Q.
Three charges $–q_{1}, + q_{2}$ and$\ - q_{3}$ are placed as shown in figure. The $x$ component of the force on $–q_{1}$ is proportional to
A) $\frac{q_{2}}{b^{2}} - \frac{q_{3}}{b^{2}}\ $sin θ
B) $\frac{q_{2}}{b^{2}} - \frac{q_{3}}{b^{2}}\ $cos θ
C) $\frac{q_{2}}{b^{2}} + \frac{q_{3}}{b^{2}}\ $sin θ
D) $\frac{q_{2}}{b^{2}} + \frac{q_{3}}{b^{2}}\ $cos θ
Explanation
(c) Force on $–q_{1\ }$due to $q_{2}\ $is
$F_{2} = \frac{kq_{1}q_{2}}{b^{2}}\ $ along X-axis
Force on $–q_{1}$ due to $–q_{2\ }$is
$F_{13} =$ $\frac{kq_{1}q_{3}}{a^{2}}\ at\ \angle\theta\ $With negative direction of Y-axis.
$\therefore\ x\ $component of force on $–q_{1}$ is
$F_{x} = F_{12} + F_{13}\sin\theta = kq_{1}\left\lbrack \ \frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}}\sin\theta \right\rbrack$
$ie,\ \ \ F_{x} \propto \left\lbrack \ \frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}}\sin\theta \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.