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29 July, 23:25

A charge 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 were given a speed of '2v' the closest distance of approach would be

(a) r2 (b) 2r (c) r (d) r4

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  1. 30 July, 02:19
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    In the given process, the kinetic energy is converted into electric potential energy

    For closest distance of r

    1 / 2 m v² = k Qq / r, [ m is mass of the charged particle ]

    For velocity 2v

    K E = 1 / 2 m (2v) ²

    = 1/2 m 4v²

    4 x1/2 mv²

    K E = 4 x k Qq / r

    If R be new closest distance

    K E = k Qq / R

    4 x k Qq / r = k Qq / R

    4 R = r
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