Energy stored in a capacitor

CIE A-Level Physics· 22.3· 15 min read

1. Origin and Derivation of Energy Stored★★☆☆☆⏱ 5 min

When charging a capacitor, the battery does work to move negative charge from the positive plate to the negative plate, separating charge against electrostatic attraction. This work done is stored as electrostatic potential energy in the capacitor.

📘 Definition

Stored Energy in a Capacitor

$W$ or $E$

Total work done by the power source to separate charge onto the capacitor's plates, stored as potential energy that can be fully released when the capacitor discharges.

🔬 Derivation

Goal:

Derive the general formula for stored energy

Starting from:

Work done to add a small charge $d q$ at instantaneous voltage $v$

1
For a capacitor with charge $q$ , voltage $v = \frac{q}{C}$ . Work to add $d q$ :
2
$dW = v\ dq = \frac{q}{C} dq$
3
Integrate from 0 to total final charge $Q$ :
4
$W = \int_0^Q \frac{q}{C} dq = \frac{1}{C} \left[\frac{q^2}{2}\right]_0^Q$

Result:

This gives the base formula: $W = \frac{Q ^{2}}{2 C}$ , which can be rewritten using $Q = C V$ to get the three equivalent forms below.

Exam tip:

You may be asked to derive this formula in Paper 2, so memorize the integration step-by-step.

2. Calculations Using the Three Equivalent Formulas★★★☆☆⏱ 6 min

We can rearrange the base formula using $Q = C V$ to get three different forms, so you can pick the one that matches the given values in the question: $W = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{1}{2} \frac{Q ^{2}}{C}$ .

📐 Worked Example

A 500 μF capacitor is charged to 12 V. Calculate the energy stored.

1
We are given $C$ and $V$ , so use $W = \frac{1}{2} C V^{2}$ . Convert C to farads:
2
$C = 500 \times 10^{-6} = 5 \times 10^{-4}\ F$
3
Substitute values:
4
$W = \frac{1}{2} \times 5 \times 10^{-4} \times (12)^2 = 0.5 \times 5 \times 10^{-4} \times 144$
5
Final result: $W = 3.6 \times 1 0^{- 3}$ J = 3.6 mJ

✓ Quick check

Pick the most efficient formula for this problem: What is the energy stored in a 2 μF capacitor holding $4 \times 1 0^{- 4}$ C of charge?

Which formula requires the least extra calculation?
- A. $W = \frac{1}{2} C V^{2}$
- B. $W = \frac{1}{2} \frac{Q ^{2}}{C}$
- C. $W = \frac{1}{2} Q V$
Reveal answer
B —
Correct! Q and C are given directly, so this formula needs no extra steps. Using A or C requires calculating V first, adding unnecessary work and increasing error risk.

3. Energy in Combinations and Dielectrics★★★☆☆⏱ 7 min

Total energy stored in any combination of capacitors (series or parallel) is always the sum of the energy stored in each individual capacitor. When a dielectric is inserted into a capacitor, energy changes depend on whether the capacitor is connected to a battery or isolated.

📐 Worked Example

A 2 μF and 4 μF capacitor are connected in parallel across a 6 V battery. Find the total stored energy.

1
Calculate total capacitance for parallel:
2
$C_{total} = C_1 + C_2 = 2 + 4 = 6\ \mu F = 6 \times 10^{-6}\ F$
3
Use the energy formula for total capacitance:
4
$W_{total} = \frac{1}{2} C_{total} V^2 = 0.5 \times 6 \times 10^{-6} \times 6^2$
5
Final result: $W_{t o t a l} = 108 \times 1 0^{- 6}$ J = 108 μJ. This matches summing energy of each individual capacitor, as expected.

Exam tip:

Always state whether the capacitor is isolated or connected when explaining energy changes for dielectric questions.

4. Common Pitfalls

Wrong move:

Forgetting the factor of $\frac{1}{2}$ and using $W = C V^{2}$ or $W = Q V$

Why:

Voltage increases linearly from 0 to V during charging, so average voltage is $\frac{V}{2}$ , not V

Correct move:

Always include the factor of $\frac{1}{2}$ in any energy calculation for capacitors

Wrong move:

Using capacitance in microfarads directly without converting to farads

Why:

Energy calculations in joules require capacitance in base units (farads), so the result will be 10⁶ times too large if you use μF

Correct move:

Always convert prefixes: 1 μF = 1×10⁻⁶ F, 1 nF = 1×10⁻⁹ F before substituting

Wrong move:

Using reciprocal addition for total energy in series capacitors

Why:

Some students incorrectly copy the series capacitance rule for energy, but energy is additive always

Correct move:

Total energy = sum of individual energies for both series and parallel combinations: $W_{t o t a l} = W_{1} + W_{2}$

Wrong move:

Assuming energy always increases when a dielectric is inserted

Why:

Energy change depends on whether Q or V is held constant, which is determined by the circuit connection

Correct move:

Check for connection: isolated (Q constant) → W decreases, connected to battery (V constant) → W increases

5. Quick Reference Cheatsheet

Formula / Rule	When to use
$\frac{1}{2} C V^{2}$	Given C and V
$\frac{1}{2} Q V$	Given Q and V
$\frac{1}{2} Q^{2} / C$	Given Q and C
$W_{t o t a l} = W_{1} + W_{2}$	Any series/parallel combination
W decreases	Dielectric inserted into isolated (disconnected) cap
W increases	Dielectric inserted into battery-connected cap

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 12
MCQ: Energy stored change after dielectric
2023 · 22
Calculation: Total energy parallel caps
2024 · 13
MCQ: Energy for given charge and cap

Going deeper

SyllabusCIE 9702 A-Level Physics SyllabusCheck section 22.3 for official learning outcomes

What's Next

Understanding energy stored in capacitors is a critical foundation for the next topic of capacitor charging and discharge through resistors, which is one of the most frequently tested topics in CIE A-Level Physics Paper 2. The energy concept also connects to electric field energy, a key idea in advanced electromagnetism, and is used to solve practical problems involving flash circuits, defibrillators, and timing circuits. Mastery of this sub-topic also helps you correctly answer common multiple choice questions about energy changes when capacitor configurations are altered, which appear almost every year in Paper 1.