Q 12-01-001JEE MainTop questionMedium
Point charges $+4\ \mu\text{C}$ and $+1\ \mu\text{C}$ are $30\ \text{cm}$ apart. At what distance from the $4\ \mu\text{C}$ charge, on the line joining them, is the electric field zero?
Answer: (C) $20\ \text{cm}$
Between two like charges the fields oppose, so set their magnitudes equal at distance $x$ from the $4\ \mu\text{C}$ charge:
$$\frac{k(4)}{x^2} = \frac{k(1)}{(30 - x)^2} \;\Rightarrow\; \frac{2}{x} = \frac{1}{30 - x} \;\Rightarrow\; x = 20\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics