Q 12-02-103JEE MainJEE Main 2023 (8 Apr, Shift 2)Easy
Electric potential at a point $P$ due to a point charge of $5\times10^{-9}$ C is $50$ V. The distance of $P$ from the point charge is (Assume $\dfrac1{4\pi\epsilon_0}=9\times10^9\ \text{N m}^2\,\text{C}^{-2}$)
Answer: (D) $90\ \text{cm}$
$r=\dfrac{kq}{V}=\dfrac{9\times10^9\times5\times10^{-9}}{50}=0.9\ \text{m}=90$ cm.
Solution by Sreeraj P, M.Sc Physics