The dielectric slab is introduced between the plates of a parallel plate charged capacitor. Which one of the following quantities will not change?

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 $\epsilon_0 = 9 \times 10^{-12} \ Fm^{-1}$]

$A$ capacitor of capacitance $15 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.5$,dielectric strength $30 \, MV/m$,and potential difference $V = 30 \, V$. The area of the plate is ....... $\times 10^{-4} \, m^{2}$.

$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is $E_0$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections,the field between the plates shall be

$A$ parallel plate capacitor filled with oil of a dielectric constant $3$ between the plates has capacitance $C$. If the oil is removed,then the capacitance of the capacitor will be