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?

Respuesta :

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:

[tex]F=k\frac{q_1q_2}{r^2}[/tex]

Where:

[tex]\begin{gathered} k=\text{ electric constant} \\ q_1,q_2=\text{ charges} \\ r=\text{ distance between charges} \end{gathered}[/tex]

The charge of a proton is:

[tex]q_p=1.606\times10^{-19}C[/tex]

Now, we plug in the values:

[tex]F=(9\times10^9\frac{Nm^2}{C^2})\frac{(1.606\times10^{-19}C)(1.606\times10^{-19}C)}{(1\times10^{-15}m)^2}[/tex]

Solving the operations:

[tex]F=232.13N[/tex]

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

[tex]F_g=G\frac{m_1m_2}{r^2}[/tex]

Where;

[tex]\begin{gathered} G=\text{ gravitational constant} \\ m_1,m_2=\text{ masses} \\ r=\text{ distance between masses} \end{gathered}[/tex]

The mass of a proton is given by:

[tex]m_p=1.67\times10^{-27}kg[/tex]

Now, we plug in the values:

[tex]F_g=(6.67\times10^{-11}\frac{Nm^2}{kg^2})\frac{(1.67\times10^{-27}kg)(1.67\times10^{-27}kg)}{(1\times10^{-15}m)^2}[/tex]

Solving the operations:

[tex]F_g=1.86\times10^{-34}N[/tex]

Now, we determine the ratio between the forces:

[tex]\frac{F}{F_g}=\frac{232.13N}{1.86\times10^{-34}N}=1.25\times10^{36}[/tex]

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

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