Q 12-02-170JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
A parallel plate capacitor having capacitance $12\ \text{pF}$ is charged by a battery to a potential difference of $10\ \text{V}$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is
Answer: (C) $508\ \text{pJ}$
With the battery disconnected the charge stays fixed and $U = \dfrac{Q^2}{2C}$, so $U$ falls by the factor $K$.
$$U_i = \tfrac12CV^2 = \tfrac12\times12\times10^{-12}\times100 = 600\ \text{pJ},\qquad U_f = \frac{600}{6.5} = 92.3\ \text{pJ}$$
Work done by the capacitor on the slab $= 600 - 92.3 \approx 508\ \text{pJ}$.
Solution by Sreeraj P, M.Sc Physics