Q. An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r.$ The coulomb force $\overrightarrow{F}$ between the two is $\left( where\ K = \frac{1}{4\pi\varepsilon_{0}} \right)$
A) $- K\frac{e^{2}}{r^{3}}\widehat{r}$
B) $K\frac{e^{2}}{r^{3}}\overrightarrow{r}$
C) $- K\frac{e^{2}}{r^{3}}\overrightarrow{r}$
D) $K\frac{e^{2}}{r^{3}}\widehat{r}$
Explanation
(c) $\overrightarrow{F} = \ - k\frac{e^{2}}{r^{2}}\widehat{r} = \ - k.\frac{e^{2}}{r^{3}}\overrightarrow{r}$ $\because\left\lbrack \widehat{r} = \frac{\overrightarrow{r}}{r} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.