Capacitor Energy Calculator — Joules Stored, and the Capacitance or Voltage to Store Them

Background.

A charged capacitor stores energy in the electric field between its plates, and the amount is U = ½CV². This page computes it, and — because the textbook version of this problem almost always runs the other way — it also solves backwards: give it a target energy and a working voltage and it returns the capacitance you need, or give it an energy and a capacitance and it returns the voltage required. All three are the same equation rearranged, so whichever direction you pick, the full set comes back and the numbers always agree.

The square is the important part. Energy rises with the square of voltage but only linearly with capacitance, which means voltage is always the stronger lever. Doubling the capacitance doubles the stored energy; doubling the voltage quadruples it. That is why energy-storage designs push the voltage as high as the dielectric will tolerate rather than simply piling on farads, and it is also why a capacitor charged to twice its rated voltage is not twice as stressed but four times as energetic when it fails.

The default here is the classic case from OpenStax University Physics: a heart defibrillator that must deliver 400 joules at 10,000 volts. Run it forwards — 8 µF at 10 kV — and you get 400 J. Run it backwards in capacitance mode and you get the textbook's own answer, 8.00 µF, from C = 2U/V². Those 400 joules sit on 0.08 coulombs of charge, and they stay there after the supply is switched off. A capacitor is not a resistor; nothing bleeds the energy away unless something is provided to bleed it.

That leads to the safety point, which this page states beside every result rather than hiding in an FAQ. Treat any charged capacitor as live. Large or high-voltage parts can hold a dangerous charge for hours after power-down, and the standard practice is to discharge through a suitably rated bleeder resistor and then verify with a meter before touching anything. What this page deliberately does not do is tell you whether a particular number of joules is dangerous. Those thresholds live in product-safety standards — IEC 61010-1 for laboratory and measurement equipment, IEC 62368-1 for IT and audio-video equipment — they depend on voltage, accessibility and equipment class, and they are not published openly. Rather than invent a threshold, the page reports the energy and points you at the standards that classify it.

One more thing worth knowing before you use the result for a design. Charging a capacitor from a fixed voltage source through a resistor wastes exactly as much energy in the resistor as it stores in the capacitor — ½CV² each — and that is true for any resistance, large or small. Making the resistor smaller makes the charge faster, not more efficient. Getting past 50 % round-trip efficiency requires an inductive or switching converter, not a better resistor. And everything here assumes an ideal part: no equivalent series resistance, no leakage and no dielectric absorption, so the energy shown is what an ideal capacitor would give back, not what a real electrolytic will return.

What is capacitor energy calculator?

The energy stored in a capacitor is the work done to move charge from one plate to the other against the potential difference that builds up as you do it. Because that potential difference itself grows with the charge already moved — v = q/C — the work is an integral rather than a simple product: U = ∫₀^Q (q/C) dq = Q²/2C. Substituting Q = CV gives the three equivalent published forms, which OpenStax prints together as U_C = ½V²C = ½Q²/C = ½QV. The factor of one half is the signature of that integral, and it is what people drop when they compute QV and get twice the right answer. Physically the energy resides in the electric field in the dielectric, not on the plates, which is why the same expression can be rewritten as an energy density ½ε₀E² multiplied by the volume of the field. Practically, the capacitor is the fastest energy-delivery device available: it holds less energy per kilogram than any battery by orders of magnitude, but it can release what it has in microseconds, which is why capacitors and not batteries drive camera flashes, defibrillators, spot welders and pulsed lasers. A supercapacitor closes part of the energy-density gap and is specified in watt-hours for that reason, which is why this page reports that unit alongside joules.

How to use this calculator.

1. 01Pick what you want to solve for. Energy is the default; capacitance and voltage are the inverse problems, and they are the ones most textbook questions actually ask.
2. 02Enter the capacitance and choose its unit — picofarads through farads. In capacitance mode this field is ignored.
3. 03Enter the voltage the capacitor is charged to, not its rating. In voltage mode this field is ignored.
4. 04Enter the target energy in joules if you are solving for capacitance or voltage. In energy mode this field is ignored.
5. 05Read the energy in whichever magnitude suits your scale — joules for banks, millijoules for ordinary electronics, microjoules for RF parts, watt-hours for supercapacitors.
6. 06Check the charge figure if you are sizing a discharge path: Q ÷ the current your load draws gives the order of magnitude of the discharge time.
7. 07Before working on the hardware, discharge through a rated bleeder resistor and confirm with a meter. The energy shown is still there after the power is off.

The formula.

Start from what charging actually involves. To move a small amount of charge dq onto a plate that is already at potential v = q/C, the work needed is v dq. Integrating from an empty capacitor to a final charge Q gives U = ∫₀^Q (q/C) dq = Q²/2C. That integral is where the factor of one half comes from, and skipping it — computing QV instead of ½QV — is the single most common error in this topic and doubles the answer.

Substituting Q = CV into Q²/2C gives U = ½CV², and substituting only once gives U = ½QV. OpenStax prints all three together as Equation 8.10, and this page computes ½CV² then derives the other two, so they agree by construction rather than by luck. A test asserts all three agree to nine or more decimal places.

Working the default: C = 8 µF = 8×10⁻⁶ F and V = 10,000 V. V² is 1×10⁸. Multiply: 8×10⁻⁶ × 1×10⁸ = 800, and half of that is 400 J. In watt-hours that is 400 ÷ 3600 = 0.1111 Wh. The charge is Q = CV = 8×10⁻⁶ × 10⁴ = 0.08 C, which is 80,000 µC. Checking the second form: ½QV = ½ × 0.08 × 10,000 = 400 J, the same number.

The inverse modes are one line each. Capacitance from energy and voltage is C = 2U/V², which for 400 J at 10 kV gives 2 × 400 ÷ 10⁸ = 8×10⁻⁶ F — the 8.00 µF that OpenStax's defibrillator example states. Voltage from energy and capacitance is V = √(2U/C), which for 400 J in 8 µF gives √(10⁸) = 10,000 V. The square root is why negative energies are refused rather than allowed to produce a NaN.

On the exponents, checked against the algebra rather than asserted: U ∝ V² means multiplying the voltage by two multiplies the energy by four, and U ∝ C¹ means doubling the capacitance only doubles the energy. Both directions are pinned by tests. Charge behaves differently again — Q = CV is linear in both, so doubling the voltage doubles the charge while quadrupling the energy.

Rounding happens once, at the end, to ten decimal places. That precision is generous for joules but not for every unit: a 100 pF capacitor at 5 V stores 1.25 nanojoules, which is zero to ten decimals in joules and 0.00125 in microjoules. That is exactly why four magnitudes are reported. At the other end, a 3000 F supercapacitor at 2.7 V stores 10,935 J, or 3.0375 Wh.

What this does not model: equivalent series resistance, leakage, dielectric absorption, temperature and DC-bias derating. It also assumes the energy is fully recoverable, which no real dielectric manages. And it says nothing about the energy lost in getting the charge there — from a fixed supply through a resistor, that loss equals the stored energy exactly, whatever the resistance, capping resistive charging at 50 % efficiency.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax (Rice University), University Physics Volume 2, §8.3 'Energy Stored in a Capacitor'. Primary source for the equation implemented here, printed as Equation 8.10: 'U_C = 1/2 V^2 C = 1/2 Q^2/C = 1/2 QV'. The same section works the heart-defibrillator example that this page uses as its default — given U_C = 4.00×10² J at V = 1.00×10⁴ V, 'C = 2U_C/V² = 8.00 µF' — which this calculator's capacitance mode reproduces exactly. Retrieved 2026-07-29.
2. [2]OpenStax (Rice University), University Physics Volume 2, §8.1 'Capacitors and Capacitance'. Source of C = Q/V — 'the ratio of the maximum charge Q that can be stored in a capacitor to the applied voltage V across its plates' — and of the farad as 'one coulomb per one volt'. This relation is how the charge output is derived in all three modes. Retrieved 2026-07-29.
3. [3]Engineering LibreTexts / James M. Fiore, Introduction to Circuit Analysis, §6.1.2 'Capacitance and Capacitors'. Independent second authority, consulted to check the expression rather than to derive it: 'W = 1/2 CV², where W is energy in joules, C is capacitance in farads, and V is voltage in volts', plus C = Q/V. Checked on a case separate from the worked example — 100 µF at 12 V gives 7.2 mJ and 1200 µC, which this calculator returns. Retrieved 2026-07-29.
4. [4]MIT OpenCourseWare, 6.002 Circuits and Electronics, Spring 2007 (Prof. Anant Agarwal). Course materials on energy storage in the capacitor element and on the energy balance of resistive charging, which is the basis for the 50 % efficiency limit discussed above. Landing page verified to load and to list the course number, term and instructor. Retrieved 2026-07-29.