The force between two point charges placed in a material medium of dielectric constant $K$ is $F$. If the material is removed,then the force between them becomes . . . . . . .

$A$ parallel plate capacitor has plates of area $A$ separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $k(x)=K(1+\alpha x)$ where $x$ is the distance measured from one of the plates. If $(\alpha d) << 1$,the total capacitance of the system is best given by the expression:

If a dielectric is inserted into a charged capacitor (battery removed),then the quantity that remains constant is

$A$ capacitor is charged by a battery and the energy stored is $U$. Now,the battery is removed and the distance between the plates is increased to four times its original value. The new energy stored becomes:

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is completely filled with a dielectric material of constant $K$,then the potential difference across the other capacitor will become:

$A$ 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)