Coulomb's Law: Practice Problem & Solution
7. 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) (where $K = \frac{1}{4\pi\epsilon_0}$)
Solution Explained:
To solve this problem, we apply the core principles of Coulomb's Law. Understanding the underlying formula is key to arriving at the correct answer below:
The Coulomb force is attractive, so $\vec{F} = -K \frac{e^2}{r^2} \hat{r}$. Multiplying numerator and denominator by $r$ (since $\hat{r} = \frac{\vec{r}}{r}$) gives $\vec{F} = -K \frac{e^2}{r^3} \vec{r}$.
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