$A$ parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

The space between the plates of a parallel plate capacitor is filled with a dielectric whose dielectric constant varies with distance $x$ as per the relation: $K(x) = K_0 + \lambda x$ (where $\lambda$ is a constant). The capacitance $C$ of the capacitor would be related to its vacuum capacitance $C_0$ by the relation:

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.

In one design of a capacitor,thin sheets of metal of area $80 \ mm \times 80 \ mm$ sandwich between them a piece of paper whose thickness is $40 \ \mu m$. The relative permittivity of the paper is $4.0$ and its dielectric strength is $20 \ MVm^{-1}$. Calculate the maximum charge that can be put on the capacitor. [Permittivity of free space $\epsilon_0 = 9 \times 10^{-12} \ Fm^{-1}$]

$A$ 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 . . . . . . .

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: