When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :- 

A dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$. The induced charge $q^{\prime}$ on the surface of slab is given by

Write the capacitance of parallel plate capacitor with medium of dielectric of dielectric constant $\mathrm{K}$.

Putting a dielectric substance between two plates of condenser, capacity, potential and potential energy respectively

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2\,\mu \,F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu \,F$