Q 12-01-107JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
Two particles $A$ and $B$ having charges $20\ \mu\text{C}$ and $-5\ \mu\text{C}$ respectively are held fixed with a separation of $5$ cm. At what position should a third charged particle be placed so that it does not experience a net electric force?
Answer: (A) At $5$ cm from $-5\ \mu\text{C}$ on the right side
For unlike charges the null point lies outside, on the side of the smaller charge (right of $B$). Let it be $x$ from $B$:
$$\frac{20}{(5 + x)^2} = \frac{5}{x^2} \Rightarrow 2x = 5 + x \Rightarrow x = 5\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics