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An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The coulomb force $\vec{F}$ between the two is:
$\left(\right.$ where $\left.K=\frac{1}{4 \pi \varepsilon_0}\right)$
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$\left(\right.$ where $\left.K=\frac{1}{4 \pi \varepsilon_0}\right)$
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Verified Answer
The correct answer is:
$-K \frac{e^2}{r^3} \hat{r}$
The charge on hydrogen nuclear $q_1$
$\begin{aligned}
& =+v e \\
& \therefore \text { Coulomb's force } F=K \frac{q_1 q_2}{r^2} \\
& =K \frac{(+e)(-e)}{r^2}=-\frac{k e^2}{r^2} \hat{r}
\end{aligned}$
$\begin{aligned}
& =+v e \\
& \therefore \text { Coulomb's force } F=K \frac{q_1 q_2}{r^2} \\
& =K \frac{(+e)(-e)}{r^2}=-\frac{k e^2}{r^2} \hat{r}
\end{aligned}$
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