Physics - Capacitor Question with Solution | TestHub

Equivalent capacitance for three capacitors $C_1,C_2$ and $C_3$ in series is given by $\begin{array}{rl}& \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{C_2{C}_3+C_3{C}_1+C_1{C}_2}{C_1{C}_2{C}_3}\\ & \Rightarrow C_{eq}=\frac{C_1{C}_2{C}_3}{C_1{C}_2+C_2{C}_3+C_3{C}_1}\\ & \Rightarrow C_{eq}=\frac{\mathrm{C}(2\mathrm{C})(3\mathrm{C})}{(3\mathrm{C})(2\mathrm{C})+(2\mathrm{C})+(3\mathrm{C})\mathrm{C}}=\frac{6}{11}\mathrm{C}\\ & \Rightarrow \text{ Charge on capacitors }\left(C_1,C_2\&C_3\right)\text{ in series }\\ & =C_{eq}V=\frac{6}{11}CV\end{array}$ $=C_{eq}V=\frac{6}{11}CV$ Charge on capacitor $C_4=C_{4}V=4CV$ $\frac{\text{ Charge on }{C}_2}{\text{ Charge on }{C}_4}=\frac{\left(\frac{6C}{11}\right)V}{(4C)V}\times \frac{1}{4}=\frac{3}{22}$