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

###### FREE Expert Solution

E → electrostatic potential energy

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

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

Initial distance of  62 pm:

1 pm = 10-12 m

d = 6.2x10-11 m

Einitial = 3.712x10-18 N•m = 3.712x10-18 J

Final distance of 1.0 nm:

###### Problem Details

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