The electrical force between two protons is given by:
[tex]F_e = k \frac{e^2}{r^2} [/tex]
where
[tex]k=8.99 \cdot 10^9 Nm^2C^{-2}[/tex] is the Coulomb's constant
[tex]e=1.6 \cdot 10^{-19}C[/tex] is the proton charge
r is the separation between the two protons
The gravitational force between the two protons is given by:
[tex]F_g=G \frac{m^2}{r^2} [/tex]
where
[tex]G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}[/tex] is the gravitational constant
[tex]m=1.67 \cdot 10^{-27} kg[/tex] is the proton mass
r is the separation between the two protons
If we divide the electric force by the gravitational force, we get
[tex]\frac{F_e}{F_g}= \frac{k}{G} \frac{e^2}{m^2}=1.2 \cdot 10^{36}[/tex]
which means that the electric force between the two protons is [tex]1.2 \cdot 10^{36}[/tex] times greater than the gravitational force.
Moreover, the two protons have same electric charge, and the electrostatic force between two same-sign charges is repulsive, while the gravitational force is always attractive: therefore, the correct answer is
The electrical force is 1.2 × 1036 times greater than the gravitational force, but only the gravitational force is attractive.
