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

` sqrt (lambdaq)/( 2 varepsilon_0m)` .

Explanation

As both have same charge tharge threfore repulion will take place when charge reaches from `P` to `O`. If we provide such a velocity that it come to `O` then it will not retum to `P` For miimum velocity it should reach `O`
` V_P ( KQ)/( sqrt ( R^2 + 3 R^2)) = (KQ)/(2R) & V_0 = (KQ)/R`
From consevation of mechanical enrgy ` Delta U + Delta =0 rArr - delta K = Delta U`
`1/2 mv^2 = q (Delta V)`
`1/2 mv^2 = q (V_0 -V_P ) rArr 1/2 mv^2 = q [(Kq)/R - (KQ)/(2R)]`
`1/2 mv^2 = (KQq)/(2R) rArr v= sqrt (Qq)/(Rm)` put
`v= sqrt ( 1/(4 pi in_0) xx ( lambda(2 pi R)q)/(mR)) rArr v= sqrt (( almbdaq)/(2 in_0 m))`.