Figure shows three circuits, each consisting of a switch and two capacitors, initially charged as indicated. In which circuit the charge on the left hand capacitor will change after the closing of the switch is :

In (1), the potential of the capacitors are :

$$\begin{gathered} V_L=\frac{6q}{2c}=\frac{3q}{C} \mathrm{and} \\ {V}_e=\frac{3q}{C}. As V_L= V_R, \mathrm{and} \mathrm{so} \mathrm{there} \mathrm{is} \mathrm{no} f\mathrm{low} \mathrm{of} \mathrm{charge} \mathrm{in} \\ \mathrm{between} \mathrm{the} \mathrm{capacitors}. \\ \mathrm{In} \left(2\right)V_L=\frac{6q}{3C}=\frac{2q}{C} and V_R=\frac{3q}{C}. As V_R > V_L \end{gathered}$$ 

and so charge will below towards left capacitor

In  (3) , $V_L=\frac{6q}{3C}=\frac{2q}{C} and V_R=\frac{3q}{2C}. As V_L>V_C, charge will flow to the right capacitor .$