Question

# Two electrons in an atom are separated by 1.5 x 10‾¹º m, the typical size of an atom. What is the electric force between them?

Solution

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Electric force $F$ between charges $q_A$ and $q_B$ at a distance $r$ from each other is given as:

$$$F = \dfrac{k q_A q_B}{r^2}$$$

where $k = 9 \cdot 10^9 ~\mathrm{\dfrac{N m^2}{C^2}}$ is Coulomb constant.

We are told that the distance $r$ between the electrons is $r = 1.5 \cdot 10^{-10} ~\mathrm{m}$ and we know that charge of an electron $q_e$ is equal to negative value of the elementary charge $q_e = - e = - 1.6 \cdot 10^{-19} ~\mathrm{C}$. We can plug in this data into equation $(1)$, a general equation for force between the two charges and find that the force between these two electrons is equal to:

\begin{align*} F &= \dfrac{k q_e q_e }{r^2} \\ F &= \dfrac{k e^2 }{r^2} \\ F &= 9 \cdot 10^9 ~\mathrm{\dfrac{N m^2}{C^2}} \cdot \dfrac{ (1.6 \cdot 10^{-19})^2 ~\mathrm{C^2} }{(1.5 \cdot 10^{-10})^2 ~\mathrm{m^2} } \end{align*}

$\boxed{F = 1.024 \cdot 10^{-8} ~\mathrm{N}}$

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