Physics - Capacitor Question with Solution | TestHub

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) $\begin{array}{rl}\text{ and }4{\mathrm{V}}_1& =\frac{4}{3}{\mathrm{V}}_2\\ \therefore {\mathrm{V}}_2& =3{\mathrm{V}}_1...(ii)\end{array}$ From (i) and (ii), ${\mathrm{V}}_1=3.5\mathrm{V}\text{ and }{\mathrm{V}}_2=10.5\mathrm{V}$ Thus charge of ${\mathrm{C}}_3$, ${\mathrm{q}}_3={\mathrm{C}}_3{\mathrm{V}}_2=4\times 10.5=42\mathrm{V}\text{. }$