Question

Three charges $-q_1$ , $q_2$ and $-q_3$ are placed as shown in the figure. The $x$ -component of the force on $-q_1$ is proportional to

• A. $\frac{q_2}{b^2}-\frac{q_3}{a^2}sin\theta$
• B. $\frac{q_2}{b^2}-\frac{q_3}{a^2}cos\theta$
• C. $\frac{q_2}{b^2}+\frac{q_3}{a^2}sin\theta$
• D. $\frac{q_2}{b^2}+\frac{q_3}{a^2}cos\theta$

Answer 1

Answer: (C)

$x$ component of force on $-q\Rightarrow$


$F_{1}cos(90-\theta )+F_2$
$=\frac{K{q}_1{q}_3}{a^2}sin\theta +\frac{K{q}_1{q}_2}{b^2}$