The electrostatic force between two charges kept in air is $F$. If $30 \%$ of the space between the charges is filled with a medium,then the electrostatic force between the charges becomes $\frac{F}{2.56}$. The dielectric constant of the medium is

The capacity of an air condenser is $2.0 \, \mu F$. If a medium is placed between its plates,the capacity becomes $12 \, \mu F$. The dielectric constant of the medium will be:

Between the plates of a parallel plate capacitor of plate area $A$ and capacity $0.025 \mu F$,a metal plate of area $A$ and thickness equal to $1/3$ of the separation between the plates of the capacitor is introduced. If the capacitor is charged to $100 \ V$,then the amount of work done to remove the metal plate from the capacitor is (in $\mu J$)

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \,F$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \,V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now,a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates,with the voltage source remaining connected to the capacitor.

$A$ parallel plate capacitor has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.