Formula & Calculator
Energy Stored in a Capacitor
The electrical energy stored in the electric field of a charged capacitor.
Interpretation
Energy stored in a capacitor is E = ½ C·V² – it depends on the capacitance and the square of the voltage.
This energy is stored in the electric field between the plates.
Example: C=100µF, V=50V → E = 0.5 × 100e-6 × 2500 = 0.125 joules.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| E | Energy | J |
| C | Capacitance | F |
| V | Voltage | V |
What it means
The energy stored in a capacitor is given by E = ½ C V², where C is the capacitance and V is the voltage across it. This energy is stored in the electric field between the plates. The formula shows that the energy depends on the square of the voltage, so doubling the voltage quadruples the stored energy. Capacitors are used as energy storage elements in power supplies, flash tubes, and dynamic memory cells. In circuits, the stored energy can be released rapidly, making capacitors useful for pulse power applications. The energy density of capacitors is much lower than batteries, but they offer much faster charge/discharge rates. Understanding the energy stored is crucial for sizing capacitors in DC‑DC converters and for ensuring safety, as charged capacitors can deliver a dangerous shock. Example: A 100µF capacitor charged to 50V stores E = 0.5 * 100e-6 * 50² = 0.5 * 100e-6 * 2500 = 0.125 joules (125 mJ).
Worked example
Energy Stored in a Capacitor – Practical Example
Real‑World| Parameter | Value |
|---|---|
| C | 1000 µF = 1×10⁻³ F |
| V | 12 V |
| Formula | E = ½·C·V² |
Common mistakes
- Units: C in farads, V in volts → E in joules.
- Energy stored: It is proportional to the square of the voltage – doubling voltage quadruples energy.
- Instantaneous: This is the energy stored at a given voltage.
- Alternative: E = ½·Q·V = Q²/(2C) – useful when charge is known.
- Sign: Energy is always positive.
Applications
Energy stored in a capacitor is E = ½·C·V², which depends on capacitance and the square of voltage. This energy is stored in the electric field between the plates and is released during discharge. Engineers use this formula to design energy‑storage capacitors, to calculate the energy available for pulsed applications, and to ensure that capacitors are safely discharged. In power supplies, the stored energy is used to smooth rectified waveforms. In defibrillators, it is the energy delivered to the patient. By understanding the energy relationship, professionals can select capacitors with adequate energy rating and predict performance. This formula is essential for capacitor selection and safety.
- Capacitor selection for energy storage applications
- Power supply filtering and bulk capacitance sizing
- Pulsed power systems (defibrillators, flash lamps)
- Energy harvesting and storage circuits
- Safety discharge calculations for service personnel
Frequently Asked Questions
The energy stored is E = ½ C V², where C is capacitance and V is the voltage across the capacitor.
Joules (J) when C is in farads and V in volts.
Energy is proportional to V², so doubling voltage quadruples stored energy.
E = ½ Q² / C, since Q = CV.
In the electric field between the plates.
For a fixed voltage, inserting a dielectric increases C and thus increases stored energy (E = ½ CV²).
Limited by the breakdown voltage; energy rating is often given in Joules or watt-seconds.
In flash photography, power supply filtering, and energy storage for pulsed power applications.
Energy is stored (Joules); power is the rate of energy transfer (P = dE/dt = V·I).
Common errors include: 1) forgetting the factor 1/2, 2) using the wrong voltage (peak vs RMS for AC), 3) using the formula for a discharging capacitor incorrectly, 4) applying it to a capacitor with non-linear dielectric, and 5) confusing energy with power.