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)

$\frac{\textit{q}_{2}}{b^{2}} + \frac{\textit{q}_{3}}{a^{2}} \text{sin} \theta$

Answer 2

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 $\left(q_{1} q_{2}\right)$ $\text{Force on} \left(- \left(\text{q}\right)_{1}\right) \text{due to} \left(\left(- \text{q}\right)_{3}\right)$ $= \frac{\left(\left(\right. \textit{q}\right)_{1} \left(\left.\right) \left(\right. \textit{q}\right)_{3} \left.\right) }{4 \pi \left(\epsilon \right)_{0} a^{2}}$ $\textit{F}_{2} = \frac{\textit{q}_{1} \textit{q}_{3} }{4 \pi \epsilon _{0} a^{2}}$ as shown $F_{2}$ > makes an angle of $\left(\left(\text{90}\right)^{0} - \theta \right)$ 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 $\propto \left[\right. \frac{\textit{q}_{2}}{b^{2}} + \frac{\textit{q}_{3}}{a^{2}} \text{sin} \theta \left]\right.$ .