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
• B. Move to the origin and remain at rest
• C. Move to infinity
• D. Execute oscillatory but not simple harmonic motion

Answer 1

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

$F_{net}\Rightarrow 2Fcos\theta =2\frac{1}{4\pi {ϵ}_0}\frac{-qQ}{(a^2+x^2)}\times \frac{x}{(a^2+x^2{)}^{1/2}}$
i.e., $F_{net}=-\frac{1}{4\pi {ϵ}_0}.\frac{2qQx}{(a^2+x^2{)}^{3/2}}$
As the restoring force $F_{net}$ is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmonic.