A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is then halved and a dielectric material of dielectric constant K is introduced between the plates, what will happen to
(i) Potential difference across the plates
(ii) Energy stored by the capacitor.
Justify your answer in each case.

Before the dielectric is inserted ,the space between the plates is presumably filled with air ,there fore initial capacitance is given as  C = eA/d and let us conisder the V as a potential diffrence and E as energy stored

where  A = aera of plates and d = distance between plates

it is given that  seperation is halved i.e d' =d/2  and dielectric material of constant K is introduced , there fore the  capacitance is give as ,  C' =keA/d' =2 keA/d

                          C' =2keA/d = 2kC

potential diffrence across the plates is given as , v = charge/capacitance i.e  v = q/C'

                                                                        V'= q/2kC =V/2k

from above formula it is clear that potential diffrence across the plates reduced by the factor  2k

now come to the potential energy stored by capacitor ,i.e  E = 1/2$\times$

                                                                               E' =1/2$\times$(2kC)$\times$V/4k^2 = E/2k

Energy stored by the capacitor is also reduce by the factor 2k