In the circuit shown,the steady state voltage across capacitor $C$ is a fraction of the battery e.m.f. $V$. The fraction is decided by:

When we touch the terminals of a high voltage capacitor,even after the power supply has been disconnected,the capacitor has a tendency to:

For the given circuit,find the voltage across the $2\ \mu F$ capacitor. (in $V$)

Each plate of a parallel plate capacitor has a charge $q$ on it. The capacitor is now connected to a battery. Now,

Three identical capacitors $C_1, C_2$ and $C_3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{r}$. The cell electromotive force (emf) $V_0=8 \,V$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged,$S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium,the charge on $C_3$ is found to be $5 \mu C$. The value of $\varepsilon_{r}$ is:

$100$ capacitors,each having a capacity of $10\,\mu F$,are connected in parallel and are charged by a potential difference of $100\,kV$. The energy stored in the capacitors and the cost of charging them,if electrical energy costs $108\;paise\;per\;kWh$,will be