A circular ring of radius R with uniform positive charge density `lambda` per unit length is located in the y-z plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the point P `(Rsqrt3, 0, 0)` on the positive x-axis directly towards O, with an initial speed v. Find the smallest (non-zero) value of the speed v such that the particle does not return to P.

Explanation

`V_(x)=(kq)/(sqrt(R^(2)+x^(2)))=(kq)/(sqrt(R^(2)+(sqrt(3)R)^(2)))=(kq)/(2R)`
P.E. of charge q at `P(sqrt3,0,0)` is `U_(P)=Vq=(kq^(2))/(2R)`
P.E. of charge q at the centre of ring
`U_(0)=V_(0)q=(kq^(2))/(R)`
In order that the charged particle does not return to P, it must just cross the centre O and thereafter it will be repelled on the other side
Applying energy conservation principle
`(KE)_(p)+U_(P)=U_(0)`
`implies (1)/(2)mv^(2)+(kq^(2))/(2R)=(kq^(2))/(R)impliesV=sqrt((kq^(2))/(mR))=sqrt((q^(2))/(4piepsilon_(0)mR))`