$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)

In a parallel plate capacitor setup,the plate area of the capacitor is $2 \, m^{2}$ and the plates are separated by $1 \, m$. If the space between the plates is filled with a dielectric material of thickness $0.5 \, m$ and area $2 \, m^{2}$ (see figure),the capacitance of the setup will be $......... \, \varepsilon_{0}$. (Dielectric constant of the material $= 3.2$) (Round off to the nearest integer).

$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

$A$ capacitor is connected to a $10\,V$ battery. The charge on the plates is $10\,\mu C$ when the medium between the plates is air. The charge on the plates becomes $100\,\mu C$ when the space between the plates is filled with oil. The dielectric constant of the oil is:

The outer sphere of a spherical air capacitor is earthed. For increasing its capacitance: