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NEET PHYSICSMedium

Two equal negative charges of charge -q are fixed at the points (0, a) and (0, -a) on the Y-axis. A positive charge Q is released from rest at the point (2a, 0) on the X-axis. The charge Q will:

A

execute simple harmonic motion about the origin.

B

move to the origin and remain at rest.

C

move to infinity.

D

execute oscillatory but not simple harmonic motion.

Step-by-Step Solution

The positive charge QQ is attracted by both fixed negative charges q-q. Due to the symmetry of the setup, the vertical components of the forces cancel out, while the horizontal components add up, creating a net restoring force directed towards the origin.

The magnitude of this net force at a distance xx from the origin is given by the superposition of Coulomb forces: Fnet=2kQqx(x2+a2)3/2F_{net} = \frac{2kQqx}{(x^2 + a^2)^{3/2}}.

For a particle to execute Simple Harmonic Motion (SHM), the restoring force must be directly proportional to the displacement (FxF \propto -x). In this case, the force is proportional to xx only when xax \ll a (small oscillations). However, the charge is released at x=2ax = 2a, which is not small compared to aa. Since the relationship between force and displacement is non-linear for this amplitude, the motion will be oscillatory (repetitive back and forth through the equilibrium point) but not simple harmonic.

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