A parallel plate capacitor each with plate area A and separation ‘d’ is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change if any, will take place in
(i) charge on the plates
(ii) electric field intensity between the plates,
(iii) capacitance of the capacitor
Justify your answer in each case

Explanation

(a) Initial capacitance `C_(o)=(epsi_(0)A)/(d),` potential difference=V
(i) Initial charge `q_(0)=C_(0)V=(epsi_(0)A)/(d)V`
`therefore` When battery is disconnected the charge on the capacitor remains unchanged and equal to `q=q_(0)=(epsi_(0)A)/(d)V`.
(ii) Iniital electric field between the plates, `E_(0)=(sigma)/(epsi_(0))=(q//A)/(epsi_(0))=(q)/(Aepsi_(0))`
After introduction of dielectric, the permuttivity of medium become `Kepsi_(0)`,
So final electric field between the plates, `E=(q)/(AK epsi_(0))=(E_(0))/(K)`
i.e., electric field reduces to `(1)/(K)` times.
(iii) After introduction of dielectric, the capacitance becomes `KC_(0)`.