Problem: What is the change in potential energy if the distance separating two protons is increased to from 62 pm to 1.0 nm?

E → electrostatic potential energy

$\boxed{\mathbf{∆}\mathbf{E}\mathbf{=}{\mathbf{E}}_{\mathbf{final}}\mathbf{-}{\mathbf{E}}_{\mathbf{initial}}}$

$\boxed{\mathbf{E}\mathbf{=}{\mathbf{k}}_{\mathbf{e}}\frac{{\mathbf{Q}}_{\mathbf{1}}\mathbf{·}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{d}}}$

Initial distance of  62 pm:

1 pm = 10^-12 m

$\mathbf{d}\mathbf{=}\mathbf{62}\mathbf{ }\cancel{\mathbf{pm}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{ }\mathbf{m}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{pm}}}$

d = 6.2x10^-11 m

$$\begin{gathered} \mathbf{E}\mathbf{=}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\frac{\mathbf{N}\mathbf{·}{\mathbf{m}}^{\mathbf{2}}}{{\mathbf{C}}^{\mathbf{2}}}\mathbf{)}\frac{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{ }\mathbf{C}\mathbf{)}\mathbf{·}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{ }\mathbf{C}\mathbf{)}}{\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{ }\mathbf{m}\mathbf{)}} \\ \mathbf{E}\mathbf{=}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\frac{\mathbf{N}\mathbf{·}{\mathbf{m}}^{\cancel{\mathbf{2}}}}{\cancel{{\mathbf{C}}^{\mathbf{2}}}}\mathbf{)}\frac{\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{38}}\mathbf{ }\cancel{{\mathbf{C}}^{\mathbf{2}}}\mathbf{)}}{\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{ }\cancel{\mathbf{m}}\mathbf{)}} \end{gathered}$$

E_initial = 3.712x10^-18 N•m = 3.712x10^-18 J

Final distance of 1.0 nm: