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 was given a speed 2v, the closest distance of approach would be:

• A. r
• B. 2r
• C. r/2
• D. r/4

Answer 1

Answer: (D)

r/4

As initially q is at infinite distance from Q, thus $P.E_i=0$

For case 1: the closest distance given is r and velocity of q is v initially.

Applying conservation of energy, $P.E_i+K.E_i=P.E_f+K.E_f$

$0+\frac{1}{2}m{v}^2=\frac{KQq}{r}+0$

$⟹$$\frac{1}{2}m{v}^2=\frac{KQq}{r}$ ................(1)

For case 1: Let the closest distance be r' and velocity of q is 2v initially.

Applying conservation of energy, $P.E_i+K.E_i=P.E_f+K.E_f$

$0+\frac{1}{2}m(2v{)}^2=\frac{KQq}{r^{\prime }}+0$

$⟹$$\frac{1}{2}m(2v{)}^2=\frac{KQq}{r^{\prime }}$ ................(2)

Solving (1) and (2), $r^{\prime }=\frac{r}{4}$