What charges will flow after shorting of the switch $S$ in the circuit illustrated in figure through sections $1$ and $2$ in the directions indicated by the arrows?

The charge distribution, when the switch is closed, is shown here.

In figure, the two capacitors are connected in series and hence their equivalent capacitance is given by

$C_S=\frac{C_1{C}_2}{\left(C_1+C_2\right)}$

If $q$ is charged on wither of the capacitors, then

$q=\left(\frac{C_1{C}_2}{C_1+C_2}\right)E$

For the closed loop $LKPOL$, applying Kirchhoff's Loop Law, we get

$E-\frac{q_2}{C_2}=0 \left\{\because -△V=0\right\}$

$\Rightarrow q_2=C_{2}E$                                         ….(1)

For the closed loop $MLONM$, applying Kirchhoff's Loop Law, we get

$-\frac{q_1}{C_1}+\frac{q_2}{C_2}-E=0$                         ……(2)

Substituting the value of $q_2$ from equation (1) in equation (2), we get

$$\begin{gathered} -\frac{q_1}{C_1}+\frac{C_{2}E}{C_2}-E=0 \\ \Rightarrow q_1=0 \end{gathered}$$

So, charge flowing through $1$ is given by

$△q_1=C_{2}E$

and the charge that flows through section $2$ is given by

$△q_2=-q_1-q=-\left(\frac{C_1{C}_2}{C_1+C_2}\right)E$