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

Total charge on the ring `= lambda 2 pi R = 2 pi R lambda = Q`
`r = sqrt(R^(2) + (sqrt(3) R)^(2)) = 2R`
If the particle crosses the centre `O`, it will not come back to `P` due to repulsive force. The value of `v`will be minimum if speed of particle at origin is zero.
Applying energy conservation `P` and `O`
`(1)/(2) mv_(min)^(2) + qV_(p) = qV_(o)`
`(1)/(2) mv_(min)^(2) = q(V_(o) - V_(p)) = q . (1)/(4 pi in_(0)) [ (Q)/(R) - (Q)/(2R)]`
`= (Qq)/(8 pi in_(0) R) = (2 pi R lambda q)/(8 pi in_(0) R) = (lambda q)/(4 in_(0))`
`v_(min)^(2) = (lambda q)/(2 m in_(0))`
`v_(min) = sqrt((lambda q)/(2 m in_(0)))`