Electrostatics Question 208
Question: An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force $ \overrightarrow{F} $ between the two is (Where $ K=\frac{1}{4\pi {\varepsilon _{0}}} $ ) [CBSE PMT 2003]
Options:
A) $ -K\frac{e^{2}}{r^{3}}\hat{r} $
B) $ K\frac{e^{2}}{r^{3}}\vec{r} $
C) $ -K\frac{e^{2}}{r^{3}}\vec{r} $
D) $ K\frac{e^{2}}{r^{2}}\hat{r} $
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Answer:
Correct Answer: C
Solution:
$ \overrightarrow{F}=-k\frac{e^{2}}{r^{2}}\hat{r}=-k.\frac{e^{2}}{r^{3}}\overrightarrow{r} $
$ ( \because \hat{r}=\frac{\overrightarrow{r}}{r} ) $
