Q 12-02-096JEE MainJEE Main 2023 (11 Apr, Shift 1)Easy
A parallel plate capacitor of capacitance $2\ \text{F}$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$. The ratio $\dfrac{E_2}{E_1}$ is
Answer: (C) $1:2$
The charge $Q$ is shared by capacitance $2C$: $E_2=\dfrac{Q^2}{2(2C)}=\dfrac{E_1}{2}$. Ratio $1:2$.
Solution by Sreeraj P, M.Sc Physics