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 . . . . . . .

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

$A$ capacitor of capacitance $2 \mu F$ is charged to $50 V$ and then disconnected from the source. Later,the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by $25 \%$ of its initial value,then the dielectric constant of the dielectric material is

$A$ dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $x = 0$ to $x = 3d$:

The plates of a parallel plate capacitor are charged up to $100\,V$. $A$ $2\,mm$ thick dielectric plate is inserted between the plates. To maintain the same potential difference,the distance between the capacitor plates is increased by $1.6\,mm$. The dielectric constant of the plate is:

Half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,then the new capacitance will be: