The correct answer is A because:

The time for the pd to decrease by half $(\frac{V}{2})$ is 53 ms
The energy stored E is defined by the equation
- E = $\frac{1}{2}$ QV = $\frac{1}{2}$ CV²
Since the charge Q will decrease when discharging, but capacitance C stays the same, we can use the equation

Therefore, when V becomes $\frac{V}{2}$ then ${(\frac{V}{2})}^{2}$ = $\frac{V^{2}}{4}$
- $\frac{1}{2} C \frac{V^{2}}{4} = \frac{E}{4}$
After another 53 ms, this will decrease to
- ${(\frac{V}{2})}^{2} = \frac{V^{2}}{16}$
- $\frac{1}{2} C \frac{V^{2}}{16} = \frac{E}{16}$
After another 53 ms, this will decrease to
- ${(\frac{V}{8})}^{2} = \frac{V^{2}}{64}$
- $\frac{1}{2} C \frac{V^{2}}{64} = \frac{E}{64}$
Therefore, after 53 × 3 = 159 ms the energy stored has fallen to $\frac{E}{64}$

Capacitor Charge & Discharge (AQA A Level Physics)

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