Electric Charges and Fields - Result Question 9
9. 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
[2003]
(a) $K \frac{e^{2}}{r^{3}} \vec{r}$
(b) $K \frac{e^{2}}{r^{2}} \hat{r}$
(c) $-K \frac{e^{2}}{r^{3}} \hat{r}$
(d) $-K \frac{e^{2}}{r^{3}} \vec{r}$ $(.$ where $.K=\frac{1}{4 \pi \varepsilon_0})$
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Answer:
Correct Answer: 9. (d)
Solution:
- (d) Charges (-e) on electron and (e) on proton exert a force of attraction given by
Force $=(K) \frac{(-e)(e)}{r^{2}} \hat{r}=\frac{-K e^{2}}{r^{3}} \vec{r} \quad(\because \hat{r}=\frac{\vec{r}}{|r|})$
Magnitude of Coulomb force is given by $\frac{1}{4 \pi \varepsilon_0} \frac{q_2 q_2}{r^{2}}$, but in vector form $\vec{F}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^{3}} \vec{r}$
