An electron is moving around the nucleus of a | vedclass

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(C) The Coulomb force between the nucleus (charge $+e$) and the electron (charge $-e$) is given by Coulomb's Law: $\overrightarrow{F} = K \frac{q_1 q_

Vedclass · Accepted Answer

$ - K\frac{e^2}{r^3}\vec{r}$

Vedclass · Answer

(C) The Coulomb force between the nucleus (charge $+e$) and the electron (charge $-e$) is given by Coulomb's Law: $\overrightarrow{F} = K \frac{q_1 q_2}{r^2} \hat{r}$.Substituting $q_1 = +e$ and $q_2 = -e$,we get $\overrightarrow{F} = K \frac{(+e)(-e)}{r^2} \hat{r} = -K \frac{e^2}{r^2} \hat{r}$.Since the unit vector $\hat{r} = \frac{\vec{r}}{r}$,we can substitute this into the expression:$\overrightarrow{F} = -K \frac{e^2}{r^2} \left( \frac{\vec{r}}{r} \right) = -K \frac{e^2}{r^3} \vec{r}$.

An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The Coulomb force $\overrightarrow{F}$ between the two is (where $K = \frac{1}{4\pi\varepsilon_0}$):

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