Question

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.50 mm away.

Answer 1

Let a = the initial acceleration 
Let m = the mass of the protons = 1.67x10^-27 kg 
Let k = Coulomb's constant = 8.99x10^9 N•m²/C² 
Let q = the charge of a proton = 1.60x10^-19 C 
Let G = the gravitational constant = 6.67x10^-11 m³/kg•s² 
Let r = the distance between the protons = 4.00x10^-3 m 
Let F_e = the electrostatic force between the two protons = kq²/r² 
Let F_g = the gravitational force between the two protons = Gm²/r² 
The net force is: F = F_e - F_g: 
F = (kq² - Gm²)/r² 
Using F = ma, the initial acceleration is: 
a = (kq²/m - Gm)/r² 
I will leave the calculation to you.