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Question

Two equal negative charge – q are fixed at the fixed 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
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B
Move to the origin and remain at rest
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C
Move to infinity
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D
Execute oscillatory but not simple harmonic motion
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Solution

The correct option is D Execute oscillatory but not simple harmonic motion
By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO. Under the action of this force charge Q will move towards O. If at any time charge Q is at a distance x from O. Net force on charge Q

Fnet2F cos θ=214πϵ0qQ(a2+x2)×x(a2+x2)1/2
i.e., Fnet=14πϵ0.2qQx(a2+x2)3/2
As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmonic.

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