Energy Stored in a Capacitor

Equivalent forms

U = \frac{Q^2}{2C}

U = \tfrac{1}{2} Q V

The factor of 1/2 is the universal signature of energy stored in a linear restoring system — the same 1/2 appears in (1/2)kx², (1/2)mv², and (1/2)LI².

Symbol	Name	SI unit	Dimension	Range
$U$	Stored energyoutput Electrostatic energy stored in the capacitor	J	M·L²·T⁻²	0 – 100
$C$	Capacitance Capacitance of the capacitor	F	M⁻¹·L⁻²·T⁴·I²	0.000001 – 0.001
$V$	Voltage Potential difference across the capacitor	V	M·L²·T⁻³·I⁻¹	0 – 500

Unit systems

Symbol	SI	CGS	Imperial
$U$	J	erg	ft·lbf
$C$	F	cm	F
$V$	V	statV	V

Where it holds

Valid for linear

(Q = CV)

capacitors at any voltage below dielectric breakdown. Fails for nonlinear dielectrics or once breakdown ionizes the medium.

Dimensional analysis

$[U] = [C][V]^{2} \to (F)(V^{2}) = (C/V)(V^{2}) = C\cdot V = J \checkmark$

Discovery

Alessandro Volta / Michael Faraday · 1745

Following the invention of the Leyden jar in 1745, 18th-century experimenters realized that storing charge stored energy. The modern energy formula was formalized after Faraday's capacitance concept (1830s).

Try this

How does a camera flash store enough energy to light up a room for a millisecond?

A 220 μF capacitor is charged to 300 V. How much energy does it store, and how much power is delivered if it discharges in 2 ms?

Research status: stable

Real-world applications

Camera flash units and defibrillators
DRAM memory cells storing 1 bit as charge
Pulsed lasers and railgun energy banks
Power-factor correction in AC grids

Common misconceptions

Energy is not stored on the plates — it lives in the electric field between them
$U = QV$ is wrong by a factor of 2 — the average voltage during charging is V/2
Connecting two charged capacitors in parallel loses half the energy as heat/radiation, even with zero resistance

Experimental verification

Discharge a known capacitor through a resistor and integrate

I^{2}R dt

over time; the result matches

(1/2)CV^{2} to

within instrumentation error. Calorimetric measurements in flash-discharge experiments confirm the same.

Derivation

Charge up incrementally: at intermediate charge q the voltage is

v = q/C

.

Work to add dq is

dW = v dq = (q/C) dq

.

Integrate from 0 to Q:

U = Q^{2}/(2C) = (1/2) C V^{2}

.

Limiting cases

V \to 0

⟶

U \to 0

No voltage means no charge separation, hence no stored energy.

V \to 2V

⟶

U \to 4U

Energy scales quadratically with voltage — doubling V quadruples U.

C \to \infty

⟶

U \to \infty

(at fixed V)An infinite-capacity reservoir at finite voltage holds unlimited energy — ideal supercapacitor limit.

What if…

What if the capacitance is halved while keeping V fixed?

Stored energy halves. U scales linearly with C at fixed V.

What if charge Q is held constant while a dielectric

(\kappa = 3)

is inserted?

Energy decreases by factor $\kappa = 3$ because the voltage drops to $V/\kappa$ ; the dielectric is pulled in, doing work on the system.

What if two identical charged capacitors are connected in parallel?

Total voltage stays the same, total capacitance doubles, but only half the original energy survives — the rest is dissipated as heat or EM radiation.

1

Camera flash energy

Given ·

C:: 0.00022
V:: 300

Find · U

Steps

Identify: $C = 220 \mu F = 2.2\times 10^{-4} F$ , $V = 300\,\mathrm{V}$
Apply $U = (1/2) C V^{2}$
Compute $V^{2} = 90 000$
Multiply: $U = 0.5 \times 2.2\times 10^{-4} \times 9\times 10^{4} = 9.9\,\mathrm{J}$
Power if released in 2 ms: $P = U/\Delta t = 9.9 / 0.002 \approx 4.95\,\mathrm{kW}$

Result ·

U = (1/2)(220\times 10^{-6})(300^{2}) = 9.9\,\mathrm{J}

2

Energy after voltage doubles

Given ·

C:: 0.0001
V:: 50

Find · U when V doubles to 100 V

Steps

Initial: $U_{1} = (1/2)(10^{-4})(50^{2}) = 0.125\,\mathrm{J}$
After $V \to 2V$ : $U_{2} = (1/2)(10^{-4})(100^{2}) = 0.5\,\mathrm{J}$
Ratio $U_{2}/U_{1} = 4$ — quadratic scaling confirmed

Result ·

U_{1} = 0.125\,\mathrm{J}

,

U_{2} = 0.5\,\mathrm{J}

— energy quadruples

parallel plate capacitorCoulomb's Law→Energy Stored in an Inductor→