$A$ 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:

The function of a dielectric in a capacitor is

$A$ parallel plate capacitor has an area of $6 \, cm^2$ and a separation of $3 \, mm$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$,and $K_3 = 14$. The dielectric constant of a material which,when fully inserted in the above capacitor,gives the same capacitance would be:

$A$ parallel plate capacitor having air as the dielectric medium is charged by a potential difference of $V$ volt. After disconnecting the battery,the distance between the plates of the capacitor is increased using an insulated handle. As a result,the potential difference between the plates . . . . . . .

$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \, pF$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $\frac{x}{4} \, pF$. The value of $x$ is $............$

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).