Question

Two protons are 11.86 fm apart. (1 fm= 1 femtometer = 1 x 10-15 m.) What is the ratio of the electric force to the gravitational force on one proton due to the other proton?

Answer 1

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Firstly, we need to write the formulas for both the gravitational force and electric force. Our gravitational force is:

$F_g=G\frac{m_1m_2}{d^2}$

And the electric force is:

$F_e=k\frac{q_1q_2}{d^2}$

We can see that these forces have almost equal formulas. What we want is Fe/Fg. Before this, we can simplify the forces, as both particles have the same charge and mass. We're left with the following:

$F_g=G\frac{m^2}{d^2}$

And

$F_e=k\frac{q^2}{d^2}$

By dividing both, we get

$\frac{F_e}{F_g}=\frac{(\frac{kq^2}{d^2})}{(\frac{Gm^2}{d^2})}=\frac{kq^2}{d^2}*\frac{d^2}{Gm^2}$

We have d^2 on the numerator and denominator. We can elimante the distance then, as it is different from zero. We have the following:

$\frac{F_e}{F_g}=\frac{kq^2}{Gm^2}$

We can then replace our values with the constants. k is Coulomb's constant, q is the charge of a proton, G is Newton's constant, and m is the mass of a proton. We finally get

$\frac{F_e}{F_g}=\frac{(9*10^9)*(1.6*10^{-19})^2}{(6.67*10^{-11})*(1.67*10^{-27})^2}=1.2386*10^{36}$

So, the electric force is 1.2386*10^36 times higher than the gravitational. The most interesting about this, is that it doesn't depend on the distance the two of them are apart.