Q 12-02-027NEETJEE MainTop questionMedium
A capacitor is charged by a battery and then disconnected. A dielectric slab of constant $K$ is now inserted to fill the gap. Which of the following is true?
Answer: (D) The charge stays the same, and the energy becomes $\dfrac{1}{K}$ times
Disconnected, so $Q$ is fixed. $C$ becomes $KC$, $V = \dfrac{Q}{C}$ becomes $\dfrac{V}{K}$, and $U = \dfrac{Q^2}{2C}$ becomes $\dfrac{U}{K}$. (The slab is pulled in, and the field does work.)
Solution by Sreeraj P, M.Sc Physics