A fully charged capacitor of capacitance 2.0 mF discharges through a 15 kΩ resistor - AQA - A-Level Physics - Question 21 - 2022 - Paper 2

To determine the fraction of stored energy that remains in the capacitor after discharging, we can use the formula for the energy stored in a capacitor:

E = rac{1}{2} C V^2

Where:

• $E$ is the energy stored,
• $C$ is the capacitance (2.0 mF = 2.0 \times 10^{-3} F), and
• $V$ is the voltage across the capacitor.

As the capacitor discharges through a resistor, the voltage across it decreases exponentially as represented by: $V(t) = V_0 e^{-\frac{t}{RC}}$

In the given problem, we need to find the time constant $\tau = RC$:

• Resistance $R = 15 kΩ = 15 \times 10^{3} Ω$
• Capacitance $C = 2.0 \times 10^{-3} F$

Calculating the time constant: $\tau = RC = (15 \times 10^{3})(2.0 \times 10^{-3}) = 30 s$

After 1.0 minute (60 s), we can find the fraction of the voltage that remains: $V(60s) = V_0 e^{-\frac{60}{30}} = V_0 e^{-2}$

The energy at $t = 60 s$ will therefore be: $E(60s) = \frac{1}{2} C (V_0 e^{-2})^2 = \frac{1}{2} C V_0^2 e^{-4}$

Thus, the fraction of the stored energy that remains is: $\text{Fraction remaining} = \frac{E(60s)}{E(0)} = e^{-4}$

Since e^{-4} is a common mathematical constant, the closest answer that applies is: $e^{-4} \approx \frac{1}{e^2}$

Thus, the correct option from the choices provided is D: $\frac{1}{e^2}$.