Question

The energy and capacity of a charged parallel plate
capacitor are U and Crespectively. Now a dielectric
slab of 4-6 is inserted in it then energy and
capacity becomes . (Assuming charge on plates
remains constant)
(1) 6U, 6c
(2) U,C
(3)U/6,6C
(4) U, 6C​

Answer 1

Answer:

option 3

Explanation:

as we insert dielectric of dielectric constant K(say)

energy stored reduces by k i.ie.m U'=U/k

and capacitance increases by k i.e., C'=KC