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Capacitance and stored electric potential energy

Key idea: H2 Physics lessons on point-charge interactions, uniform-field motion and capacitor energy.

  • GCE A-Level H2 Physics 2027

Build the idea

Learn the idea

Big question: How does a capacitor store energy as charge separates?

Capacitance C = Q/V measures charge stored per potential difference. During charging, V rises in proportion to Q for a fixed capacitor, so the area under a V–Q graph gives stored energy: E = 1/2QV = 1/2CV² = Q²/(2C). The factor one half appears because voltage grows from zero to its final value.

Interpret capacitance as a ratio

Capacitance C = Q/V tells how much charge magnitude is stored on either plate per potential difference. For a fixed linear capacitor, C is constant while Q and V change together; capacitance is not the amount of charge currently stored.

The unit is the farad: 1 F = 1 C V⁻¹. Rearrange Q = CV only after checking that charge, capacitance and potential difference refer to the same capacitor.

Check your understanding: A 4.0 µF capacitor has a potential difference of 12 V. What charge magnitude is stored?

Q = CV = (4.0×10⁻⁶)(12) = 4.8×10⁻⁵ C, or 48 µC.

Accumulate energy as voltage rises

Moving a small charge dQ onto the plates requires work V dQ. Since V grows from zero to its final value Q/C, the work is the triangular area under the V–Q graph: U = ½QV.

Using Q = CV gives U = ½CV² = Q²/(2C). The factor one half arises because the charging voltage is not at its final value throughout. Energy is associated with the electric field between the plates.

Check your understanding: Why is stored energy not simply QV?

During charging, voltage rises from zero to V, so the average voltage for a linear capacitor is V/2.

Energy stored by a capacitorA voltage–charge graph forms a triangle whose area is the work done in charging.area = ½QVgradient = 1/CQV
Scroll diagram horizontally to read all labels.
U is the area under V against Q: ½QV = ½CV² = Q²/(2C).

Key ideas to keep

  • Capacitance is a component property, not the instantaneous charge.
  • Energy is area under V against Q, not Q against V unless axes are handled carefully.
  • Disconnecting a charged capacitor keeps Q fixed in the ideal model, not necessarily V.

See the reasoning

Worked example

Recover charge and capacitance from stored energy

Question: A capacitor stores 18 mJ at 30 V. Find C and Q using two equivalent energy equations.

  1. Step 1: Choose the form containing C and V

    Why: Energy and voltage are given.

    Working: From U = ½CV², C = 2U/V².

  2. Step 2: Calculate capacitance

    Why: Convert 18 mJ before substituting.

    Working: C = 2(18×10⁻³)/30² = 4.0×10⁻⁵ F = 40 µF.

  3. Step 3: Find and cross-check charge

    Why: Q = CV and ½QV should return the given energy.

    Working: Q = (40×10⁻⁶)(30) = 1.20×10⁻³ C.

Answer: From U = ½CV², C = 2U/V² = 4.0 × 10⁻⁵ F = 40 μF. Then Q = CV = 1.20 × 10⁻³ C. Check: ½QV = 18 mJ.

Check: ½(1.20×10⁻³)(30) = 18 mJ, matching the question.

Use a hint if needed

Practise with support

Try this

A 20 μF capacitor holds 0.50 mC. Find V and U.

Hint: Convert both prefixes before using C = Q/V.

Check your answer

V = Q/C = 25 V. U = ½Q²/C = 6.25 × 10⁻³ J.

Now work without the hint

Practise independently

Your turn

Derive the three capacitor-energy forms from the V–Q graph and C = Q/V.

Check your answer

For constant C, the V–Q graph is a straight line from the origin, so its area is U = ½QV. Substituting V = Q/C gives U = ½Q²/C; substituting Q = CV gives U = ½CV².

Avoid these traps

Common mistakes

Common mistake

A capacitor stores charge on only one plate.

What is wrong with this reasoning?

Show better thinking

Equal and opposite charges reside on its two plates; Q denotes the magnitude on either plate.

Common mistake

Stored energy is QV because every increment of charge crosses the final p.d.

What is wrong with this reasoning?

Show better thinking

The p.d. rises during charging, so energy is the triangular V–Q area ½QV for constant capacitance.

Write for the examiner

Exam guidance

Draw the linear V–Q graph and use its triangular area when explaining capacitor energy.

Exam-style practice [6 marks]

A 15 μF capacitor is at 20 V. Calculate Q and U, and explain the factor ½.

Plan before you answer

  • Find Q from CV.
  • Find energy with one equivalent form.
  • Explain the triangular graph area.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

Q = CV = 3.0 × 10⁻⁴ C. U = ½CV² = 3.0 × 10⁻³ J. The charging p.d. grows linearly from zero to V, so the area under the V–Q graph is triangular.

Come back in three days

Check what stayed with you

Recall question

At fixed capacitance, voltage doubles. State the charge and energy factors.

Check the answer

Q = CV doubles. U = ½CV² increases by a factor of four.

Keep moving

Try this next

Use the longer mixed questions to connect the ideas, calculations and diagrams from this topic.

Open Electric Fields structured practice

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. Topic 14 states no explicit exclusions. Coulomb-force, point-field and point-potential equations are used for point charges in free space or air. Electric field and force are vectors; potential and potential energy are scalars. E = ΔV/d is restricted to a uniform field, while E = −dV/dr is the local negative potential gradient.

  • GCE A-Level H2 PhysicsTopic 14(j) / Topic 14(k) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027