#
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 $

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

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

**
**