At what distance is the electrostatic force between two protons equal to the weight of one proton?

The force of repulsion between two protons is equal to the weight experienced by a proton.

Weight of an object is defined as the force by which the Earth attracts the body.The expression for the weight is given as,

$F=mg$

Here, F is the weight of the object, m is the mass of the object and g is the acceleration due to gravity.

The electrostatic force between two protons is,

$F_e=\frac{K{q}^2}{r^2}----------(1.1)$

Here, is the electrostatic force constant $(K=9\times {10}^{9}N-m^2/C^2)$ , q is the charge on protons $\left(q=1.6\times {10}^{-19}C\right)$, and r is the separation between the protons.

The weight of one proton is,

$F=m_{p}g----------(1.2)$

Here, m_p is the mass on a proton $\left(m_p=1.67\times {10}^{-27}kg\right)$, and g is the acceleration due to gravity $\left(g=9.8m/s^2\right)$.

Since, the electrostatic force between the protons equals the weight of one proton.

Equating equation (1.1) and (1.2),

$m_{p}g=\frac{K{q}^2}{r^2}$

The expression for the separation between the protons is given as,

$r=\sqrt{\frac{K{q}^2}{m_{p}g}}$

Substituting all known values,

$$\begin{gathered} r=\sqrt{\frac{\left(9\times {10}^{9}N-m^2/C^2\right)\times {\left(1.6\times {10}^{-19}C\right)}^2}{\left(1.67\times {10}^{-27}kg\right)\times \left(9.8m/s^2\right)}} \\ =0.119m \end{gathered}$$

Hence, at a distance of 0.119 m the electrostatic force between two protons equal to the weight of one proton.