Question

You have one electron directly over another electron. The force between the electrons is a repulsiveforce of 9.11 x 10-30 Newtons. If the charge of each electron is 1.6 x 10-19 Coulombs, what is the distancebetween the electrons?

Answer 1

Given:

The force of repulsion between the electrons is,

$F=9.11\times10^{-30}\text{ N}$

The charge of each electron is,

$e=1.6\times10^{-19}\text{ C}$

To find:

The distance between the electrons

Explanation:

The force between two electrons at a distance 'r' is,

$\begin{gathered} F=\frac{e^2}{4\pi\in_0r^2} \\ Here,\text{ }\frac{1}{4\pi\in_0}=9\times10^9\text{ N.m}^2.C^{-2} \end{gathered}$

Substituting the values, we get,

$\begin{gathered} 9.11\times10^{-30}=9\times10^9\times\frac{(1.6\times10^{-19})^2}{r^2} \\ r^2=\frac{(1.6\times10^{-19})^2\times9\times10^9}{9.11\times10^{-30}} \\ r^2=25.29 \\ r=5.03\text{ m} \end{gathered}$

Hence, the distance between the electrons is 5.03 m.