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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 - (where $K = \dfrac{1}{4\pi\varepsilon_0}$)
A
$K\dfrac{e^2}{r^2}\hat{r}$
B
$-K\dfrac{e^2}{r^3}\hat{r}$
C
$K\dfrac{e^2}{r^3}\vec{r}$
D
$-K\dfrac{e^2}{r^3}\vec{r}$
Detailed Solution
Take the nucleus (charge $+e$) at the origin and the electron (charge $-e$) at position $\vec{r}$.
Magnitude of the Coulomb force: $F = K\dfrac{e \cdot e}{r^2} = K\dfrac{e^2}{r^2}$
The force on the electron is attractive, i.e. directed towards the nucleus, along $-\hat{r}$.
$\vec{F} = K\dfrac{e^2}{r^2}(-\hat{r})$
Using $\hat{r} = \dfrac{\vec{r}}{r}$: $\vec{F} = -K\dfrac{e^2}{r^3}\vec{r}$
Magnitude of the Coulomb force: $F = K\dfrac{e \cdot e}{r^2} = K\dfrac{e^2}{r^2}$
The force on the electron is attractive, i.e. directed towards the nucleus, along $-\hat{r}$.
$\vec{F} = K\dfrac{e^2}{r^2}(-\hat{r})$
Using $\hat{r} = \dfrac{\vec{r}}{r}$: $\vec{F} = -K\dfrac{e^2}{r^3}\vec{r}$