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  • Charges and are at points A and B of a right angle triangle <img alt="OAB" src="http://ent

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 \frac{x_{2}}{x_{1}}
   
Option: 2 \frac{x_{1}^{3}}{x_{2}^{3}}
Option: 3 \frac{x_{1}}{x_{2}}  
Option: 4 \frac{x_{2}^{2}}{x_{1}^{2}}

Answers (1)

best_answer

\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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Deependra Verma

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