Charges Q1 and Q2 are at points A and B of a right angle triangle OAB… | JEE Main 2020 (6 Sep, Shift 1)

Q35JEE Main 2020 · 6 SEP · SHIFT 1Medium

Electric Field & Field Lines

Charges Q1 and Q2 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

$\frac{Q_1}{Q_2}$ is proportional to:

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Answer: (c)

At $O$ : $E_1 = \dfrac{kQ_1}{x_1^2}$ along $OA$ and $E_2 = \dfrac{kQ_2}{x_2^2}$ along $OB$ (in magnitude).
Hypotenuse $AB$ has direction $(x_2, -x_1)$ , so a perpendicular has direction $(x_1, x_2)$ , i.e. $\dfrac{E_2}{E_1} = \dfrac{x_1}{x_2}$ .
$\dfrac{Q_2 x_1^2}{Q_1 x_2^2} = \dfrac{x_1}{x_2} \Rightarrow \dfrac{Q_1}{Q_2} = \dfrac{x_1}{x_2}$
So the answer is (c) $\dfrac{x_1}{x_2}$ .

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