At a particular instant a proton exerts an electric force of $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{5}\mathbf{.}\mathbf{76}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{13}}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{N}$on an electron. How far apart are the proton and the electron?

The given data can be listed below as,

• The proton exerts an electric force of$(0,5.76\times {10}^{-13},0)\hspace{0.33em}N$

The coulomb’s law states that the force exerted by an object is inversely proportional to their distance’s square and directly proportional to the product of their charges.

The equation of the force gives the distance amongst the proton and the electron.

From Coulomb’s law, the electric force on the electron by the proton is expressed as:

$F=k\frac{q_1{q}_2}{r^2}$

Here, F is the magnitude of the electric force that is $5.76\times {10}^{-13}\hspace{0.33em}\mathrm{N}$; k is the coulomb’s constant that is about $8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2,q_1\text{and}{q}_2$are the charges of the proton and the electron that are $+1.602\times {10}^{-19}\mathrm{C}\text{and}-1.602\times {10}^{-19}\mathrm{C}$and r is the distance amongst them.

Substituting the values in the above equation, we get-

role="math" localid="1658117645354" $$\begin{gathered} 5.76\times {10}^{-13}\mathrm{N}=8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{\left(+1.602\times {10}^{-19}\mathrm{C}\right)\times \left(-1.602\times {10}^{-19}\mathrm{C}\right)}{(r{)}^2} \\ {r}^2=\frac{\left(8.99\times {10}^9{\mathrm{N}}^2{\mathrm{m}}^2/{\mathrm{C}}^2\right)\times \left(+1.602\times {10}^{-19}\mathrm{C}\right)\times \left(-1.602\times {10}^{-19}\mathrm{C}\right)}{\left(5.76\times {10}^{-13}\mathrm{N}\right)} \\ {r}^2=4.005\times {10}^{-16}\hspace{0.33em}{m}^2 \\ r=2.001\times {10}^{(}-8)\hspace{0.33em}m \end{gathered}$$

Thus, the proton and the electron are$2.001\times {10}^{-8}m$apart.