Energy stored in a capacitor

Charging a capacitor stores energy in it — energy a camera flash or a defibrillator dumps out in a fraction of a second.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §26.4.

Before you start

What you need first

Topic 12 — $C = Q/V$ and $Q = CV$.
Topic 9 — moving charge through a voltage costs energy.
Energy in joules; area of a triangle.

What you'll be able to do

Use $U = \tfrac12 CV^2$ for the stored energy.
Switch to the forms $U = \tfrac12 QV$ and $U = \dfrac{Q^2}{2C}$.
See why the energy is the area under the $Q$–$V$ line.

The result

Energy stored

Charging a capacitor means pushing charge onto the plates against a voltage that keeps rising. The total energy stored works out to:

$$ U = \tfrac12 CV^2 $$

where:
Symbol	Meaning	SI unit
U	energy stored in the capacitor	J
C	capacitance	F
V	voltage across it	V

Three forms, one energy. Using $Q = CV$ you can also write $U = \tfrac12 QV = \dfrac{Q^2}{2C}$. Pick whichever matches what you are given.

Why the factor of ½?

As you add charge, the voltage grows in proportion: $Q = CV$ is a straight line. The energy is the area under that line — a triangle, whose area is $\tfrac12 \times \text{base} \times \text{height} = \tfrac12 QV$.

The last bit of charge costs the full voltage, but the first bit cost almost nothing — so you pay the average, half the final voltage. That is where the $\tfrac12$ comes from.

$Q$ rises in step with $V$; the shaded triangle (area $\tfrac12 QV$) is the stored energy.

📐 Worked example 1

Energy from C and V

A $C = 5.0\ \mu\text{F}$ capacitor is charged to $V = 12$ V. Find the energy stored.

1Use $U = \tfrac12 CV^2$ (square the voltage: $12^2 = 144$):

$$ U = \tfrac12 (5.0\times10^{-6})(12)^2 = \tfrac12 (5.0\times10^{-6})(144) $$

2Work it out:

$$ U \approx 3.6\times10^{-4}\ \text{J} $$

📐 Worked example 2

Energy from Q and C

A $C = 10\ \mu\text{F}$ capacitor holds $Q = 50\ \mu\text{C}$. Find the energy stored.

1Here $Q$ and $C$ are given, so use $U = Q^2/2C$:

$$ U = \frac{(50\times10^{-6})^2}{2(10\times10^{-6})} $$

2Work it out:

$$ U \approx 1.25\times10^{-4}\ \text{J} $$

Same energy whichever form you use — choose the one that fits the given quantities.

✏️ Try it yourself

A camera-flash capacitor of $C = 200\ \mu\text{F}$ is charged to $V = 300$ V.

(a) Find the energy it stores.
(b) If it were charged to twice the voltage, how would the energy change?

(a) Step 1. $U = \tfrac12 CV^2 = \tfrac12 (200\times10^{-6})(300)^2.$ Step 2. $U = \tfrac12 (200\times10^{-6})(90000) \approx 9.0\ \text{J}$ (enough for a bright flash). (b) Energy goes as $V^2$, so doubling the voltage gives four times the energy — about $36$ J.

Common mistakes

Mistake	Fix
Forgetting to square $V$.	$U = \tfrac12 CV^2$ — the voltage is squared.
Dropping the $\tfrac12$.	The energy is the triangle area, so the factor of $\tfrac12$ is always there.
Mixing the three forms.	$\tfrac12 CV^2$, $\tfrac12 QV$, $Q^2/2C$ are equal — don't combine pieces of different ones.
Leaving µF or µC unconverted.	Convert to F and C before computing.

Recap — the whole topic on one screen

Idea	What you own now
Stored energy	$U = \tfrac12 CV^2 = \tfrac12 QV = \dfrac{Q^2}{2C}$.
The ½	Energy is the area under the $Q$–$V$ line (a triangle).
Scaling	Energy goes as $V^2$: double the voltage → four times the energy.

Next topic

Dielectrics

Slide an insulator into the gap and the capacitance jumps. Next we see how a dielectric boosts $C$ — and lets a capacitor hold more charge and energy.

→ Topic 15 · Dielectrics

Mistake	Fix
Forgetting to square \(V\).	\(U = \tfrac12 CV^2\) — the voltage is squared.
Dropping the \(\tfrac12\).	The energy is the triangle area, so the factor of \(\tfrac12\) is always there.
Mixing the three forms.	\(\tfrac12 CV^2\), \(\tfrac12 QV\), \(Q^2/2C\) are equal — don't combine pieces of different ones.
Leaving µF or µC unconverted.	Convert to F and C before computing.

Idea	What you own now
Stored energy	\(U = \tfrac12 CV^2 = \tfrac12 QV = \dfrac{Q^2}{2C}\).
The ½	Energy is the area under the \(Q\)–\(V\) line (a triangle).
Scaling	Energy goes as \(V^2\): double the voltage → four times the energy.