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.....$V$.

$A$ capacitor,$C_1 = 6 \ \mu F$ is charged to a potential difference of $V_0 = 5 \ V$ using a $5 \ V$ battery. The battery is removed and another capacitor,$C_2 = 12 \ \mu F$ is inserted in place of the battery. When the switch $S$ is closed,the charge flows between the capacitors for some time until an equilibrium condition is reached. What are the charges ($q_1$ and $q_2$) on the capacitors $C_1$ and $C_2$ when the equilibrium condition is reached?

The two plates of a parallel plate capacitor are given charges $Q_1$ and $Q_2$. The capacity of the capacitor is $C$. When the switch $(S)$ is closed,mark the correct statement. (Assume both $Q_1$ and $Q_2$ to be positive)

In the circuit shown,the charge on the $5\, \mu F$ capacitor is $........\mu C$.

Show that for a combination of capacitors in series or parallel,the total energy stored is the sum of the energies stored in individual capacitors.

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100 \, V$ $D.C.$ source. Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between its plates. Capacitor $C_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards,the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $......... \, V$. (Assuming dielectric constant $K = 10$)