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 $\mathrm{O}$ is perpendicular to the hypotenuse, then $Q_{1} / Q_{2}$ is proportional to :
Correct Option: , 3
(3) Electric field due charge $Q_{2}, E_{2}=\frac{k Q_{2}}{x_{2}^{2}}$
Electric field due charge $Q_{1}, E_{1}=\frac{k Q_{1}}{x_{1}^{2}}$
From figure,
$\tan \theta=\frac{E_{2}}{E_{1}}=\frac{x_{1}}{x_{2}} \Rightarrow \frac{k Q_{2}}{x_{2}^{2} \times \frac{k Q_{1}}{x_{1}^{2}}}=\frac{x_{1}}{x_{2}}$
$\Rightarrow \frac{Q_{2} x_{1}^{2}}{Q_{1} x_{2}^{2}}=\frac{x_{1}}{x_{2}} \Rightarrow \frac{Q_{2}}{Q_{1}}=\frac{x_{2}}{x_{1}}$ or, $\frac{Q_{1}}{Q_{2}}=\frac{x_{1}}{x_{2}}$
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