18. 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 2v, the closest distance of approach would be
Energy is conserved in the phenomenon
Initially, 1/2 mv^2 = kqQ/ r ......(i)
Finally , 1/2 m (2v)^2 = kqQ/r1 ......(ii)
From eqns (i) and (ii) , we get
1/4 = r1/r ⇒r1 = r/4
Energy is conserved in the phenomenon
Initially, 1/2 mv^2 = kqQ/ r ......(i)
Finally , 1/2 m (2v)^2 = kqQ/r1 ......(ii)
From eqns (i) and (ii) , we get
1/4 = r1/r⇒ r1 = r/4