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<<a) along the y-axis, the net force acting on the particle is proportional to

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$- y$

$- \frac{1}{y}$

$\frac{1}{y}$

$y$

answer is B.

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Detailed Solution

$\begin{array}{l} F_{net} = 2F cos θ \\ F_{net} = \frac{2kq (\frac{q}{2})}{{(\sqrt{y^{2} {+a}^{2}})}^{2}} . \frac{y}{\sqrt{y^{2} {+a}^{2}}} \\ F_{net} = \frac{{kq}^{2} y}{a^{3}} \Rightarrow F_{n e t} \propto y. \end{array}$ ; $y < < a$

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