For capacitors in series C\(_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2}\) C\(_{\text{eq}} = \frac{5 \times 15}{5 + 15} = \frac{75}{20}\) = 3.75μF Charge on the series branch Q = Ceq \times V =3.75 \times 24 = 90 μC. Potential difference across the 5 µF capacitor V\(_{5} = \frac{Q}{C} = \frac{90}{5}\) = 18 V