A proton and an electron are separated by $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathbf{m}$ the radius of a typical atom. Calculate the magnitude of the electric force that the proton exerts on the electron and the magnitude of the electric force that the electron exerts on the proton.

Coulomb's law is the most fundamental in electrostatics. This law determines the force one charged particle exerts over the other and vice versa. The physical expression of this law is:

$\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, r is the distance between two charged particles, and q is the charges.

As it is known that the charge of the electron is${\mathrm{q}}_1=-1.6\times {10}^{-19}\mathrm{C}$

It is also known the charge of the proton isdata-custom-editor="chemistry" ${\mathrm{q}}_2=+1.6\times {10}^{-19}\mathrm{C}$

It is given that the separation between the electron and proton is data-custom-editor="chemistry" $\mathrm{r}=1\times {10}^{-10}\mathrm{m}$.

It is known that the force can be determined as:

data-custom-editor="chemistry" $\begin{array}{rcl}\left|\mathrm{F}\right|& =& \left|\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}\right|\\ & =& 9\times {10}^9\times \frac{{\left(1.6\times {10}^{-19}\right)}^2}{{\left(1\times {10}^{-10}\right)}^2}\\ & =& 2.304\times {10}^{-8}\mathrm{N}\end{array}$

As it is known that the charge of the electron is${\mathrm{q}}_1=-1.6\times {10}^{-19}\mathrm{C}$

It is also known the charge of the proton is${\mathrm{q}}_2=+1.6\times {10}^{-19}\mathrm{C}$

It is given that the separation between the electron and proton is$\mathrm{r}=1\times {10}^{-10}\mathrm{m}$

It is known that the force can be determined as:

$\begin{array}{rcl}\left|\mathrm{F}\right|& =& \left|\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}\right|\\ & =& 9\times {10}^9\times \frac{{\left(1.6\times {10}^{-19}\right)}^2}{{\left(1\times {10}^{-10}\right)}^2}\\ & =& 2.304\times {10}^{-8}\mathrm{N}\end{array}$