Question

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 d and dielectric constant k is now placed between the plates. What change, if any, will take place in: 

(i) charge on plates? 

(ii) electric field intensity between the plates? 

(iii) capacitance of the capacitor? Justify your answer in each case.

Answer 1

(i) The charge Q =CV, V = same, C = increases; there, charge on plates increases. 

(ii) A electric field E  = V/d, and V = constant, d = constant; therefore, electric field strength remains the same. 

(iii) The capacitance of capacitor increases as K >1.