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 $q_{0} = q/2$ 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:

Options:

-y

$\frac{-1}{y^2}$

$\frac{-1}{y}$

Correct Answer:

Explanation:

$F_{net} = 2Fcos\theta = \frac {2.k. q.q_0/2}{a^2+y^2} \times \frac{y}{\sqrt(a^2+y^2)} = \frac{kq^2y}{(a^2+y^2)^\frac{3}{2}}$

Since y << a

$\Rightarrow F = \frac{kq^2y}{a^3}$

$\Rightarrow F \propto y$

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: