Question

Two protons are 7.818 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

We are given that two protons are 7.1818 fm apart. To determine the electric force between them we need to use the following formula:

`F=k(q_1q_2)/(r^2)`

Where:

`\begin{gathered} k=\text{ electric constant} \\ q_1,q_2=\text{ charges} \\ r=\text{ distance between charges} \end{gathered}`

The charge of a proton is:

`q_p=1.606*10^(-19)C`

Now, we plug in the values:

`F=(9*10^9(Nm^2)/(C^2))((1.606*10^(-19)C)(1.606*10^(-19)C))/((1*10^(-15)m)^2)`

Solving the operations:

`F=232.13N`

Now, we need to determine the gravitational force. We will use the following formula:

`F_g=G(m_1m_2)/(r^2)`

Where;

`\begin{gathered} G=\text{ gravitational constant} \\ m_1,m_2=\text{ masses} \\ r=\text{ distance between masses} \end{gathered}`

The mass of a proton is given by:

`m_p=1.67*10^(-27)kg`

Now, we plug in the values:

`F_g=(6.67*10^(-11)(Nm^2)/(kg^2))((1.67*10^(-27)kg)(1.67*10^(-27)kg))/((1*10^(-15)m)^2)`

Solving the operations:

`F_g=1.86*10^(-34)N`

Now, we determine the ratio between the forces:

`(F)/(F_g)=(232.13N)/(1.86*10^(-34)N)=1.25*10^(36)`

This means that the electric force is 1.25 by ten to the 36th times larger than the gravitational force.