A circular ring of radius R with uniform positive charge density `lambda` per unit length is located in the y z plane with its center at the origin O. A particle of mass m and positive charge q is projected from that point `p( - sqrt(3) R, 0,0)` on the negative x - axis directly toward O, with initial speed V. Find the smallest (nonzero) value of the speed such that the particle does not return to P ?

Answer

2

Explanation

Total charge in the right is `Q=(2piR)lambdaQ=(2piR)lambda`
Potential due to a ring at a distance of x from its centre on its axis is given by `V(x)=(1)/(4piepsi_0).(Q)/(sqrt(R^(2)+x^(2)))`
And at the centre is `V_("centre")=(1)/(4piepsi_0)(Q)/(R),` Using the above formula
`V_(p)=(1)/(4piepsi_0).(2piRlambda)/(sqrt(R^(2)+3R^(2)))=(lambda)/(4epsi_0)" "," "V_(0)=(1)/(4piepsi_0).(2piRlambda)/(R)=(lambda)/(2epsi_0)," "V_(0)gtV_(p)`
PD between points O and P is `V=V_(0)-V_(p)=(lambda)/(2epsi_0)-(lambda)/(4epsi_0)=(lambda)/(4epsi_0)`
`therefore" "(1)/(2)mv^(2)gt=qV" "or" "vgt=sqrt((2qV)/(m))" "or" "vgt=sqrt((2qV)/(m))" "or" "vgt=sqrt((qlambda)/(2epsi_(0)m))`