Answer
$\frac{F_{el}}{F_g} = 1.5\times 10^{10}$
Work Step by Step
We can find the magnitude of the electric field:
$F = \vert q \vert~E$
$E = \frac{F}{\vert q \vert}$
$E = \frac{3.0\times 10^{-6}~N}{2.0\times 10^{-9}~C}$
$E = 1500~N/C$
We can find the magnitude of the electrostatic force on a proton:
$F_{el} = q~E$
$F_{el} = (1.6\times 10^{-19}~C)(1500~N/C)$
$F_{el} = 2.4\times 10^{-16}~N$
We can find the magnitude of the gravitational force on the proton:
$F_g = ma$
$F_g = (1.67\times 10^{-27}~kg)(9.8~m/s^2)$
$F_g = 1.6\times 10^{-26}~N$
We can find the ratio $\frac{F_{el}}{F_g}$:
$\frac{F_{el}}{F_g} = \frac{2.4\times 10^{-16}~N}{1.6\times 10^{-26}~N} = 1.5\times 10^{10}$
