The capacitance of parallel capacitors
\(=\frac{2 \times 4}{2+4}=\frac{4}{3} \mu \mathrm{F}\)
If \(\mathrm{V}_1\) and \(\mathrm{V}_2\) are the p.d. s across \(\mathrm{C}_1\) and parallel capacitors, then
\(\mathrm{V}_1+\mathrm{V}_2=14\) ...(i)
\(\begin{aligned}
\text { and } 4 \mathrm{~V}_1 & =\frac{4}{3} \mathrm{~V}_2 \\
\therefore \quad \mathrm{V}_2 & =3 \mathrm{~V}_1 \quad ...(ii)
\end{aligned}\)
From (i) and (ii),
\(\mathrm{V}_1=3.5 \mathrm{~V} \text { and } \mathrm{V}_2=10.5 \mathrm{~V}\)
Thus charge of \(\mathrm{C}_3\),
\(\mathrm{q}_3=\mathrm{C}_3 \mathrm{~V}_2=4 \times 10.5=42 \mathrm{~V} \text {. }\)