1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A parallel plate capacitor is charged by a battery.The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates. How will (i) its capacitance, (ii) electric field between the plates and (iii) energy stored in the capacitor be affected ? Justify your answer giving necessary mathematical expressions for each case.

Open in App
Solution

## The capacitance without dielectric is C=Aϵ0dWhen dielectric slab is inserted, the capacitance becomes, C′=AKϵ0d=KC where K be the dielectric constant. i)Thus, the capacitance will increase K times of the initial.ii) As the battery is disconnected so the charge on capacitor remains constant.Since, Q=CV so potential V will decrease and also E=V/d so the field E will also decrease.iii) Stored energy, U=Q22C . As charge Q is constant and C is increasing so energy will decrease.

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Dielectrics
PHYSICS
Watch in App
Join BYJU'S Learning Program