question_answer

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 [IIT 1984; Bihar MEE 1995; MP PMT 1996]

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

Correct Answer: D

Solution :

           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 \[{{F}_{net}}\Rightarrow 2F\cos \theta =2\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-qQ}{({{a}^{2}}+{{x}^{2}})}\times \frac{x}{{{({{a}^{2}}+{{x}^{2}})}^{{1}/{{}}\;2}}}\] i.e., \[{{F}_{net}}=-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2qQx}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{{3}/{{}}\;2}}}\] As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmonic.