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)

Let a particle of charge q having velocity $v$ approaches Q upto a closest distance r and if the velocity becomes $2v,$ the closest distance will be $r$ . The law of conservation of energy yields, kinetic energy of particle = electric potential energy between them at closest distance of approach. Or $\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r}$ Or $\frac{1}{2}m{v}^2=k\frac{Qq}{r}$ ...(i) $\left(k=constant=\frac{1}{4\pi {\epsilon }_0}\right)$ and $\frac{1}{2}m{(2v)}^2=k\frac{Qq}{r}$ ?(ii) Dividing Eq. (i) by Eq. (ii), $\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{(2v)}^2}=\frac{\frac{kQq}{r}}{\frac{kQq}{r}}$ $\Rightarrow$ $\frac{1}{4}=\frac{r}{r}$ $\Rightarrow$ $r=\frac{r}{4}$