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 remains connected . A dielectric slab of thickness of and dielectric constant K is now placed between the plates . What change , if any , will take place in :
electric filed intensity between the plates ? Justify your answer .

Explanation

Given that plate area of either plate of parallel plate capacitor = A distance between the plates = d and potential difference between the plates = V
Hence initially capacitance ` C = (in_(0) A)/(d)` and charge on plates Q = CV
As the battery remains connected throughout the potential difference between the plates remains unchanged (V. = V) On placing a dielectric slab of the thickness .d. and dielectric constant .K. between the plates
New charge on plates `Q. = C. V.. = KCV = KQ`