Q. 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) $\frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}}\cos\theta$
Explanation
(c) $F_{2} =$ Force applied by $q_{2}$ on $- q_{1}$
$F_{3} =$ Force applied by $( - q_{3})$ on $- q_{1}$
$x$-component of Net force on $- q_{1}$ is
$F_{x} = F_{2} + F_{3}\sin\theta = k\frac{q_{1}q_{2}}{b^{2}} + k.\frac{q_{1}q_{3}}{a^{2}}\sin\theta$
$\Rightarrow F_{x} = k.q_{1}\left\lbrack \frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}}\sin\theta \right\rbrack \Rightarrow F_{x} \propto \left\lbrack \frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}}\sin\theta \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.