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 $


$F_{1} \, cos\left(\right.90 \, -\theta \left.\right) \, + \, F_{2}$
$= \, \frac{K q_{1} q_{3}}{a^{2}}sin\theta +\frac{K q_{1} q_{2}}{b^{2}}$