Ask Questions, Get Answers


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 (a) charge on the plate (b) electric field intensity between the plates (c) capacitance of the capacitor? Justify your answer in each case.

Please log in or register to answer this question.

Related questions