PHYSICS Force of repulsion question?

Question

If the force of repulsion between two protons is equal to the weight of the protron, how far apart are the protrons?
Mass of protron is: 1.6 x 10(-27) kg and the Charge of the protron is: 1.6 x 10(-19) C
PLEASE SHOW WORK AND PLEASE HELP!!!

Answer 1

If two Protons were to touch ,then the distance between them would be aprox 2 Femto.
The Atomic Elctrostatic Gravity Force between them would be as per the Following equation;

Force =Ko x Mp xMp /diameter^2

Force =Ko x(1.6726231 x 10^-27)^2 / (2 x10-15)^2
Where Ko is the universal electrostaic gravity constant and Mp is the Proton mass.

The force between them is equal to 1.944379284 x 10^8 Newtons. This is a tremendous force that exist between two Protons touching at a proximity of 2 x10^-15 meters of distance apart.

Therefore , when micromasses such as Proton to Proton interact, the Eledtrostatic Gravity Force produced is of a greater magnetude compared to large mass structure Gravity interaction. For this reason Force between Protons and between Neutrons in the Nucleus of the Atom is called the "Strong Force."
This Force is not a repulsive Force.Otherwise the Nucleus of an Atom would not stay contained and would fly apart due to repulsion.

Answer 2

f=Q1Q2E/r^2; f=GM1M2/R^2; we are given Q1,Q2 & M1,M2 thus we get the value of E (epsilon) by equating the two equations as 6.69*10^-27. force = weight means f=M*g M you are given g is 9.8m/s^2. thus we get r= 1.04*10^-19mts