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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
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Verified Answer
The correct answer is:
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$
The force body diagram
$F_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_3}{a^2}$ $\mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1 \mathrm{q}_3}{\mathrm{~b}^2}$
$\mathrm{F}_{\mathrm{X}}=\mathrm{F}_1 \sin \theta+\mathrm{F}_2=\frac{\mathrm{q}_1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_1}{\mathrm{~b}_2}\right] \Rightarrow \mathrm{F}_{\mathrm{X}} \propto\left(\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_2}{\mathrm{~b}^2}\right)$
$F_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_3}{a^2}$ $\mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1 \mathrm{q}_3}{\mathrm{~b}^2}$
$\mathrm{F}_{\mathrm{X}}=\mathrm{F}_1 \sin \theta+\mathrm{F}_2=\frac{\mathrm{q}_1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_1}{\mathrm{~b}_2}\right] \Rightarrow \mathrm{F}_{\mathrm{X}} \propto\left(\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_2}{\mathrm{~b}^2}\right)$
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