Q 12-02-109JEE MainJEE Main 2022 (24 Jun, Shift 2)Easy
If the charge on a capacitor is increased by $2\ \text{C}$, the energy stored in it increases by $44\%$. The original charge on the capacitor is (in C)
Answer: (A) 10
$U = \dfrac{Q^2}{2C}$, so
$$\frac{(Q+2)^2}{Q^2} = 1.44\ \Rightarrow\ Q + 2 = 1.2Q\ \Rightarrow\ Q = 10\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics