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
When the capacitor is disconnected from the battery, charge $Q$ remains constant.
On inserting a dielectric of constant $K$:
$C' = KC$
$V' = \frac{Q}{C'} = \frac{V}{K}$ → decreases
$E' = \frac{V'}{d} = \frac{E}{K}$ → decreases
$U' = \frac{1}{2} C' V'^2 = \frac{1}{2} KC \left(\frac{V}{K}\right)^2 = \frac{U}{K}$ → decreases
∴ The quantities that change are (A), (C), and (D).