Three charges $-q_{1}, +q_{2}$ and $-q_3$ are placed as shown in the figure. The x-component of the force on $-\alpha_1$ is proportional to
Text Solution
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F 2 = Force applied by $\alpha_{2}$ on $-q_1$
F 3 = Force applied by $\left(-q_{3}\right)$ on – $\alpha_{1}$
x-component of Net force on $-q_1$ is
F x = F 2 + F 3 sin θ θ $=k \frac{q_1 q_2}{b^2} + k \frac{q_1 q_3}{a^2} \sin \hat{\theta}$
⇒ ⇒ $F_x = k \left[ \frac{q_1 q_2}{b^2} + \frac{q_1 q_3}{a^2} \sin \theta \right]$
⇒ ⇒ $F_x = k \cdot q_1 \left[ \frac{q_2}{b^2} + \frac{q_3}{a^2} \sin \theta \right]$ ⇒ ⇒ $F_x \propto \left(\frac{q_2}{b^2} + \frac{q_3}{a^2} \sin \theta \right)$
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