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
Concept. For like charges the null point lies between them, nearer the smaller charge, where the magnitudes are equal.
Step 1. Equate.
$$\frac{4}{x^2} = \frac{9}{(50 - x)^2} \;\Rightarrow\; \frac{50 - x}{x} = \frac32$$
Step 2. Solve. $x = 20$ cm.
Answer. $20$ cm (Option B).
Solution by Sreeraj P, M.Sc Physics