Before connecting the circuit shown in figure, all capacitors were uncharged. The circuit now is in steady state. Now the switch S is closed. During the time the steady state is reached again,

 

Column I		Column II
I	Charge on C1	P	Increases
II	Charge on C2	Q	Decreases
III	Charge on C3	R	Remains same
IV	Charge on C4	S	Becomes zero

	I	II	III	IV
1	Q	R	P	Q
2	P	R	Q	Q
3	Q	S	R	Q
4	R	P	Q	Q

Before closing the switch, let the charges on the capacitors be as shown.

Applying KVL in the upper loop,
$-\text{10}+\frac{q_1}{2}-10=0$

$\Rightarrow q_1=40\text{μC}$

Applying KVL in the lower loop,

$\text{10}-\frac{q_2}{2}+10-\frac{q_2}{2}-\frac{q_2}{2}=0$

$\text{20}=\frac{3q_2}{2}\Rightarrow q_2=\frac{40}{3}\text{μC}$
After the switch S is closed, let the charge distribution be as shown below.

Applying KVL for upper loop,
$-\text{10}+\frac{q_3}{2}-10=0$

$\Rightarrow q_3=40\text{μC}$

Applying KVL for lower left loop,

$\text{10}-\frac{q_4}{2}=0\Rightarrow q_4=20\text{μC}$

Applying KVL for lower right loop,

$\text{10}-\frac{q_5}{2}-\frac{q_5}{2}=0\Rightarrow q_5=10\text{μC}$

Since, $q_1=q_3=40\text{μC}$ ,  so, charge on C1 remains same
Since, $q_2=\frac{40}{3}\text{μC}$ , $q_4=20\text{μC}$ , charge on C2 increases, 
Since, $q_2=\frac{40}{3}\text{μC}$, $q_5=10\text{μC}$, charge on C3 and C4 decreases.