A 1.0 µF capacitor is charged for 20 s using a constant current of 10 µA - AQA - A-Level Physics - Question 20 - 2019 - Paper 2

The energy ( E ) stored in a capacitor can be calculated using the formula:

$E = \frac{1}{2} C V^2$

However, we need to find the voltage ( V ) first. The voltage across a capacitor can be found using:

$V = \frac{Q}{C}$

where:

C = capacitance (1.0 µF = 1.0 x 10⁻⁶ F)

Q = charge calculated above (2 x 10⁻⁴ C)

Substituting the values: $V = \frac{2 \times 10^{-4} ext{ C}}{1.0 \times 10^{-6} ext{ F}} = 200 ext{ V}$

Now, we can substitute V back into the energy formula:

$E = \frac{1}{2} \times (1.0 \times 10^{-6}) \times (200)^2 = \frac{1}{2} \times (1.0 \times 10^{-6}) \times 40000 ext{ V}^2$

Calculating this gives: $E = 2.0 \times 10^{-2} ext{ J}$

The calculated energy transferred to the capacitor is $2.0 \times 10^{-2} \text{ J}$, which corresponds to option C in the question.