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(m_1m_2)/(d^2)`

And the electric force is:

`F_e=k(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(m^2)/(d^2)`

And

`F_e=k(q^2)/(d^2)`

By dividing both, we get

`(F_e)/(F_g)=(((kq^2)/(d^2)))/(((Gm^2)/(d^2)))=(kq^2)/(d^2)*(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:

`(F_e)/(F_g)=(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

`(F_e)/(F_g)=((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.