Q 12-01-130JEE MainJEE Main 2020 (6 Sep, Shift 1)Medium
Charges $Q_1$ and $Q_2$ are at points A and B of a right-angled triangle OAB (right angle at O). The resultant electric field at point O is perpendicular to the hypotenuse; then $Q_1/Q_2$ is proportional to:
Answer: (C) $\dfrac{x_1}{x_2}$
Take O as origin with A at $(0, x_1)$ and B at $(x_2, 0)$. The fields at O (for positive charges) are
$$\vec E_1 = -\frac{kQ_1}{x_1^2}\hat j, \qquad \vec E_2 = -\frac{kQ_2}{x_2^2}\hat i$$
The hypotenuse AB is along $(x_2, -x_1)$. The resultant is perpendicular to it when
$$\left(-\frac{kQ_2}{x_2^2}\right)x_2 + \left(-\frac{kQ_1}{x_1^2}\right)(-x_1) = 0 \Rightarrow \frac{Q_1}{x_1} = \frac{Q_2}{x_2}$$
$$\frac{Q_1}{Q_2} = \frac{x_1}{x_2}$$
Solution by Sreeraj P, M.Sc Physics