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 $Q_2$
$=\frac{k{Q}_2}{x_2^2}$
$E_1=\frac{k{Q}_1}{x_1^2}$
From diagram
$\tan \theta =\frac{E_2}{E_1}=\frac{x_1}{x_2}$
$\frac{k{Q}_2}{x_2^2\times \frac{k{Q}_1}{x_1^2}}=\frac{x_1}{x_2}$
$\frac{Q_2{x}_1^2}{Q_1{x}_2^2}=\frac{x_1}{x_2}$
$\frac{Q_2}{Q_1}=\frac{x_2}{x_1}$
$\frac{Q_1}{Q_2}=\frac{x_1}{x_2}$