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Q.
Three charges $-q_{1},+q_{2}$ and $-q_{3}$ are placed as shown in the figure. The $x$ -component of the force on $-q_{1}$ is proportional to
Electric Charges and Fields
Solution:
Force on $\left(-q_{1}\right)$ due to $q_{2}=\frac{-q_{1} q_{2}}{4 \pi \varepsilon_{0} b^{2}}$
$\therefore F_{1}=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} b^{2}}$ along $\left(q_{1} q_{2}\right)$
Force on $\left(-q_{1}\right)$ due to $\left(-q_{3}\right)=\frac{\left(-q_{1}\right)\left(-q_{3}\right)}{4 \pi \varepsilon_{0} a^{2}}$
$F_{2}=\frac{q_{1} q_{3}}{4 \pi \varepsilon_{0} a^{2}}$ as shown
$F_{2}$ makes an angle of
$\left(90^{\circ}-\theta\right)$ with $\left(q_{1} q_{2}\right)$
Resolved part of $F_{2}$ along $q_{1} q_{2}$
$=F_{2} \cos \left(90^{\circ}-\theta\right)=\frac{q_{1} q_{3} \sin \theta}{4 \pi \varepsilon_{0} a^{2}} \text { along }\left(q_{1} q_{2}\right)$
$\therefore $ Total foroe on $\left(-q_{1}\right)=\left[\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} b^{2}}+\frac{q_{1} q_{3} \sin \theta}{4 \pi \varepsilon_{0} a^{2}}\right]$ along $x$ -axis
$\therefore x$ -component of force $\propto\left[\frac{q_{2}}{b^{2}}+\frac{q_{3}}{a^{2}} \sin \theta\right]$
