Question

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

Answer 1

Answer:

Explanation:

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