Q 12-01-010NEETJEE MainMedium
Two point charges $+4\ \mu$C and $+9\ \mu$C are $50$ cm apart. The point on the line joining them where the electric field is zero is at a distance from the $4\ \mu$C charge of
Answer: (B) $20$ cm
$\dfrac{4}{x^2} = \dfrac{9}{(50 - x)^2} \Rightarrow \dfrac{50 - x}{x} = \dfrac{3}{2} \Rightarrow x = 20$ cm (between the charges, nearer the smaller one).
Solution by Sreeraj P, M.Sc Physics