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A circular ring of radius R with uniform...

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.

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The correct Answer is:
A, B, D
Potential energy can be found at the initial point A and final point O. The difference in potential energy has to be provided by the K.E. of the charge at A.

`V(x)=(1)/(4piepsilon_0)*(Q)/(sqrt(R^2+x^2)),` at A.
`V_O=(1)/(4piepsilon_0)*(2piRlambda)/(R)`, at O. or `V_O=(lambda)/(2epsilon_0)`
`V_P=(1)/(4piepsilon_0)(2piRlambda)/(sqrt(R^2+(sqrt3R)^2))=(lambda)/(4epsilon_0)`
Potential difference between points O and P=V
`:.` `V=V_O-V_P`
or `V=(lambda)/(2epsilon_0)-(lambda)/(4epsilon_0)` or `V=(lambda)/(4epsilon_0)`
The kinetic energy of the charged particle is converted into its potential energy at O.
`:.` Potential energy of charge (q) =qV
Kinetic energy of charged particle `=1/2mv^2`
For minimum speed of particle so that it does not return to P,
`1/2mv^2=qV` or `v^2=(2qV)/(m)=(2qxxlambda)/(mxx4epsilon_0)`
or `v=sqrt((qlambda)/(2epsilon_0m))`
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