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. $\frac{{\text{q}}_2}{b^2}-\frac{{\text{q}}_3}{a^2}\text{cos}\theta$
• B. $\frac{{\text{q}}_2}{b^2}+\frac{{\text{q}}_3}{a^2}\text{sin}\theta$
• C. $\frac{{\text{q}}_2}{b^2}+\frac{{\text{q}}_3}{a^2}\text{cos}\theta$
• D. $\frac{{\text{q}}_2}{b^2}-\frac{{\text{q}}_3}{a^2}\text{sin}\theta$

Answer 1

Answer: (B)

Force on ( $-q_1$ ) due to $q_2=\frac{q_1{q}_2}{4\pi {ϵ}_0{b}^2}$
$\therefore$ ${\text{F}}_1=\frac{{\text{q}}_1{\text{q}}_2}{4\pi {ϵ}_0{b}^2}$ along $\left(q_1{q}_2\right)$
Force on $\left(-{\mathrm{q}}_1\right)$ due to $\left(-{\mathrm{q}}_3\right)=\frac{\left(q_1\right)\left(q_3\right)}{4\pi {\epsilon }_0{a}^2}$
${\text{F}}_2=\frac{{\text{q}}_1{\text{q}}_3}{4\pi {ϵ}_0{a}^2}$ 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 $q_1{q}_2$
$=F_{2}cos\left({\left(90\right)}^0-\theta \right)$
$=\frac{{\text{q}}_1{\text{q}}_3\text{sin}\theta }{4\pi {ϵ}_0{a}^2}$ along $\left(q_1{q}_2\right)$
$\therefore$ Total force on $\left(-q_1\right)$
$=[\frac{{\text{q}}_1{\text{q}}_2}{4\pi {ϵ}_0{b}^2}+\frac{{\text{q}}_1{\text{q}}_3\text{sin}\theta }{4\pi {ϵ}_0{a}^2}]$ along x-axis
$\therefore$ x-component of force $\propto [\frac{{\text{q}}_2}{b^2}+\frac{{\text{q}}_3}{a^2}\text{sin}\theta ]$ .