Q 12-01-093JEE MainJEE Main 2022 (24 Jun, Shift 1)Easy
A vertical electric field of magnitude $4.9\times10^5\ \text{N C}^{-1}$ just prevents a water droplet of mass $0.1\ \text{g}$ from falling. The value of charge on the droplet will be: (Given $g = 9.8\ \text{m s}^{-2}$)
Answer: (B) $2.0\times10^{-9}\ \text{C}$
$$qE = mg\ \Rightarrow\ q = \frac{10^{-4}\times 9.8}{4.9\times10^{5}} = 2.0\times10^{-9}\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics