Q 12-02-131JEE MainJEE Main 2021 (27 Jul, Shift 1)Easy
Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now be:
Answer: (C) $\dfrac{3V}{K+2}$
Total charge (conserved after the battery is removed): $Q = (2C + C)V = 3CV$.
New total capacitance: $2C + KC = (K+2)C$.
$$V' = \frac{3CV}{(K+2)C} = \frac{3V}{K+2}$$
Solution by Sreeraj P, M.Sc Physics