Question

A particle carrying a charge q is shot with a speed v towards a fixed particle carrying a charge Q. It approaches Q up to a certain distance r and then returns as shown in Fig. 20.12

If charge q were moving with a speed 2v, the distance of the closest approach would be

• A. r
• B. $\frac{r}{8}$
• C. $\frac{r}{2}$
• D. $\frac{r}{4}$

Answer 1

The correct option is D $\frac{r}{4}$

Charge q will momentarily come to rest at a distance r from charge Q when all its KE is converted to PE, i.e.,
$\frac{1}{2}m{v}^2=\frac{1}{4\pi {ϵ}_0}.\frac{qQ}{r}$
Therefore, the distance of closest approach is given by
$r=\frac{qQ}{4\pi {ϵ}_0}.\frac{2}{m{v}^2}$
Thus $r\propto \frac{1}{v^2}$. Hence if v is doubled, r becomes one-fourth. Thus the correct choice is (d).