When a dielectric slab is inserted between the plates of a capacitor while it remains connected to a battery,which of the following occurs 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)

Two metallic plates of radius $r$ are placed at a distance $d$ apart,and the capacitance is $C$. If a plate of radius $r/2$ and thickness $d$ with a dielectric constant $6$ is placed between the plates of the capacitor,then its new capacitance will be

The figure shows two identical parallel plate capacitors connected to a battery and a closed switch $S$. Now,the switch is opened and a dielectric material with dielectric constant $K = 3$ is inserted into the free space between the plates of both capacitors. What is the ratio of the total electrostatic energy stored in both capacitors before and after the insertion of the dielectric?

The distance between the plates of a parallel plate capacitor is $d$. $A$ metal plate of thickness $d/2$ is introduced between the plates. The capacitance will then be

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