Charge $Q_{1} a n d Q_{2}$ are at points A and B of a right angle triangle OAB (see fig). The resultant electric field at point O is perpendicular to the hypotenuse, then $Q_{1} / Q_{2}$ is proportional to:

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$\frac{x_{2}^{}}{x_{1}^{}}$

$\frac{x_{1}^{}}{x_{2}^{}}$

$\frac{x_{2}^{2}}{x_{1}^{2}}$

$\frac{x_{1}^{3}}{x_{2}^{3}}$

answer is D.

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Detailed Solution

$\begin{array}{l} E_{1} \sin θ = E_{2} \cos θ \\ \frac{k Q_{1}}{x_{1}^{2}} \sin θ = \frac{k Q_{2}}{x_{2}^{2}} \cos θ \\ \frac{Q_{1}}{Q_{2}} = {(\frac{x_{1}}{x_{2}})}^{2} \cot θ = \frac{x_{1}}{x_{2}} \end{array}$

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