Three charges $-q_{1},+q_{2}$ and $-q_{3}$ are placed as shown in the figure. The $x=$ component of the…
- $\frac{q_{2}}{b^{2}}-\frac{q_{3}}{a^{2}} \cos \theta$
- $\frac{q_{2}}{b^{2}}+\frac{q_{3}}{a^{2}} \sin \theta$
- $\frac{q_{2}}{b^{2}}+\frac{q_{3}}{a^{2}} \cos \theta$
- $\frac{q_{2}}{b^{2}}-\frac{q_{3}}{a^{2}} \sin \theta$
Solution
\(\mathrm{F}_2=\) Force applied by \(\mathrm{q}_2\) on \(-\mathrm{q}_1\)
\(\mathrm{F}_3=\) Force applied by \(\left(-q_3\right)\) on \(-q_1\)
\(x\)-component of Net force on \(-q_1\) is
\(\begin{aligned}
& F_X=F_2+F_3 \sin \theta=K \frac{q_1 q_2}{b^2}+K \cdot \frac{q_1 q_3}{a^2} \sin \theta \\
& \Rightarrow F_X=K\left[\frac{q_1 q_2}{b^2}+\frac{q_1 q_3}{a^2} \sin\theta\right] \\
& \Rightarrow F_X=K \cdot q_1\left[\frac{q_2}{b_2}+\frac{q_3}{a^2} \sin \theta\right] \Rightarrow F_X \propto\left(\frac{q_2}{b^2}+\frac{q_3}{a^2} \sin \theta\right)
\end{aligned}\)Asked in: JEE Mains - Electrostatics - Test 1