Question

Three charges $- \text{q}_{1} , \, + \text{q}_{2}$ and $- \text{q}_{3}$ are placed as shown in the figure. The $\text{x} -$ component of the force on $- \text{q}_{1}$ is proportional to

• A. $\frac{\textit{q}_{2}}{b^{2}} - \frac{\textit{q}_{3}}{a^{2}} \text{cos} \theta $
• B. $\frac{\textit{q}_{2}}{b^{2}} + \frac{\textit{q}_{3}}{a^{2}} \text{sin} \theta $
• C. $\frac{\textit{q}_{2}}{b^{2}} + \frac{\textit{q}_{3}}{a^{2}} \text{cos} \theta $
• D. $\frac{\textit{q}_{2}}{b^{2}} - \frac{\textit{q}_{3}}{a^{2}} \text{sin} \theta $

Answer 1

Answer: (B)

Force on ( $- q_{1}$ ) due to $q_{2} = \frac{q_{1} q_{2}}{4 \pi \epsilon _{0} b^{2}}$
$\therefore $ $\textit{F}_{1} = \frac{\textit{q}_{1} \textit{q}_{2} }{4 \pi \epsilon _{0} b^{2}}$ along $\left(q_{1} q_{2}\right)$
Force on $\left(-\mathrm{q}_1\right)$ due to $\left(-\mathrm{q}_3\right)=\frac{\left(q_1\right)\left(q_3\right)}{4 \pi \varepsilon_0 a^2}$
$\textit{F}_{2} = \frac{\textit{q}_{1} \textit{q}_{3} }{4 \pi \epsilon _{0} a^{2}}$ as shown
$F_{2}$ > makes an angle of $F_2>$ makes an angle of $\left(90^{\circ}-\theta\right)$ with $\left(q_1 q_2\right)$
Resolved part of $F_{2}$ along $q_{1} q_{2}$
$= F_{2} cos \left(\left(90\right)^{0} - \theta \right)$
$ = \frac{\textit{q}_{1} \textit{q}_{3} \text{sin} \theta }{4 \pi \epsilon _{0} a^{2}}$ along $\left(q_{1} q_{2}\right)$
$\therefore $ Total force on $\left(- q_{1}\right)$
$= \left[\right. \frac{\textit{q}_{1} \textit{q}_{2}}{4 \pi \epsilon _{0} b^{2}} + \frac{\textit{q}_{1} \textit{q}_{3} \text{sin} \theta }{4 \pi \epsilon _{0} a^{2}} \left]\right.$ along x-axis
$\therefore $ x-component of force $ \propto \left[\right. \frac{\textit{q}_{2}}{b^{2}} + \frac{\textit{q}_{3}}{a^{2}} \text{sin} \theta \left]\right.$ .