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: $E = \frac{1}{2} C V^2$ We need to find the voltage (V) first, using the formula: $V = \frac{Q}{C}$ Substituting values:

• $C = 1.0 \mu F = 1.0 \times 10^{-6} F$
• We already calculated $Q = 2.0 \times 10^{-5} C$

Now, calculate V: $V = \frac{2.0 \times 10^{-5} C}{1.0 \times 10^{-6} F} = 20 V$

Finally, substituting the value of V back into the energy formula: $E = \frac{1}{2} (1.0 \times 10^{-6} F)(20 V)^2 = \frac{1}{2} (1.0 \times 10^{-6})(400) = 2.0 \times 10^{-4} J = 2.0 \times 10^{-2} J$