Energy Stored in Capacitors

A capacitor is a device that stores electrical energy by accumulating opposite charges on two conducting plates separated by an insulating material (the dielectric). When you charge a capacitor, you are essentially pumping charge from one plate to the other against an ever-increasing electric field — and the energy required to do this is stored in that field. This is analogous to stretching a spring: the more you stretch it, the harder it becomes to stretch further, and all that effort is stored as potential energy. The energy stored in a capacitor is not stored in the charges themselves, but in the electric field that exists between the plates. This is a profound idea in physics — fields can carry energy. When the capacitor discharges, that field collapses and the stored energy is released, driving current through a circuit. This is why capacitors are used in camera flashes, defibrillators, and power supplies: they can release large amounts of energy almost instantaneously. Understanding the energy stored in a capacitor requires combining the definition of capacitance (C = Q/V) with the work done to move each successive bit of charge onto the plates. Because the voltage across the capacitor increases as more charge accumulates, the total energy is not simply QV but rather one-half of it. This factor of one-half is the mathematical signature of any system where a quantity builds up proportionally against its own resistance — just like the elastic potential energy ½kx² of a spring.

Key takeaways
  • Work Done to Add a Small Charge
  • Integrating from Zero to Full Charge
  • Rewriting in Terms of Voltage
  • Energy Stored in the Electric Field

Work Done to Add a Small Charge

Suppose a capacitor already has charge q on its plates, creating a voltage v = q/C across it. To move an additional infinitesimal charge dq onto the positive plate, you must do work dW = v · dq = (q/C) dq against this voltage. Notice the voltage is not constant — it grows as q grows.

Integrating from Zero to Full Charge

To find the total work W done in charging the capacitor from q = 0 to q = Q, integrate: W = ∫₀^Q (q/C) dq = (1/C) · [q²/2]₀^Q = Q²/(2C). This integral is the area of a triangle on a v-versus-q graph, confirming the factor of ½.

Rewriting in Terms of Voltage

Since Q = CV, we can substitute to get equivalent forms: U = Q²/(2C) = ½CV² = ½QV. All three forms are correct and useful in different situations. If you know C and V, use ½CV². If you know Q and C, use Q²/(2C). The energy U is measured in joules (J).

Energy Stored in the Electric Field

For a parallel-plate capacitor with plate area A and separation d, the capacitance is C = ε₀A/d and the electric field between the plates is E = V/d. Substituting gives U = ½ε₀E²·(Ad). Since Ad is the volume between the plates, the energy density (energy per unit volume) is u = ½ε₀E². This remarkable result shows that every region of space with an electric field E contains energy — even empty space.

The Key Equation

U=12CV2=Q22C=12QVU = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV
  • U — Electric Potential Energy (J (joules)): The total energy stored in the electric field of the capacitor.
  • C — Capacitance (F (farads)): A measure of how much charge the capacitor stores per volt of potential difference.
  • V — Voltage (Potential Difference) (V (volts)): The electric potential difference between the two plates of the capacitor.
  • Q — Charge (C (coulombs)): The magnitude of charge stored on one plate of the capacitor.

Worked Example

Problem. A 470 μF capacitor is charged to a voltage of 12 V. (a) How much energy is stored in the capacitor? (b) If this energy is released in 2 ms to power a flash lamp, what is the average power delivered?

U≈33.8 mJ,P≈16.9 WU \approx 33.8 \text{ mJ}, \quad P \approx 16.9 \text{ W}

Answer. The capacitor stores approximately 33.8 mJ of energy and delivers an average power of 16.9 W during the 2 ms discharge.

Figure
Diagram
Energy Stored in a Capacitor− plate+ plate−−−−−+++++E fieldd (separation)Energy vs. VoltageVUU = ½CV²Area = stored energyU = ½CV² = Q²/2C = ½QVEnergy density: u = ½ε₀E²
Origins · 1 minMichael Faraday and how this was discovered
MF
Michael Faraday
1791–1867 · British

Faraday pioneered the experimental study of electromagnetism and electrochemistry. He introduced the concept of the electric field and demonstrated that the capacity of a capacitor depends on the geometry of its plates and the dielectric material between them. The unit of capacitance, the farad (F), is named in his honour. Faraday had almost no formal mathematical education yet became one of the greatest experimental physicists of all time. James Clerk Maxwell later translated Faraday's physical intuitions into rigorous mathematical form, producing Maxwell's equations.

  • Capacitance and Capacitors — Capacitance C = Q/V is the foundational definition used in every energy formula — you must understand capacitance before calculating stored energy.
  • Electric Potential and Voltage — Voltage (potential difference) is the key variable in U = ½CV²; understanding how voltage arises from separated charges is essential.
  • Electric Fields — The energy stored in a capacitor is actually stored in its electric field, described by the energy density u = ½ε₀E².
  • RC Circuits: Charging and Discharging — When a capacitor discharges through a resistor, the stored energy is converted to heat — connecting energy storage to circuit behaviour.