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. $\frac{q_2}{b^2}-\frac{q_3}{a^2}\cos \theta$
• 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. $\frac{q_2}{b^2}-\frac{q_3}{a^2}\sin \theta$

Answer 1

Answer: (B)

Force on $\left(-q_1\right)$ due to $q_2=\frac{-q_1{q}_2}{4\pi {\epsilon }_0{b}^2}$
$\therefore F_1=\frac{q_1{q}_2}{4\pi {\epsilon }_0{b}^2}$ along $\left(q_1{q}_2\right)$
Force on $\left(-q_1\right)$ due to $\left(-q_3\right)=\frac{\left(-q_1\right)\left(-q_3\right)}{4\pi {\epsilon }_0{a}^2}$

$F_2=\frac{q_1{q}_3}{4\pi {\epsilon }_0{a}^2}$ as shown
$F_2$ makes an angle of
$\left(90^{\circ }-\theta \right)$ with $\left(q_1{q}_2\right)$
Resolved part of $F_2$ along $q_1{q}_2$
$=F_2\cos \left(90^{\circ }-\theta \right)=\frac{q_1{q}_3\sin \theta }{4\pi {\epsilon }_0{a}^2}\text{ along }\left(q_1{q}_2\right)$
$\therefore$ Total foroe on $\left(-q_1\right)=\left[\frac{q_1{q}_2}{4\pi {\epsilon }_0{b}^2}+\frac{q_1{q}_3\sin \theta }{4\pi {\epsilon }_0{a}^2}\right]$ along $x$ -axis
$\therefore x$ -component of force $\propto \left[\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \theta \right]$