Question

SECTION C
28. A parallel plate capacitor, each with plate area A & separation d, is charged to a p.dV
The battery used to charge it remains connected. A dielectric slab of thickness d & dielec
tric constant K is now placed between the plates. What change if any, will take place in
(a) charge on the plates?
(b) electric field between the plates?
(c) capacitance of the capacitor?
[3]​

Answer 1

Explanation:

i) The charge and on the capacitor plates remain same

ii) The electric field intensity between the Capricorn plates decreases due to the introduction of electric introduction of dielectric feel creates and intrinsic electric field director opposite to virgin electric field that is why the electric field intensity decreases

iii) the capacitance of the capacitor increases due to the introduction of a dielectric dot electric field in decreases kam are therefore the capacitor can get more charge to bring back the electric field to its original value this increases the capacity of holding the charge and hence the capacitance increases