The correct answer is D because:

The energy stored, E is defined by the equations
- E = $\frac{1}{2}$ QV = $\frac{1}{2}$ CV² = $\frac{Q^{2}}{2 C}$
Since we have been given the capacitance C and charge Q, we can use the equation
- E = $\frac{Q^{2}}{2 C}$
The change in energy stored is given by the equation
- ΔE = $\frac{1}{2 C}$ ( $Q_{2}^{2} - Q_{1}^{2}$ ) where $Q_{1}$ is the initial charge and $Q_{2}$ is the final charge
Substituting in the values gives
- ΔE = $\frac{1}{2 \times (12 \times 10^{- 9})}$ ((80 × 10^–6)² – (60 × 10^–6)²) = 0.12 J

A is incorrect as 12 × 10^–6 F (12 µF) instead of 12 × 10^–9 F (12 nF) has been substituted. ( $Q_{2} - Q_{1}$ )² has been used instead of ( $Q_{2}^{2} - Q_{1}^{2}$ )

B is incorrect as 12 × 10^–6 F (12 µF) instead of 12 × 10^–9 F (12 nF) has been substituted.

C is incorrect as ( $Q_{2} - Q_{1}$ )² has been used instead of ( $Q_{2}^{2} - Q_{1}^{2}$ )

Capacitance (AQA A Level Physics)

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