Question

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 were given a speed of 2v the closest distance of approach would be?

r

2r

r/2

r/4

Answer 1

By principle of conservation of energy,

1/2mv² = kqQ/r ….(1)

Finally ,

½ m (2v²) = kqQ/r’ ….(2)

Dividing equation (1) by (2)

r’ = r/4