Figuring out the magnitude of electric repulsion

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The discussion focuses on calculating the electric repulsion between two carbon atoms separated by 0.20 nm using Coulomb's law. The distance is converted to meters, and the proportionality constant (k) is noted as 8.99 x 10^9. The charge of each carbon atom is determined by multiplying the number of protons (6) by the elementary charge (1.60 x 10^-19 C). The final calculation yields a force of 2.07 x 10^-7 N. The participant seeks confirmation of their mathematical approach and reasoning, which appears sound.
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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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