A parallel plate air capacitor is charged, and then the battery is disconnected. Now on inserting a dielectric between the plates of the capacitor, which of the following change:
(A) potential difference between the plates
(B) charge on the plates
(C) electric field between the plates
(D) energy stored in the capacitor
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
With the battery disconnected the plates are isolated, so the charge \( Q \) stays constant. Inserting a dielectric of constant \( K \) raises the capacitance to \( C' = KC \).
(A) potential difference between the plates — Changes. \( V' = Q/C' = V/K \), so it falls.
(B) charge on the plates — Does not change. There is no path for charge to enter or leave the plates.
(C) electric field between the plates — Changes. \( E' = V'/d = E/K \), so it falls.
(D) energy stored in the capacitor — Changes. \( U' = Q^{2}/2C' = U/K \), so it falls.
So (A), (C) and (D) change.