Three identical parallel conducting plates A, B and C are placed as shown. Switches ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ are opened and connect A and C to earth when closed. +Q charge is given to B.

Charge on plate  $S_2$ + Charge on plate  $B=Q$

$C_1=\frac{A{\epsilon }_0}{2d},C_2=\frac{A{\epsilon }_0}{d}$

I. Now if only $S_1$  is closed then charge on plate ‘C’ will still be zero
 $\therefore$ charge on plate  $B=0$

$\therefore$total charge + Q is on plate  $B$'

$\therefore -Q$ amount of charge will flow through  $S_1$

So as to form a capacitor C1

II. Similarly, if only S₂ is closed – Q amount of charges will flow through S₂ in both of above cases 
Charge of amount Q (sign is not taken) is flown

Now, if  $S_1$ and $S_2$ both are closed let the distribution charges on the capacitors are
Now  

$\frac{-q}{C_1}-\frac{(Q+q)}{C_2}=0$

$\Rightarrow q=\frac{-Q}{3}$

Amount of charge flown through $S_1$  and $S_2$ are $\frac{Q}{3}$ and $\frac{2Q}{3}$