Answer:
The electric force is [tex]6.2\cdot 10^{35}[/tex] times stronger than the gravitational force
Explanation:
The magnitude of the electrostatic force between two charges is given by:
[tex]F_E=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
In this problem:
[tex]q_1=1.6\cdot 10^{-19}C[/tex] (charge of the proton)
[tex]q_2=3.2\cdot 10^{-19}C[/tex] (charge of a nucleus of helium, twice the charge of a proton)
[tex]r=100 \mu m = 100\cdot 10^{-6} m[/tex]
So the electric force is
[tex]F_E=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})(3.2\cdot 10^{-19})}{(100\cdot 10^{-6})^2}=4.6\cdot 10^{-20} N[/tex]
Instead, the magnitude of the gravitational force between two objects is given by
:
[tex]F_G=G\frac{m_1 m_2}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
Here we have:
[tex]m_1=1.67\cdot 10^{-27}kg[/tex] is the mass of the proton
[tex]m_2=6.68\cdot 10^{-27}kg[/tex] is the mass of a nucleus of helium (4 times the mass of the proton)
[tex]r=100 \mu m = 100\cdot 10^{-6} m[/tex] is the separation
So the gravitational force is
[tex]F_G=(6.67\cdot 10^{-11})\frac{(1.67\cdot 10^{-27})(6.68\cdot 10^{-27})}{(100\cdot 10^{-6})^2}=7.4\cdot 10^{-56} N[/tex]
So, we see that the electric force is much stronger than the gravitational factor, by a factor of:
[tex]\frac{F_E}{F_G}=\frac{4.6\cdot 10^{-20}}{7.4\cdot 10^{-56}}=6.2\cdot 10^{35}[/tex]
