AOB is a right angled triangle. Charges $Q_1$ and $Q_2$ are placed as shown. If the resultant electric field at O is perpendicular to the hypotenuse then $\frac{Q_1}{Q_2}$ is proportional to:

$\frac{x_1}{x_2}$

The correct answer is Option (1) → $\frac{x_1}{x_2}$

$\text{Answer: }\frac{Q_1}{Q_2}\propto\frac{x_1}{x_2}$

$A(0,x_1),\ B(x_2,0).$

$\vec E_1=\frac{kQ_1}{x_1^2}(0,-1),\ \vec E_2=\frac{kQ_2}{x_2^2}(-1,0).$

$\vec R=\Big(-\frac{kQ_2}{x_2^2},-\frac{kQ_1}{x_1^2}\Big),\ \vec h=(x_2,-x_1).$

$\vec R\cdot\vec h=0\Rightarrow -\frac{kQ_2}{x_2^2}x_2+\frac{kQ_1}{x_1^2}x_1=0.$

$\Rightarrow \frac{Q_1}{Q_2}=\frac{x_1}{x_2}.$