
Three charges $-q_1,+q_2$ and $-q_3$ are placed as shown in the figure. The $x$-component of the force on…

- $\frac{\mathrm{q}_2}{\mathrm{~b}^2}-\frac{\mathrm{q}_3}{\mathrm{a}^2} \cos \theta$
- $\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$
- $\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \cos \theta$
- $\frac{\mathrm{q}_2}{\mathrm{~b}^2}-\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$
Solution

$\mathrm{F}_{\mathrm{X}}=\mathrm{F}_1 \sin \theta+\mathrm{F}_2=\frac{\mathrm{q}_1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_1}{\mathrm{~b}_2}\right] \Rightarrow \mathrm{F}_{\mathrm{X}} \propto\left(\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_2}{\mathrm{~b}^2}\right)$
Asked in: JEE Main 2003