$A$ sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

$A$ capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process

$A$ parallel plate capacitor with width $4\,cm$,length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. $A$ dielectric slab of dielectric constant $5$ having length $1\,cm$,width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of the parallel plate capacitor. The electrostatic energy of this system will be ......... $\epsilon_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)

When a dielectric material is introduced between the plates of a charged capacitor,the electric field between the plates:

$A$ parallel plate capacitor has charge $5 \times 10^{-6} \ C$. $A$ dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \ C$, then the dielectric constant of the slab is . . . . . . .

The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in the figure,will be (area of plate $= A$)