Question

What is the Coulomb force between two electrons that are on opposite sides of the atom, `2.80 xx 10^-10` `m` apart?

Answer 1

We will use Coulomb's Law: `F=(kq_1q_2)/r^2`

See below for the details of the calculation. The force acting between the two electrons is repulsive and of magnitude:

`F=2.94xx10^-9` `N`

Explanation:

`F=(kq_1q_2)/r^2`

`k` is the Coulomb's Law constant, `9xx10^9` `Nm^2C^-2`
`q_1=q_2=1.6xx10^-19` `C`, the charge on the electron
`r` is given in the question and is `2.80xx10^-10` `m`.

Substituting these values into the equation:

`F=(9xx10^9xx1.6xx0^-19xx1.6xx10^-19)/(2.80xx10^-10)^2`
`= 2.94xx10^-9` `N`