Answer
$\frac{F_g}{F_e} = 4.4\times 10^{-40}$
The gravitational force is insignificant compared to the electrostatic force.
Work Step by Step
We can find an expression for the gravitational force:
$F_g = \frac{G~M_1~M_2}{r^2}$
$F_g = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(9.1\times 10^{-31}~kg)(1.67\times 10^{-27}~kg)}{r^2}$
$F_g = \frac{1.014\times 10^{-67}~N~m^2}{r^2}$
We can find an expression for the electrostatic force:
$F_e = \frac{k~q_1~q_2}{r^2}$
$F_e = \frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)(1.6\times 10^{-19}~C)}{r^2}$
$F_e = \frac{2.304\times 10^{-28}~N~m^2}{r^2}$
We can find the ratio of the gravitational force to the electrostatic force:
$\frac{F_g}{F_e} = \frac{\frac{1.014\times 10^{-67}~N~m^2}{r^2}}{\frac{2.304\times 10^{-28}~N~m^2}{r^2}} = 4.4\times 10^{-40}$
Clearly, the gravitational force is insignificant compared to the electrostatic force.
