Electrostatics 1 Question 7
7. Two charges, each equal to $q$, are kept at $x=-a$ and $x=a$ on the $x$-axis. A particle of mass $m$ and charge $q _0=\frac{q}{2}$ is placed at the origin. If charge $q _0$ is given a small displacement $y(y<a)$ along the $y$-axis, the net force acting on the particle is proportional to
(a) $y$
(b) $-y$
(c) $\frac{1}{y}$
(d) $-\frac{1}{y}$
(2013 Main)
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Answer:
Correct Answer: 7. (b)
Solution:
$$ \begin{aligned} F _{net} & =2 F \cos \theta \\ F _{net} & =\frac{2 k q \frac{q}{2}}{\left(\sqrt{y^{2}+a^{2}}\right)^{2}} \cdot \frac{y}{\sqrt{y^{2}+a^{2}}} \\ F _{\text {net }} & =\frac{2 k \sin \theta}{\left(y^{2}+a^{2}\right)^{3 / 2}} \Rightarrow \frac{k q^{2} y}{a^{3}} \propto y \end{aligned} $$
