Energy From Charge - Electric Fields, Electrostatics & Capacitors Step by Step

Capacitor energy can be read from the charge moved onto the plates and the voltage reached, with the one-half factor coming from the rising voltage during charging. The 12 C, 6 V, and 36 J ledger is exact for an ideal capacitor; real parts have leakage, equivalent series resistance, dielectric absorption, voltage limits, tolerance, and measurement uncertainty.

A charged ideal capacitor stores energy equal to one half of charge times voltage. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Use charge and voltage

The capacitor stores 12 coulombs at 6 volts.

Q = 12\ \text{C}\qquad V = 6\ \text{V}

Energy is half charge times voltage

For an ideal capacitor, stored energy is one half of charge times voltage.

U = \tfrac{1}{2} QV

Compute stored energy

One half of 12 times 6 gives 36 joules.

U = \tfrac{1}{2}\cdot 12\ \text{C}\cdot 6\ \text{V} = 36\ \text{J}

Charge and voltage both enter the budget

The checked diagram is the middle row. Other exact rows show the same one-half charge-voltage budget without using decimals.

\begin{array}{c|c|c}Q&V&U\\6\ \text{C}&4\ \text{V}&12\ \text{J}\\12\ \text{C}&6\ \text{V}&36\ \text{J}\\18\ \text{C}&8\ \text{V}&72\ \text{J}\\\end{array}