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 $q$ were given a speed $2v$, the closest distances of approach would be

• A. r
• B. 2r
• C. r/2
• D. r/4

Answer 1

Answer: (D)

r/4

Answer 2

By principle of conservation of energy $\frac{1}{2}mv^{2} = \frac{KqQ}{r}\quad\quad\dots\left(i\right)$ Finally, $\frac{1}{2}m\left(2v\right)^{2} = \frac{KqQ}{r^{2}}\quad \quad \dots \left(ii\right)$ Equation $\left(i\right) \div \left(ii\right)$, $\frac{1}{4} = \frac{r'}{r}$ $\Rightarrow\quad r' = \frac{r}{4}.$