Q123 / 200 · NEET 2003 · Physics
An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The coulomb force $\vec{F}$ between the two is:
$\left(\right.$ where $\left.K=\frac{1}{4 \pi \varepsilon_0}\right)$
- A$K \frac{e^2}{r^2} \hat{r}$
- B$-K \frac{e^2}{r^3} \hat{r}$
- C$K \frac{e^2}{r^3} \hat{r}$
- Correct answerD$-K \frac{e^2}{r^3} \hat{r}$
Solution
The charge on hydrogen nuclear $q_1$
$\begin{aligned}
& =+v e \\
& \therefore \text { Coulomb's force } F=K \frac{q_1 q_2}{r^2} \\
& =K \frac{(+e)(-e)}{r^2}=-\frac{k e^2}{r^2} \hat{r}
\end{aligned}$
Hence, the correct answer is option: 'D'.