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) $\frac{q_{2}}{b^{2}} - \frac{q_{3}}{{a\ }^{2}}\cos\theta$
B) $\frac{q_{2}}{b^{2}} + \frac{q_{3}}{{a\ }^{2}}\sin\theta$
C) $\frac{q_{2}}{b^{2}} + \frac{q_{3}}{{a\ }^{2}}\cos\theta$
D) $\frac{q_{2}}{b^{2}} - \frac{q_{3}}{{a\ }^{2}}\sin\theta$
Explanation
(b) Force on $–q_{1}$
[image:1_image47.emf]
$\mathbf{F} = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{b^{2}}\widehat{\mathbf{i}} + \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{3}}{a^{2}}\lbrack sin\theta\widehat{\mathbf{i}\ } - cos\theta\widehat{\mathbf{j}}\rbrack$
From above, $x''''''''$ component of force is
[image:1_image264.emf]
$F_{x} = \frac{q_{1}}{4\pi\varepsilon_{0}}\left\lbrack \frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}}sin\theta \right\rbrack$
$F_{x} \propto \left\lbrack \frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}}sin\theta \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.