Find the charge and potential difference across C1 of capacitance 1 μF for the given circuit?

Correct option is D

​$$\textbf{Given:} \\\bullet C_1 = 1 \, \mu \text{F} \\\bullet C_2 = 2 \, \mu \text{F} \\\bullet C_3 = 2 \, \mu \text{F} \\\bullet \text{Voltage across the circuit } V = 4 \, \text{V}$$

$$\text{1. Capacitance in parallel:} \\C_{e1} = C_1 + C_4 \\\text{2. Capacitance in series:} \\\frac{1}{C_{e2}} = \frac{1}{C_{e1}} + \frac{1}{C_2} + \frac{1}{C_3}$$

$$\Rightarrow C_{e1} = 1 + 1 \\C_{e1} = 2 \, \mu F \\\text{By equation 2,} \\C_{e2} = \frac{2}{3} \, \mu F \\\text{Thus, the total charge flowing through the circuit} \\Q = C_{eq} V \\\Rightarrow Q = \frac{2}{3} \, \mu F \times 4 \, \text{V} \\Q = \frac{8}{3} \, \mu C \\\text{Charge across will the capacitance will be in the ratio of respective capacitance} \\C_{eq} V = C_1 V + C_2 V + C_3 V \\Q_1 = \frac{Q}{2} \\Q_1 = \frac{4}{3} \, \mu C \\\text{Now potential difference across } C_1 \text{ can be calculated as} \\V = \frac{4}{3} \times \frac{1}{C_1} \\\Rightarrow V = \frac{4}{3} \times 1 \\\Rightarrow V_1 = 4/3 \, \text{V}$$​​​​