energy?
No. Exceeding the rated working voltage causes dielectric breakdown — the insulating material between the plates ionises and conducts, usually destroying the part and often with a loud bang. Electrolytic capacitors fail catastrophically and can vent or explode; ceramic capacitors crack and short. Always derate: aim for a working voltage of 60–80% of the rated value to allow margin for transients. If you need more energy, use a larger capacitance, a higher-voltage-rated part, or capacitors in series (which adds rated voltages but reduces total capacitance).
How does capacitor energy compare to battery energy?
A typical lithium-ion 18650 cell stores around 40 kJ (11 Wh). The largest commercial 3 000 F supercapacitor at 2.7 V stores only E = 0.5 × 3000 × 2.7² ≈ 11 kJ — about a quarter as much, in a similar volume. Capacitors win on power density (how fast they can dump energy) by factors of 10–100, but lose badly on energy density. That is why capacitors handle ignition pulses, defibrillator shocks, camera flash and motor start surges, while batteries handle the long haul. Capacitors discharge in milliseconds; batteries in hours.
How do capacitors in series and parallel combine for stored energy?
In parallel the capacitances add (C_total = C₁ + C₂ + …) and so does the stored energy at a given voltage (E_total = ½ C_total V²). In series the reciprocals add (1/C_total = 1/C₁ + 1/C₂ + …), reducing total capacitance, while the rated voltage capability adds. For energy budgeting, the total stored energy of a series stack is the sum of the individual stored energies only when the voltage is distributed in inverse proportion to capacitance — for equal-value capacitors in series, each stores the same energy and the total is N × ½ C V_each² with V_total = N × V_each. Capacitor banks for pulse-power applications almost always mix series-parallel topology to hit both the rated voltage and the target energy.