Answer
$r = 0.119 m$
Work Step by Step
The gravitational force on felt by a proton near the surface of the Earth is $F = mg $, where $m = 1.67 \times 10^{27} kg $ is the mass of the proton and charge of Proton is $q = 1.60 \times 10^{-19} c $
The electrostatic force between two protons separated by a distance $F_e = kq^2/r$
We equate the gravitational force and electrostatic force to get two forces that are equal, we have $mg = kq^2/r^2$
Solve for $r$ and plug in all the values into the equation
$mg = kq^2/r^2$
$r^2 = kq^2/mg$
$r = \sqrt {\frac{kq^2}{mg}}$
$r = q \sqrt {\frac{k}{mg}}$
$r = (1.60 \times 10^{-19} c) \sqrt {\frac{(8.99 \times 10^9 N.m^2/C^2)}{(1.67 \times 10^{-27} kg)(9.8 m/s^2)}}$
$r = 0.119 m$
