Question

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
(a) The electric field in the capacitor
(b) The charge on the capacitor
(c) The potential difference between the plates
(d) The stored energy in the capacitor

Answer 1

after disconnecting battery let us assume the charge on the plates is Q and the potential difference is V. The relation connecting capacitance C, charge Q and potential difference V is given by C = Q/V.

 

when dielectric material is inserted between parallel plates the capacitance increases and the new capacitance C' is given by C' = k×C, where k is dielectric constant. Charge Q remains same. Hence potential difference reduces and the new value of potential difference is V/k .

 

Energy stored in the capacitor before inserting dielectric  = (1/2)×C×V2

 

after inserting dielectric, energy available in the capacitor = (1/2)×(kC)×(V/k)2 = k×(1/2)×C×V2

Answer 2

Answer:

this is a your answer to the question