Two charges, each equals 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 (a) y (b) -y (c) 1/y (d) -1/y

Answer: (a). Each charge q repels q₀ with F = (kqq₀)/(a² + y²); the x -components cancel. Net force along y: 2F(y)/(√(a² + y²)) ≈ (2kqq₀)/(a³)y = (kq²)/(a³)y for y a. It points away from the origin, so F ∝ +y (not restoring). So the answer is (a) y.

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Q41JEE Main 2013Medium

Motion of Charges in an Electric Field

Two charges, each equals 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 $(y \ll a)$ along the $y$ -axis, the net force acting on the particle is proportional to

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Answer: (a)

Each charge $q$ repels $q_0$ with $F = \frac{kqq_0}{a^2 + y^2}$ ; the $x$ -components cancel.
Net force along $y$ : $2F\frac{y}{\sqrt{a^2 + y^2}} \approx \frac{2kqq_0}{a^3}y = \frac{kq^2}{a^3}y$ for $y \ll a$ .
It points away from the origin, so $F \propto +y$ (not restoring).
So the answer is (a) $y$ .

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