Effective capacitance of parallel combination of two capacitors $C_{1}$ and $C_{2}$ is $10\; \mu F$. When these capacitors are individually connected to a voltage source of $1\; V,$ the energy stored in the capacitor $C_{2}$ is $4$ times that of $C_{1}$. If these capacitors are connected in series,their effective capacitance will be (in $; \mu F$)

$n$ identical droplets are charged to $V$ volt each. If they coalesce to form a single drop, then its potential will be

$A$ parallel plate capacitor is charged and then disconnected from the source of steady $E.M.F.$ The plates are then drawn apart farther. Again it is connected to the same source. Then:

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is:

$A$ $4\ \mu F$ capacitor is charged to $80\ V$ and a $6\ \mu F$ capacitor is charged to $30\ V$. When they are connected together,the energy lost by the $4\ \mu F$ capacitor is ..... $mJ$.

Find the potential difference between $A$ and $B$ in the given circuit in the steady state. (in $V$)