Question

A thin fixed ring of radius 1 m has a positive charge $1\times 10^{-5}C$ uniformly distributed over it. A particle of mass 0.9 g and having a negative charge of $1\times 10^{-6}$ C is placed on the axis at a distance of 1 cm from the centre of the ring. Show that the motion of the negatively charged particle is approximately simple harmonic. Calculate the time period of oscillations.

Answer 1

Given $Q=10^{-5}C$ $q=10^{-6}C,R=1m$ and $m=9\times 10^{-4}kg$ Electric field at a distance x from the centre on the axis of a ring is given by $E=\frac{1}{4\pi {\epsilon }_0}\frac{Qx}{(R^2+x^2{)}^{3/2}}$ If $x<<R,R^2+x^2\simeq R^2$ and $E\simeq \frac{Qx}{4\pi {\epsilon }_0{R}^3}$ Net force on negatively charged particle would be qE and towards the centre of ring. Hence, we can write $F=-\frac{Qqx}{(4\pi {\epsilon }_0)R^3}$ or acceleration $a=\frac{F}{m}=-\frac{Qqx}{(4\pi {\epsilon }_0)m{R}^3}$ as $a-x,$ motion of the particle is simple harmonic in nature. Time period of which will be given by $T=2\pi \sqrt{|\frac{x}{a}}|$ or $T=2\pi \sqrt{\frac{(4\pi {\epsilon }_0)m{R}^3}{Qq}}$ Substituting the values, we get $T=2\pi \sqrt{\frac{(9\times 10^{-4})(1{)}^3}{(9\times 10^9)(10^{-5})(10^{-6})}}$ =0.628 s