Capacitance and energy stored

Capacitance

When a p.d. V is applied, charge +Q builds up on one plate and -Q on the other. Capacitance is the charge stored per unit p.d.:

C = Q / V

The unit is the farad (F): 1 F = 1 C/V. Practical capacitors are often in µF (10^-6 F), nF or pF.

Energy stored

As a capacitor charges, V rises in proportion to Q. Work is done pushing each extra bit of charge onto the plates against the p.d. already there. The energy stored is the area under a graph of V against Q, which is a triangle:

W = (1/2)QV

Using Q = CV this becomes W = (1/2)CV² = Q² / 2C.

Why a half?

The charge is not all moved through the final p.d. V. The p.d. rises from 0 to V, so the average p.d. is V/2. If a battery supplies charge Q at e.m.f. V, it transfers QV; half is stored and half is dissipated in the circuit resistance.

Worked example

A 470 µF capacitor is charged to 12 V. Find the charge and the energy stored.

1. Q = CV = 470 x 10^-6 x 12 = 5.64 x 10^-3 C
2. W = (1/2)CV² = 0.5 x 470 x 10^-6 x 144

Answer: Q = 5.6 mC; W = 0.034 J

Check yourself

1. A 220 µF capacitor is charged to 9.0 V. Find the charge stored in mC.
   Show answer

   1.98 mC. Q = CV = 220 x 10^-6 x 9.0 = 1.98 x 10^-3 C = 1.98 mC.

2. A capacitor stores 6.0 mC at 12 V. Find its capacitance in µF.
   Show answer

   500 µF. C = Q / V = 6.0 x 10^-3 / 12 = 5.0 x 10^-4 F = 500 µF.

3. Find the energy stored in a 1000 µF capacitor charged to 6.0 V, in mJ.
   Show answer

   18 mJ. W = (1/2)CV² = 0.5 x 1.0 x 10^-3 x 36 = 0.018 J = 18 mJ.

4. A 100 µF capacitor holds 2.0 mC of charge. Find the energy stored in mJ.
   Show answer

   20 mJ. W = Q² / 2C = (2.0 x 10^-3)² / (2 x 100 x 10^-6) = 0.020 J = 20 mJ.

5. What p.d. is needed to store 0.50 J in a 2200 µF capacitor? Answer in V, to 3 s.f.
   Show answer

   21.3 V. V = sqrt(2W / C) = sqrt(1.0 / 2.2 x 10^-3) = sqrt(454.5) = 21.3 V.

6. What does the area under a graph of p.d. against charge for a capacitor represent?
   • The capacitance
   • The current
   • The energy stored
   • The time constant
   Show answer

   The energy stored. Area = V x Q, which is energy; for a triangle it is (1/2)QV.

7. The p.d. across a capacitor is doubled. What happens to the energy stored?
   • It becomes 4 times as large
   • It doubles
   • It halves
   • It stays the same
   Show answer

   It becomes 4 times as large. W = (1/2)CV², so W is proportional to V².

8. Why is the energy stored (1/2)QV rather than QV?
   • Half the charge leaks away
   • The p.d. rises from zero as the charge builds up, so the average p.d. is V/2
   • Only one plate stores charge
   • Capacitance halves while charging
   Show answer

   The p.d. rises from zero as the charge builds up, so the average p.d. is V/2. The charge is moved through an average p.d. of V/2.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 4 (WPH14) question paper, June 2024: list of data, formulae and relationships

Last updated 7 November 2026. Spotted a mistake? Tell us.