_{1}Q

_{2}/ 4πε

_{0}r

^{2}

Where Q

_{1}and Q

_{2}are the charges on

**point charges**1 and 2 respectively, ε

_{0}is the permittivity of free space (8.85x10

^{-12}Fm

^{-1})

Sometimes the equation is simplified to: F = kQ

_{1}Q

_{2}/ r

^{2}

Where k = 1 / 4πε

_{0}= 8.988x10

^{9}

This simplified version only works when the permittivity of the space the interaction is taking place in is equal to that of free space.

Here is an example, an electron and an alpha particle (helium nucleus) are seperated by 1 centimetre, the force between them is:

F = 1.6x10

^{-19}x 3.2x10

^{-19}/ 4πε

_{0}0.01

^{2}

F = 5.12x10

^{-38}/ 1.11212379x10

^{-10}x 0.01

^{2}

F = 5.12x10

^{-38}/ 1.11212379x10

^{-14}

F = 5.12x10

^{-38}/ 1.11212379x10

^{-14}

F = 4.603x10

^{-24}N, which is very small.

Suppose the seperation wasn't 1 centimetre, but 1 millimetre:

F = 1.6x10

^{-19}x 3.2x10

^{-19}/ 4πε

_{0}0.01

^{2}

F = 5.12x10

^{-38}/ 1.11212379x10

^{-10}x 0.001

^{2}

F = 5.12x10

^{-38}/ 1.11212379x10

^{-16}

F = 5.12x10

^{-38}/ 1.11212379x10

^{-16}

F = 4.603x10

^{-22}N, which is 2 orders of magnitude higher for a 1 order of magnitude reduction in seperation. This is due to the inverse square law.