The nucleus of the helium atom contains two protons that are separated by about 3.8 multiplied by 10^-15 m. How do you find the magnitude of the electrostatic force that each proton exerts on the other?

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The nucleus of the helium atom contains two protons that are separated by about 3.8 multiplied by 10^-15 m. How do you find the magnitude of the electrostatic force that each proton exerts on the other?

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Answer 1 · 2016-01-06T15:13:15

$15.956N$

Explanation:

We may use Coulomb's Law in electrostatics to find the force of repulsion (since like charges repel) between these 2 electric point charges, each having the unit quantity of charge equal to the charge of 1 electron, $1.6xx10^(-19)C$ .

$F=k(Q_1Q_2)/(r^2)$ , where $k=9xx10^9N.m^2//kg^2$ is Coulomb's constant.

$therefore F=((9xx10^9)xx(1.6xx10^(-19))^2)/(3.8xx10^(-15))^2$

$=15.956N$ .

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The nucleus of the helium atom contains two protons that are separated by about 3.8 multiplied by 10^-15 m. How do you find the magnitude of the electrostatic force that each proton exerts on the other?

1 Answer

Explanation:

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