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Charges $Q _{1}$ and $Q _{2}$ arc 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 $Q _{1} / Q _{2}$ is proportional to
$\frac{ x _{2}^{2}}{ x _{1}^{2}}$
$\frac{x_{1}^{3}}{x_{2}^{3}}$
$\frac{x_{1}}{x_{2}}$
$\frac{ x _{2}}{ x _{1}}$
Solution
$E _{2}=$ electric field due to $Q _{2}$
$=\frac{ kQ _{2}}{ x _{2}^{2}}$
$E _{1}=\frac{ k Q _{1}}{ x _{1}^{2}}$
From diagram
$\tan \theta=\frac{ E _{2}}{ E _{1}}=\frac{ x _{1}}{ x _{2}}$
$\frac{ kQ _{2}}{ x _{2}^{2} \times \frac{ k Q _{1}}{ x _{1}^{2}}}=\frac{ x _{1}}{ x _{2}}$
$\frac{ Q _{2} x _{1}^{2}}{ Q _{1} x _{2}^{2}}=\frac{ x _{1}}{ x _{2}}$
$\frac{Q_{2}}{Q_{1}}=\frac{x_{2}}{x_{1}}$
$\frac{Q_{1}}{Q_{2}}=\frac{x_{1}}{x_{2}}$
