Question

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 $Q _{ l } / Q _{2}$ is proportional to :

• A. $\frac{x_{2}^{2}}{x_{1}^{2}}$
• B. $\frac{x_{1}^{3}}{x_{2}^{3}}$
• C. $\frac{x_{1}}{x_{2}}$
• D. $\frac{ x _{2}}{ x _{1}}$

Answer 1

Answer: (C)

$E _{2}=$ electric field due to $Q _{2}$
$=\frac{ kQ _{2}}{ x _{2}^{2}}$
$E _{1}=\frac{ kQ _{1}}{ x _{1}^{2}}$
From diagram
$\tan \theta=\frac{E_{2}}{E_{1}}=\frac{x_{1}}{x_{2}}$
$\frac{ kQ _{2}}{ x _{2}^{2} \times \frac{ kQ _{1}}{ x _{1}^{2}}}=\frac{ x _{1}}{ x _{2}}$
$\frac{Q_{2} x_{1}^{2}}{Q_{1} x_{2}^{2}}=\frac{x_{1}}{x_{2}}$
$\frac{Q_{2}}{Q_{1}}=\frac{x_{2}}{x_{1}}$
$\frac{Q_{1}}{Q_{2}}=\frac{x_{1}}{x_{2}}$