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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
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B
move to the origin ad 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 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.

863177_124880_ans_dd098b65ee6e4b978497550d765bfdfb.png

Physics

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