A charged particle q is thrown with a velocity v towards another charged particle Q at rest. It approaches Q up to a closest distance r and then returns. If q is thrown with a velocity 2v, what should be the closest distance of approach?

Explanation

Let the closest distance of approach be r'. From the principle of conservation of energy we have,
kinetic energy = electrostatic potential energy.
In the first case,
`1/2 mv^(2) = 1/(4pi epsilon_0) cdot (qQ)/(r )`…………(1)
In the second case,
`1/2 m (2v)^(2) = 1/(4 pi epsilon_0) cdot (qQ)/(r')` .........(2) Dividing equation (1) by equation (2) we get,
`1/4 = (r')/(r) or, r' = r/4`.