Question 109 of 311

Charges ${Q}_{1}$ and ${Q}_{2}$ are at points $A$ and $B$ of a right-angled triangle $OAB$. The resultant electric field at point $O$ is perpendicular to the hypotenuse, then ${Q}_{1}/{Q}_{2}$ is proportional to:

A$\dfrac{{x}_{1}^{3}}{{x}_{2}^{3}}$B$\dfrac{{x}_{2}}{{x}_{1}}$C$\dfrac{{x}_{1}}{{x}_{2}}$D$\dfrac{{x}_{2}^{2}}{{x}_{1}^{2}}$

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Answer: (C) $\dfrac{{x}_{1}}{{x}_{2}}$

Using triangle properties
$\tan \text{θ}=\dfrac{{x}_{1}}{{x}_{2}} \tan \text{θ}=\dfrac{{E}_{2}}{{E}_{1}}=\dfrac{K{Q}_{2}/{x}_{2}^{2}}{K{Q}_{1}/{x}_{1}^{2}} \Rightarrow \dfrac{{x}_{1}}{{x}_{2}}=\dfrac{K{Q}_{2}/{x}_{2}^{2}}{K{Q}_{1}/{x}_{1}^{2}}$
$\dfrac{{Q}_{1}}{{Q}_{2}}=$ $\dfrac{{x}_{1}}{{x}_{2}}$