Each capacitor shown in the figure has a capacitance of $5.0 \ \mu F$. The battery has an $emf$ of $50 \ V$. If the switch $S$ is closed,what amount of charge will flow through $AB$?

$A$ body of mass $1 \,g$ and carrying a charge $10^{-8} \,C$ passes through two points $P$ and $Q$. $P$ and $Q$ are at electric potentials $600 \,V$ and $0 \,V$, respectively. The velocity of the body at $Q$ is $20 \,cm/s$. Its velocity in $m/s$ at $P$ is:

In the following circuit $C_1=12 \mu F, C_2=C_3=4 \mu F$ and $C_4=C_5=2 \mu F$. The charge stored in $C_3$ is . . . . . $\mu C$.

Two capacitors are shown in the diagram. Each is charged to potential $V_0$. If capacitor $C$ is filled with a dielectric of constant $2$ and capacitor $2C$ is filled with a dielectric of constant $3$,then the key is closed. The final voltage on the capacitor $C$ will be:

The ratio of potential difference that must be applied across parallel and series combination of two capacitors $C_1$ and $C_2$ with their capacitance in the ratio $1: 2$ so that energy stored in these two cases becomes same is

$A$ parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$. The ratio $E_2 / E_1$ is