Question

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force $\vec{F}$ between the two is

• A. $k\frac{e^2}{r^3}\vec{r}$
• B. $-k\frac{e^2}{r^3}\vec{r}$
• C. $-k\frac{e^2}{r^2}\hat{r}$
• D. $-k\frac{e^2}{r^3}\hat{r}$ (where $k=\frac{1}{4\pi {\epsilon }_0}$ )

Answer 1

Answer: (B)

Let charges on an electron and hydrogen nucleus are $q_1$ and $q_2$.
The Coulomb's force between them at a distance $r$ is,
$\vec{F}=-\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}\hat{r}$
Putting, $\frac{1}{4\pi {\epsilon }_0}=k$ (given)
$\vec{F}=-k\frac{q_1{q}_2}{r^2}\hat{r}$
Since, the nucleus of hydrogen atom has one proton, so charge on nucleus is $e$
ie, $q_2=e$ also $q_1=e$ for electron.
So, $\vec{F}=-k\frac{e\cdot e}{r^2}\hat{r}=-k\frac{e^2}{r^2}\hat{r}$
but $\hat{r}=\frac{\vec{r}}{|\vec{r}|}=\frac{\vec{r}}{r}$
Hence, $\vec{F}=-k\frac{e^2}{r^2}\cdot \frac{\vec{r}}{r}=-k\frac{e^2}{r^3}\cdot \vec{r}$
Note : Negative sign in the expression for Coulomb's force shows that force between electron and hydrogen nucleus is of attraction.