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An electron of charge ' $\mathrm{e}$ ' is moving round the nucleus of a hydrogen atom in a circular orbit of radius ' $r$ '. The coulomb force $\overrightarrow{\mathrm{F}}$ between the two is $\left(\right.$ here $\left.\mathrm{K}=\frac{1}{4 \pi \varepsilon_0}\right)$
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The correct answer is:
$-\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$
Using Coulomb's law, the force between two charges is given by
$\overrightarrow{\mathrm{F}}=\frac{\mathrm{k} \mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$
Here, charge of an electron, $q_1=-e$
Charge of the nucleus, $\mathrm{q}_2=\mathrm{e}$
$\overrightarrow{\mathrm{F}}=\frac{\mathrm{k}(-\mathrm{e}) \mathrm{e}}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}=-\frac{\mathrm{ke}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$
$\overrightarrow{\mathrm{F}}=\frac{\mathrm{k} \mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$
Here, charge of an electron, $q_1=-e$
Charge of the nucleus, $\mathrm{q}_2=\mathrm{e}$
$\overrightarrow{\mathrm{F}}=\frac{\mathrm{k}(-\mathrm{e}) \mathrm{e}}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}=-\frac{\mathrm{ke}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$
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