Q 12-02-135JEE MainJEE Main 2021 (31 Aug, Shift 2)Easy
A parallel plate capacitor of capacitance $200\ \mu\text{F}$ is connected to a battery of $200$ V. A dielectric slab of dielectric constant $2$ is now inserted into the space between plates of capacitor while the battery remains connected. The change in the electrostatic energy in the capacitor will be ______ J.
Numerical value type. Enter your answer.
Answer: 4
With the battery connected, $V$ stays $200$ V and $C$ becomes $400\ \mu\text{F}$.
$$\Delta U = \frac12(C' - C)V^2 = \frac12\times200\times10^{-6}\times(200)^2 = 4\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics