Calculate the charge on the capacitor in the steady state. (in $\mu C$)

Find the charge stored on the capacitor in the steady-state condition in $\mu C$.

$A$ capacitor of capacitance $C$ is connected to a battery of $V$ volts. Now,the distance between the plates of the capacitor is halved while keeping the charge constant,and it is again charged to $V$ volts. What is the energy supplied by the battery?

Four identical capacitors are connected as shown in the diagram. When a battery of $6 \, V$ is connected between $A$ and $B$,the total charge stored is found to be $1.5 \, \mu C$. The value of $C_1$ is ........ $\mu F$.

Four capacitors with capacitances $C_1 = 1 \ \mu F, C_2 = 1.5 \ \mu F, C_3 = 2.5 \ \mu F$ and $C_4 = 0.5 \ \mu F$ are connected as shown and are connected to a $30 \ V$ source. The potential difference between points $a$ and $b$ is: (in $V$)

In the steady state,what is the potential difference between points $A$ and $B$ in volts? (All capacitors are in $\mu F$.)