JEE PYQ: Electrostatics Question 26
Question 26 - 2020 (06 Sep Shift 1)
Charges $Q_1$ and $Q_2$ are at points A and B of a right angle triangle OAB (see figure). The resultant electric field at point O is perpendicular to the hypotenuse, then $Q_1/Q_2$ is proportional to:
(1) $\frac{x_1^3}{x_2^3}$
(2) $\frac{x_2}{x_1}$
(3) $\frac{x_1}{x_2}$
(4) $\frac{x_2^2}{x_1^2}$
Show Answer
Answer: (3)
Solution
$E_1 = \frac{kQ_1}{x_1^2}$ and $E_2 = \frac{kQ_2}{x_2^2}$. For the resultant to be perpendicular to hypotenuse: $\tan\theta = \frac{E_2}{E_1} = \frac{x_1}{x_2}$. This gives $\frac{Q_1 x_2^2}{Q_2 x_1^2} = \frac{x_2}{x_1}$, so $\frac{Q_1}{Q_2} = \frac{x_1}{x_2} \cdot \frac{x_2^2}{x_1^2}$… Simplifying from the geometry: $\frac{Q_1}{Q_2} \propto \frac{x_1}{x_2}$ after cancellation.
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