Three charges -q₁, +q₂and -q₃ are placed as shown in the figure. The x-component of the force on -q₁ is proportional to

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$\frac{q_{2}}{b^{2}} - \frac{q_{3}}{a^{2}} cosθ$

$\frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}} sinθ$

$\frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}} cosθ$

$\frac{q_{2}}{b^{2}} - \frac{q_{3}}{a^{2}} sinθ$

answer is D.

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Detailed Solution

F₂ = Force applied by q₂ on -q₁
F₃ = Force applied by (- q₃) on -q₁
x-component of net force on -q₁ is

$F_{x} = F_{2} + F_{3} sinθ = k \frac{q_{1} q_{2}}{b^{2}} + k \cdot \frac{q_{1} q_{3}}{a^{2}} sinθ \Rightarrow F_{x} = k [\frac{q_{1} q_{2}}{b^{2}} + \frac{q_{1} q_{3}}{a^{2}} sinθ] \Rightarrow F_{x} = k \cdot q_{1} [\frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}} sinθ] \Rightarrow F_{x} \propto (\frac{q_{2}}{b^{2}} + \frac{q_{3}}{a^{2}} sinθ)$