Q188 / 200 · Manipal (MET) 2013 · Physics

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

  • A
    $k \frac{e^2}{r^3} r$
  • Correct answer
    B
    $-k \frac{e^2}{r^3} r$
  • C
    $k \frac{e^2}{r^2} r$
  • D
    $-k \frac{e^2}{r^2} r$

Solution

$\mathbf{F}=-\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}=\mathbf{r}$ $\begin{aligned} & \mathbf{F}=-k \frac{q_1 q_2}{r^2}=\mathbf{r} \\ & \mathbf{F}=-k \frac{e \cdot e}{r^2} \hat{\mathbf{r}}=-k \frac{e^2}{r^2} \hat{\mathbf{r}} \\ & \hat{\mathbf{r}}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{\mathbf{r}}{r} \\ & F=-k \frac{e^2}{r^2} \cdot \frac{\mathbf{r}}{r}=-k \frac{e^2}{r^3} \mathbf{r}\end{aligned}$

Hence, the correct answer is option: 'B'.

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