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Charges \( Q_{1} \) and \( Q_{2} \) are at points \( A \) and \( B \) of a right angle triangle \( \mathrm{OAB} \) (see figure). The resultant electric field at point \( \mathrm{O} \) is perpendicular to the hypotenuse, then \( Q_{1} / Q_{2} \) is proportional to
|JEE Main-2020 (September)]
\( P \)
W
(1) \( \frac{x_{1}^{3}}{x_{2}^{3}} \)
(2) \( \frac{x_{2}^{2}}{x_{1}^{2}} \)
(3) \( \frac{x_{1}}{x_{2}} \)
(4) \( \frac{x_{2}}{x_{1}} \)
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