  Question

Two equal negative charges are fixed at the points $$(0,\pm a)$$ on the y-axis. A positive charge $$Q$$ is released from rest at the points $$(2a,0)$$ on the x-axis. The charge $$Q$$ will :

A
execute simple harmonic motion about the origin  B
move to the origin ad remain at rest  C
move to infinity  D
execute oscillatory but not simple harmonic motion  Solution

The correct option is D execute oscillatory but not simple harmonic motionBy symmetry of the 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 $$0$$. If at any time charge $$Q$$ is at a distance $$X$$ from $$O$$.Net force on charge $$Q$$,$$F_{net}=2F\cos{\theta}$$$$=2\times \dfrac{1}{4\pi\varepsilon_{0}}.\dfrac{-qQ}{(a^{2}+x^{2})}\times \dfrac{x}{(a^{2}+x^{2})^{1/2}}$$$$F_{net}=\dfrac{-1}{4\pi\varepsilon_{0}}.\dfrac{2qQx}{(a^{2}+x^{2})^{3/2}}$$As restoring force $$F_{net}$$ is not linear, motion will be oscillatory (with amplitude $$2a$$) but not simple harmonic. Physics

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