A charged particle q is shot towards another charged particle Q which is fixed with a speed v.it approaches Q upto a closest distance r and then returns.if q was given a speed 2v,the closest distance of approach would be
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Expert's answer
2018-09-26T10:20:09-0400
All the kinetic energy is being completely converted into potential energy of system of two particles. In the first case: 1/2 mv^2 =kQq/r In the second case: 1/2 m〖(2v)〗^2 =kQq/r' Thus, r’ =r/4
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