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 find the fraction of the stored energy remaining after a certain time during the discharge of a capacitor, we use the formula:

$$E(t) = E_0 e^{-t/RC}$$

where:

• $E(t)$ is the energy at time $t$,
• $E_0$ is the initial energy,
• $R$ is the resistance, and
• $C$ is the capacitance.

First, we calculate the time constant ($au$) using:

$au = R imes C$

Given values:

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

Calculating the time constant:

$au = (15 \times 10^{3}) \times (2.0 \times 10^{-3}) = 30 ext{ seconds}$

Next, we know that after 1.0 minute (60 seconds), the fraction of energy remaining is:

rac{E(t)}{E_0} = e^{-t/RC} = e^{-60/30} = e^{-2}

Therefore, the fraction of the stored energy remaining after 1.0 minute is:

$\frac{1}{e^{2}}$

Thus, the correct choice is:

B) rac{1}{e^2}.