A circular ring of radius R with uniform positive charge density λ per unit length is fixed 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(√3R,0,0) on the positive X-axis directly towards O, with initial velocity v. The smallest value of the speed v such that the particle does not return to P is √λqxε0m. Find x.
Ring and Charge particle has a positive charge so both will repel each other.
But when Velocity of projection of charge q is given in such a way that it just cross the center of the ring then it will never come back because of repulsion force.
Potential at distance r from circumference of curve is V=kλ2πRr
Initially, r2=(√3R)2+R2⇒r=2R
finally r=R (particle at center)
From conservation of energy:
Loss in KE = Gain in Electrical potential energy.
so, m(v21−v22)2=kλ2πRR−kλ2πR2R
m(v21−0)2=kλ2π1−kλ2π2
mv212=kλ2π2 K=14πϵ0
⇒v=√λq2ϵom
On comparing from question x=2