Physics - Electrostatics Question with Solution | TestHub

Force on ($-q_1$) due to$q_2=\frac{q_1{q}_2}{4\pi {\epsilon }_0{b}^2}$

$\therefore$${\mathit{\text{F}}}_1=\frac{{\mathit{\text{q}}}_1{\mathit{\text{q}}}_2}{4\pi {\epsilon }_0{b}^2}$along$\left(q_1{q}_2\right)$

$\text{Force on}\left(-{\text{q}}_1\right)\text{due to}\left({-\text{q}}_3\right)$$=\frac{{( \mathit{\text{q}}}_1{\left)\right( \mathit{\text{q}}}_3)}{4\pi {\epsilon }_0{a}^2}$

${\mathit{\text{F}}}_2=\frac{{\mathit{\text{q}}}_1{\mathit{\text{q}}}_3}{4\pi {\epsilon }_0{a}^2}$as shown

$F_2$> makes an angle of$\left({\text{90}}^0-\theta \right)$with$\left(q_1{q}_2\right)$

Resolved  part of $F_2$  along  $q_1{q}_2$

$=F_2\cos \left({90}^0-\theta \right)$

$=\frac{{\mathit{\text{q}}}_1{\mathit{\text{q}}}_3\text{sin}\theta }{4\pi {\epsilon }_0{a}^2}$  along  $\left(q_1{q}_2\right)$

$\therefore$   Total force on $\left(-q_1\right)$

              $=[\frac{{\mathit{\text{q}}}_1{\mathit{\text{q}}}_2}{4\pi {\epsilon }_0{b}^2}+\frac{{\mathit{\text{q}}}_1{\mathit{\text{q}}}_3\text{sin}\theta }{4\pi {\epsilon }_0{a}^2}]$along x-axis

$\therefore$   x-component of force $\propto [\frac{{\mathit{\text{q}}}_2}{b^2}+\frac{{\mathit{\text{q}}}_3}{a^2}\text{sin}\theta ]$.