A capacitor of capacitance 120 µF is charged and then discharged through a 20 kΩ resistor - AQA - A-Level Physics - Question 21 - 2017 - Paper 2

The charge on the capacitor after a time ( t ) can be calculated using the formula:

$Q(t) = Q_0 e^{-t/\tau}$

where ( Q_0 ) is the initial charge.

To find the fraction of the initial charge remaining, we can determine it as follows:

$\frac{Q(t)}{Q_0} = e^{-t/\tau}$

Substituting ( t = 4.8 ) seconds and ( \tau = 2.4 ) seconds:

$\frac{Q(t)}{Q_0} = e^{-4.8/2.4} = e^{-2}$

Now calculating:

$e^{-2} \approx 0.1353$

Thus, the fraction of the original charge remaining is approximately ( 0.14 ).

Based on the calculation, the fraction of the original charge that remains after 4.8 seconds is approximately 0.14. Therefore, the correct answer is:

A: 0.14