A charged capacitor stores energy: a camera flash or a defibrillator dumps it in a fraction of a second. How much? The answer comes from the same idea as work as the area under a graph.
Why it works
Charge the capacitor a little at a time. When it already holds charge $q$, its voltage is $v = q/C$, and moving the next small bit $\Delta q$ across takes work $v\,\Delta q$. At first $v = 0$ and it costs nothing; near the end each bit costs almost $V$. On a graph of $v$ against $q$, the work is the area under a straight line from $0$ to $(Q, V)$: a triangle with area $\tfrac{1}{2}QV$.
Key idea
$$U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}$$ Use whichever form matches what you know. At fixed $C$, doubling the voltage stores four times the energy.
Worked examples
Worked example: From C and V
How much energy does a $100\,\mu\text{F}$ capacitor store at $12\,\text{V}$?
- $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(100 \times 10^{-6})(12)^2 = 7.2 \times 10^{-3}\,\text{J} = 7.2\,\text{mJ}$.
Answer: $7.2\,\text{mJ}$
Worked example: From Q and V
A capacitor holds $30\,\mu\text{C}$ at $50\,\text{V}$. Find its stored energy.
- $U = \tfrac{1}{2}QV = \tfrac{1}{2}(30 \times 10^{-6})(50) = 7.5 \times 10^{-4}\,\text{J} = 0.75\,\text{mJ}$.
Answer: $0.75\,\text{mJ}$
Worked example: Double the voltage
A capacitor stores $5.0\,\text{mJ}$. How much does it store at twice the voltage?
- $U = \tfrac{1}{2}CV^2$ with $C$ fixed: twice $V$ gives $2^2 = 4$ times $U$.
- $U = 4 \times 5.0 = 20\,\text{mJ}$. (The charge doubled too, and so did the average voltage it was moved through.)
Answer: $20\,\text{mJ}$
Common mistake
Using $U = QV$, as if every bit of charge were moved through the full voltage $V$. The voltage builds up from zero as the capacitor charges, so on average each bit crosses only $V/2$. Check: the energy is a triangle under the $V$–$q$ graph, not a rectangle.
A battery charging a capacitor through a resistor gives out $QV$, but the capacitor keeps only $\tfrac{1}{2}QV$; the other half becomes heat in the resistance while the current flows.