Question

Positive charge Q is distributed uniformly over a ircular ring of radius R. A particle having having a mass ma and a negative charge q, is placed on its axis at a distanc ex from the centre. Find the force on the particle. Assuming x<< R, Find the time period of oscillation of the particle if it is released from there.

Answer 1

$dq=\frac{Q}{2\pi R}.Rd\theta =\frac{Q}{2\pi }.d\theta dE=\frac{qdq}{4\pi {ϵ}_0},\frac{1}{r^2}=\frac{q.\frac{Q}{2\pi }}{4\pi {ϵ}_0}.\frac{1}{R^2+x^2}$

or $dE\cos \theta =\frac{qQd\theta }{8{\pi }^2{ϵ}_0}\frac{x}{(R^2+x^2{)}^{\frac{3}{2}}}\text{Total force}=\int df\cos \theta =\frac{qQ}{8{\pi }^2{ϵ}_0}.\frac{x}{(R^2+x^2{)}^{\frac{3}{2}}}\int d\theta =\frac{xQq}{4\pi {ϵ}_0(R^2+x^2{)}^{\frac{3}{2}}}(\because \int d\theta =2\pi )x<<RF=\frac{Qqx}{4\pi {ϵ}_0{R}^3}=m{w}^{2}xorm{\left(\frac{2\pi }{T}\right)}^2=\frac{Qq}{4\pi {ϵ}_0{R}^3}T={\left[\frac{16{\pi }^3{ϵ}_{0}m{R}^3}{Qq}\right]}^{\frac{1}{2}}$