$A$ parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed $v$. If $x$ is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to $x^\alpha$, where $\alpha$ is . . . . . . .

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)

$A$ 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?

In the figure shown,after the switch $S$ is turned from position $A$ to position $B$,the energy dissipated in the circuit in terms of capacitance $C$ and total charge $Q$ is

In the circuit shown in the figure,the total charge is $750\, \mu C$ and the voltage across capacitor $C_{2}$ is $20\, V$. If $C_{3} = 8\, \mu F$,then the charge on capacitor $C_{2}$ is $....\mu C$.

In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is