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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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$- \frac{1}{y}$

–y

$\frac{1}{y}$

answer is A.

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Detailed Solution

$F_{n e t} = 2 F \cos θ$

$F_{n e t} = \frac{2 k q (\frac{q}{2})}{{(\sqrt{y^{2} + a^{2}})}^{2}} . \frac{y}{\sqrt{y^{2} + a^{2}}} F_{n e t} = \frac{2 k q (\frac{q}{2}) y}{{(y^{2} + a^{2})}^{3 / 2}} \Rightarrow \frac{k q^{2} y}{a^{3}} s o, F \infty y$

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