Question

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

• A. $\frac{q_2}{a^2}+\frac{q_3}{b^2}sin\theta$
• B. $\frac{q_2}{a^2}+\frac{q_3}{b^2}cos\theta$
• C. $\frac{q_2}{b^2}+\frac{q_3}{a^2}sin\theta$
• D. $\frac{q_2}{b^2}+\frac{q_3}{a^2}cos\theta$

Answer 1

The correct option is C $\frac{q_2}{b^2}+\frac{q_3}{a^2}sin\theta$

$F_{12}=\frac{1}{4\pi {ϵ}_0}\frac{q_3{q}_1}{a^2}[sin\theta \hat{i}+cos\theta (-\hat{j}\left)\right]$
$F_x=\frac{1}{4\pi {ϵ}_0}\frac{q_2{q}_1}{b^2}\left(\widehat{(}i\right))+\frac{1}{4\pi {ϵ}_0}\frac{q_3{q}_1}{a^2}sin\theta \widehat{(}i)=\frac{1}{4\pi {ϵ}_0}{q}_1[\frac{q_2}{b^2}+\frac{q_3}{a^2}sin\theta ]\left(\hat{i}\right)$
$\therefore F_x\propto [\frac{q_2}{b^2}+\frac{q_3}{a^2}sin\theta ]$