What is the repulsive electrical force between two protons 4.0 x 10^{-15} m apart from each other in an atomic nucleus?

This is Giancoli Answers with Mr. Dychko. The repulsive force between two protons is Coulomb's constant 8.988 times 10 to the nine Newton meters squared per Coulomb squared, multiply it by each charge and they're both 1.6 times 10 to the minus 19 Coulombs. So we take that number, squared, and divided by the distance between them at 4.0 times 10 to the minus 15 meters squared. Then we get about 14 Newtons as the repulsive force between two protons.

COMMENTS

By joseotilio25 on Sun, 2/28/2016 - 5:28 PM

how I know the values of q1 and q2 because they dont mention in the exercice

By Mr. Dychko on Thu, 3/10/2016 - 4:33 AM

Hi joseotillio25, thanks for the question. Since the problem says we're working with protons, that means the charge on each is the elementary charge of $1.6 \times 10^{-19} \textrm{ C}$ .

All the best,
Mr. Dychko

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What is the repulsive electrical force between two protons 4.0×10−15 m4.0 \times 10^{-15} \textrm{ m}4.0×10−15 m apart from each other in an atomic nucleus?

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What is the repulsive electrical force between two protons $4.0 \times 10^{-15} \textrm{ m}$ apart from each other in an atomic nucleus?