A circuit containing a capacitor is set up as shown - Scottish Highers Physics - Question 18 - 2017

The charge (Q) stored in a capacitor is given by the formula:

$Q = C \cdot V$

Given the capacitance, C = 30 µF = 30 \times 10^{-6} F, and the voltage across the capacitor, V = 6 V:

$Q = (30 \times 10^{-6} F) \cdot (6 V) = 180 \times 10^{-6} C = 1.8 \times 10^{-4} C$

The energy (U) stored in a capacitor is given by:

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

Substituting the values:

$U = \frac{1}{2} \cdot (30 \times 10^{-6} F) \cdot (6 V)^2 = \frac{1}{2} \cdot (30 \times 10^{-6}) \cdot 36$

Calculating this:

$U = 540 \times 10^{-6} J = 5.4 \times 10^{-4} J$

Thus, the maximum energy stored in the capacitor is approximately 5.4 × 10⁻⁴ J, which corresponds to option A.