Calculate stored capacitor energy in joules from capacitance and voltage, or solve for capacitance or voltage, with F, mF, µF, nF, pF, V, mV, and kV support.
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Capacitor Energy Formula
The energy stored in a capacitor is calculated using the following primary formula:
Where E is energy in joules (J), C is capacitance in farads (F), and V is voltage in volts (V). This equation can also be expressed in two equivalent forms using the relationship Q = CV, where Q is charge in coulombs:
For an ideal capacitor with constant capacitance, these three expressions agree. Choose the form that uses the quantities you know. The Q²/(2C) relation describes a capacitor; it should not be used as a general battery-energy formula.
Why the 1/2 Factor Exists
The factor of 1/2 in the capacitor energy formula is not arbitrary. It arises from the physics of how charge accumulates on a capacitor. As charge builds on the plates, the voltage across the capacitor increases linearly from 0 to its final value V. The energy delivered at any instant is the product of the current voltage and the incremental charge added. Integrating this from zero to full charge gives E = (1/2)CV^2, not CV^2. In other words, the average voltage during charging is V/2, so the total energy is half of what it would be if the full voltage were present from the start.
Energy Density and the Electric Field
The energy stored in a capacitor physically resides in the electric field between its plates. The volumetric energy density of that field is u = (1/2) * permittivity * E_field^2, where u is energy per unit volume (J/m^3), E_field is the electric field strength (V/m), and permittivity is the product of vacuum permittivity (8.854 x 10^-12 F/m) and the dielectric constant. A higher dielectric constant means more energy stored in the same physical space. For a parallel-plate capacitor with plate area A and separation d, the total stored energy equals u multiplied by the volume A*d, which reduces back to (1/2)CV^2.
Capacitance and Voltage to Joules Reference Table
| Capacitance | Voltage (V) | Energy (J) |
|---|---|---|
| 10 uF | 24 | 0.003 |
| 22 uF | 24 | 0.006 |
| 47 uF | 12 | 0.003 |
| 100 uF | 12 | 0.007 |
| 220 uF | 12 | 0.016 |
| 470 uF | 12 | 0.034 |
| 1000 uF | 12 | 0.072 |
| 2200 uF | 12 | 0.158 |
| 4700 uF | 12 | 0.338 |
| 10 mF | 12 | 0.720 |
| 0.1 F | 5 | 1.250 |
| 0.5 F | 5 | 6.250 |
| 1 F | 2.7 | 3.645 |
| 1 F | 5 | 12.500 |
| 2.7 F | 2.7 | 9.842 |
| 5 F | 2.7 | 18.225 |
| 10 F | 2.7 | 36.450 |
| 10 F | 5 | 125.000 |
| 50 F | 2.7 | 182.250 |
| 100 F | 2.7 | 364.500 |
| * Rounded to 3 decimals. Energy computed using E = 0.5 * C * V^2. | ||
Real-World Capacitor Energy by Application
The table below contains illustrative capacitance and voltage inputs, with energy calculated using E = (1/2)CV². The application names provide context; these are not verified specifications for a particular commercial device or recommended designs.
| Application | Example C | Example V | Stored Energy |
|---|---|---|---|
| DRAM refresh (per cell) | 30 fF | 1 V | 15 fJ |
| MLCC bypass (digital IC) | 100 nF | 3.3 V | 0.54 uJ |
| Camera flash (disposable) | 120 uF | 330 V | 6.5 J |
| Camera flash (professional) | 1000 uF | 400 V | 80 J |
| External defibrillator | 100 uF | 5000 V | 1,250 J |
| Implantable defibrillator (ICD) | 120 uF | 750 V | 33.75 J |
| Supercapacitor (bus regen braking) | 63 F | 125 V | 492 kJ |
| UPS ride-through module | 5.5 F | 48 V | 6.3 kJ |
| Railgun pulse capacitor bank | 0.26 F | 22,000 V | 63 MJ |
| * Illustrative inputs only. Stored energy is not delivered pulse energy or a clinical/device specification. Actual designs require manufacturer data. | |||
Capacitors vs. Batteries: Energy Storage Comparison
Farads and joules relate to each other through the energy formula, but the resulting energy values are small compared to batteries at equivalent weight and volume. This difference reflects a fundamental tradeoff between energy density and power density.
Capacitors and batteries have different energy and power characteristics. Supercapacitors are often used for short bursts and repeated cycling, while batteries are commonly used for longer energy delivery. Energy density, power, charging time and lifetime depend on the specific technology and operating conditions; use the relevant manufacturer data for a comparison.
Dielectric Materials and Stored Energy
The dielectric material between a capacitor’s plates directly affects how many joules it can store. Inserting a dielectric with relative permittivity (dielectric constant) of value k multiplies the capacitance by k while maintaining the same plate geometry. Since E = (1/2)CV^2, a higher dielectric constant at the same voltage means proportionally more stored energy.
Dielectric properties depend on material, construction, temperature, frequency and electric field. High-permittivity ceramic capacitance can vary with temperature and DC bias. Use the part’s effective capacitance and applicable voltage rating rather than assuming nominal capacitance applies under every condition.
Practical Considerations for Capacitor Energy
The formula E = (1/2)CV^2 gives the ideal stored energy under perfect conditions. Several real-world factors reduce the usable energy available from a charged capacitor.
Equivalent series resistance (ESR) dissipates energy as heat, and leakage can reduce stored charge over time. Capacitance can also depend on voltage and temperature. Voltage derating and operating limits are part-specific: consult the manufacturer’s data rather than applying one percentage to every capacitor. The usable energy over a voltage range is ½ C(Vstart² − Vend²), before circuit losses, when capacitance is approximately constant.
Farads to Joules FAQ
Can you convert farads directly to joules?
No. Farads measure capacitance (charge stored per volt) and joules measure energy. They are different physical quantities. To convert between them, you need a third variable: voltage. The relationship is E = (1/2)CV^2. Without knowing the voltage, a capacitance value in farads cannot be expressed in joules.
How much energy does a 1-farad capacitor store?
For an ideal constant 1 F, energy at 2.7 V is 3.645 J; at 5 V it is 12.5 J; at 12 V it is 72 J. Each calculation applies only if the capacitor is rated for that voltage and retains that capacitance under those conditions.
Why does doubling the voltage quadruple the energy?
Energy depends on V squared. Doubling voltage from V to 2V means the energy becomes (1/2)C(2V)^2 = (1/2)C*4V^2 = 4 times the original energy. This is why high-voltage capacitors store disproportionately more energy than low-voltage ones of the same capacitance, and why voltage rating is often the more important specification for energy storage applications.
What happens to the energy when a capacitor discharges?
The stored electrostatic energy converts to other forms depending on the discharge circuit. In a resistive load, it becomes heat. In an LC oscillator, it cycles between electric field energy (in the capacitor) and magnetic field energy (in the inductor). In a camera flash, it becomes light and heat. In a defibrillator, it becomes the electrical pulse delivered to the patient’s heart. Some energy is always lost to ESR heating and conductor resistance during discharge.
