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A particle of mass m and charge -Q is constrained to move along the axis of a ring of radius R .The ring carries a uniform charge density $\;+ \lambda\;$ along its length . Initially the particle is in the centre of the ring . Now it is displaced slightly along the axis of ring it's time period of oscillation is

$(a)\;2 \pi\;\sqrt{\large\frac{\in_{0}\;m\;R^2}{\lambda\;Q}}\qquad(b)\;\pi\;\sqrt{\large\frac{2 \in_{0}\;m\;R^2}{\lambda\;Q}}\qquad(c)\;2 \pi\;\sqrt{\large\frac{2 \in_{0}\;m\;R^2}{\lambda\;Q}}\qquad(d)\; 4\pi\;\sqrt{\large\frac{2 \in_{0}\;m\;R^2}{\lambda\;Q}}$

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Answer : (c) $\;2 \pi\;\sqrt{\large\frac{ 2 \in_{0}\;m\;R^2}{\lambda\;Q}}$
Explanation :
Electric potential on the axis of ring at a distance x is =$\; \large\frac{k\;q}{(x^2+R^2)^{\large\frac{1}{2}}}$
and Electric field on the axis of ring is $\;E_{x}=-\large\frac{\partial V}{\partial x } \hat{i}$
$E_{x}=-(-\large\frac{1}{2}\;\large\frac{k\;q\;2\;x}{(x^2R^2)^{\large\frac{3}{2}}})$
$E_{x}=\large\frac{k\;q\;x}{(x^2+R^2)^{\large\frac{3}{2}}}=\large\frac{\lambda R x}{2 \in_{0} (x^2+R^2)^{\large\frac{3}{2}} }$
Restoring force $F_{res}=-QE_{x}$
$=-\large\frac{Q \lambda R x }{2 \in_{0} (x^2+R^2)^{\large\frac{3}{2}} }$
Since x < < R
$F_{res} \approx -\large\frac{Q \lambda R x }{2 \in_{0} R^3}$
$F_{res}\;\propto \; -x\;.$ Thus the particle perform simple harmonic motion
$a=\large\frac{F_{res}}{m}=-\large\frac{Q \lambda R x}{m R^3 2 \in_{0}}=-w^2 x$
$w=\sqrt{\large\frac{Q \lambda R}{2 \in_{0} m R^3}}$
$T=\large\frac{2 \pi}{w}=2 \pi \sqrt{\large\frac{2 \in_{0} m R^2}{Q \lambda}}$
answered Feb 18, 2014 by yamini.v
edited Aug 22, 2014 by meena.p
 

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