# 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 'O' is perpendicular to the hypotenuse, the $\frac{Q_{1}}{Q_{2}}$ is proportional to   Option: 1     Option: 2 Option: 3   Option: 4

\begin{aligned} &\mathrm{E}_{2}=\text { electric field due to } \mathrm{Q}_{2}\\ &=\frac{\mathrm{kQ}_{2}}{\mathrm{x}_{2}^{2}}\\ &\mathrm{E}_{1}=\frac{\mathrm{kQ}_{1}}{\mathrm{x}_{1}^{2}}\\ &\text { From diagram }\\ &\tan \theta=\frac{E_{2}}{E_{1}}=\frac{x_{1}}{x_{2}}\\ &\frac{\mathrm{kQ}_{2}}{\mathrm{x}_{2}^{2} \times \frac{\mathrm{kQ}_{1}}{\mathrm{x}_{1}^{2}}}=\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}\\ &\frac{Q_{2} x_{1}^{2}}{Q_{1} x_{2}^{2}}=\frac{x_{1}}{x_{2}}\\ &\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}\\ &\frac{Q_{1}}{Q_{2}}=\frac{x_{1}}{x_{2}} \end{aligned}

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