The charge deposited on \(4 \mu \mathrm{F}\) capacitor in the circuit is

As the capacitors \(4 \mu \mathrm{F}\) and \(2 \mu \mathrm{F}\) are connected in parallel and are in series with \(6 \mu \mathrm{F}\) capacitor, their equivalent capacitance is
\(\frac{(2+4) \times 6}{2+4+6}=3 \mu \mathrm{F}\)
Charge in the circuit,
\(\mathrm{Q}=3 \mu \mathrm{F} \times 12 \mathrm{~V}=36 \mu \mathrm{C}\)

Since, the capacitors \(4 \mu \mathrm{F}\) and \(2 \mu \mathrm{F}\) are connected in parallel, therefore potential difference across them is same.
\(\Rightarrow \frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{4}{2} \text { or } \mathrm{Q}_{1}=2 \mathrm{Q}_{2}\)
\(\text {Also, } \quad Q_{1}=Q_{1}+Q_{2} \)
\(\therefore 36 \mu C=2 Q_{2}+Q_{2} \quad \text { or } \quad Q_{2}=\frac{36 \mu C}{3}=12 \mu C \)
\(Q_{1}=Q-Q_{2}=36 \mu C-12 \mu C \)
\(=24 \mu C=24 \times 10^{-6} C\)