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Sagot :
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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