# Solve this following

Question:

Charges $\mathrm{Q}_{1}$ and $\mathrm{Q}_{2}$ arc at points $\mathrm{A}$ and $\mathrm{B}$ of a right angle triangle OAB (see figure). The resultant electric field at point $\mathrm{O}$ is perpendicular to the hypotenuse, then $\mathrm{Q}_{1} / \mathrm{Q}_{2}$ is proportional to :

1. $\frac{x_{2}^{2}}{x_{1}^{2}}$

2. $\frac{x_{1}^{3}}{x_{2}^{3}}$

3. $\frac{X_{1}}{X_{2}}$

4. $\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}$

Correct Option: , 3

Solution:

$\mathrm{E}_{2}=$ electric field due to $\mathrm{Q}_{2}$

$=\frac{\mathrm{kQ}_{2}}{\mathrm{x}_{2}^{2}}$

$\mathrm{E}_{1}=\frac{\mathrm{kQ}_{1}}{\mathrm{x}_{1}^{2}}$

From diagram

$\tan \theta=\frac{E_{2}}{E_{1}}=\frac{x_{1}}{x_{2}}$

$\frac{\mathrm{kQ}_{2}}{\mathrm{x}_{2}^{2} \times \frac{\mathrm{kQ}_{1}}{\mathrm{x}_{1}^{2}}}=\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}$

$\frac{Q_{2} x_{1}^{2}}{Q_{1} x_{2}^{2}}=\frac{x_{1}}{x_{2}}$

$\frac{Q_{2}}{Q_{1}}=\frac{X_{2}}{X_{1}}$

$\frac{Q_{1}}{Q_{2}}=\frac{X_{1}}{X_{2}}$

Ans. (3)