Figuring out the magnitude of electric repulsion

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SUMMARY

The magnitude of electric repulsion between two carbon nuclei separated by 0.20 nm is calculated using Coulomb's law. The proportionality constant (k) is 8.99 x 10^9 N m²/C², and the charge of each carbon nucleus, derived from its six protons, is 9.60 x 10^-19 C. Substituting these values into the formula F = (k |q1| |q2|)/r² yields a force of 2.07 x 10^-7 N. The calculations and reasoning presented are confirmed to be accurate.

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Homework Statement


[/B]
In a certain organic molecule, the nuclei of 2 carbon atoms are separated by a distance of .20nm, What is the magnitude of the electric repulsion between them?

Now, I know the distance between them, but it has to be converted into meters:

.20nm = .20 X 10 ^-9m

And k is a proportionality constant, equaling:

8.99 X 10 ^9.

Homework Equations



So I am using Coulomb's law for this which states

F = (K|q1| |q2|)/r^2

The Attempt at a Solution



The only unknowns are q1 and q2. In order to solve for those, I have to know the magnitude of their charge. Since there are 6 protons in a Carbon atom, in order to get the charge I multiply 6 by e, which equals:

1.60X10^-19 C

I have solved for all the unknowns, so all that is left is to plug in all the variables and solve for the F.

F = (8.99 X 10 ^9 * (6*1.60X10^-19)^2)/(.20 X 10 ^-9m)^2

This gives me the final answer of:

2.07x10^-7 N

I wanted to verify that my math and reasoning behind my calculations were correct. Thanks!
 
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Reasoning looks pretty good to me.
 

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