First slide

Potential energy

Question

A charged particle q is shot towards another charged particle Q which is fixed, with a speed $ν$ . It approaches Q upto a closest distance r and then returns. If q were given a speed 2 $ν$ , the closest distances of approach would be

Moderate

A

r
B

2r
C

r/2
D

r/4

Solution

Charge q will momentarily come to rest at a distance r from charge Q when all it's kinetic energy converted to potential energy i.e. $\frac{1}{2} {mv}^{2} = \frac{1}{4 {πε}_{0}} . \frac{qQ}{r}$
Therefore the distance of closest approach is given by $r = \frac{qQ}{4 {πε}_{0}} . \frac{2}{{mv}^{2}} \Rightarrow r \propto \frac{1}{v^{2}}$

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