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Q.

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 q₀=q / 2 is placed at the origin. If charge q₀ 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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a

y

b

-1/y

c

-y

d

1/y

answer is D.

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Detailed Solution

From the figure shown here, the net force is
$\begin{array}{l} F_{net} = 2 Fcos θ = 2 \times \frac{k \times q \times (q / 2)}{(\sqrt{{a^{2} + y^{2})}^{2}}} \times \frac{y}{\sqrt{a^{2} + y^{2}}} \\ F_{net} = \frac{{kq}^{2} y}{a^{3}} (y ≪ a) \end{array}$
From the above relation, we see that the net force is directly proportional to y.

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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 q0=q / 2 is placed at the origin. If charge q0 is given a small displacement (y<<a) along the y-axis, the net force acting on the particle is proportional to: