If two charges $q_1$ and $q_2$ are separated by a distance $d$ and placed in a medium of dielectric constant $K$, what will be the equivalent distance between the charges in air for the same electrostatic force?

$A$ parallel plate capacitor is charged by a battery and the battery remains connected. $A$ dielectric slab of constant $K$ is inserted between the plates and then taken out. What happens to the electric field between the plates?

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

$A$ capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,an additional charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

$A$ capacitor has capacitance $C_0$ when there is no dielectric between its plates. Two slabs of dielectric constant $K_1$ and $K_2$ respectively,with area equal to the area of the plates but thickness half of the distance between the plates,are placed in between the plates. Then the new capacitance is

The capacitance of a parallel plate capacitor is $2.5 \mu F$. When it is half filled with a dielectric as shown in the figure,its capacitance becomes $5 \mu F$. The dielectric constant of the dielectric is