In an arrangement of capacitors shown $C_1=C_2=C_3=C$. The emf of the battery connected in the circuit is $E$. When the switch $S$ is closed, then

$$\begin{gathered} \mathrm{Initial} \mathrm{capacitance} \mathrm{of} \mathrm{the} \mathrm{circuit} \\ \left(\mathrm{C}\parallel \mathrm{C}\right)\mathrm{in} \mathrm{series} \mathrm{with} \mathrm{C} \\ \mathrm{C} \mathrm{equivalent} =\frac{\mathrm{C}\times 2\mathrm{C}}{3\mathrm{C}}=\frac{2\mathrm{C}}{3} \\ \mathrm{charge} \mathrm{stored} {\mathrm{Q}}_0=\frac{2\mathrm{CE}}{3} \\ \mathrm{finally} \mathrm{when} \mathrm{the} \mathrm{other} \mathrm{capacitor} \mathrm{is} \mathrm{short} \mathrm{circuited} \mathrm{then} \\ \mathrm{equivalent} \mathrm{capcitance} \mathrm{becomes} 2\mathrm{C} \\ \mathrm{and} \mathrm{charge} \mathrm{stored} 2\mathrm{CE} \\ \mathrm{charge} \mathrm{is} \mathrm{increasing} \\ \mathrm{Hence} \mathrm{a} \mathrm{charge} \mathrm{of} \left(2\mathrm{CE}-\frac{ 2\mathrm{CE}}{3}\right)=\frac{4\mathrm{CE}}{3}\mathrm{flows} \mathrm{from} \mathrm{the} \mathrm{positive} \mathrm{terminal} \\ \mathrm{of} \mathrm{the} \mathrm{battery} \mathrm{to} \mathrm{the} \mathrm{system}. \end{gathered}$$