If electrostatic potential energy between the two electrons is $7.67\times {10}^{-19} \mathrm{J}$. What is their distance of separation (in $\stackrel{\mathrm{o}}{\mathrm{A}}$) in vacuum?

$\begin{array}{l}\mathrm{Q}=K\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}\text{ or }7.67\times {10}^{-12}=\frac{{\left(1.6\times {10}^{-12}\right)}^2}{4\times 3.14\times 8.85\times {10}^{-12}\times \mathrm{d}}\\ K=\frac{1}{4\mathrm{\pi }{\in }_0}\\ \therefore d=3\times {10}^{-10} \mathrm{m}=3\text{Å}\end{array}$he permittivity of free space (a vacuum) is a physical constant equal to approximately 8.85 x 10^-12 farad per meter (F/m).