Are these two calculations of electrostatic force correct?

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SUMMARY

The calculations presented in the discussion reveal that the electrostatic force required to push two uranium nuclei together is approximately 2,000 lbs, based on Coulomb's Law, with each nucleus containing 92 protons and a separation distance of 15 femtometers. Additionally, the force needed to separate the electrons from the protons in a copper penny, modeled as two penny-sized disks 1.55 mm apart, exceeds the weight of the Earth, necessitating energy equivalent to two centuries of human energy consumption at a rate of 13 trillion watts. The calculations highlight the immense forces at play at atomic scales, challenging initial assumptions about the magnitude of these forces.

PREREQUISITES
  • Coulomb's Law for electrostatic force calculations
  • Understanding of atomic structure, specifically protons and electrons
  • Basic knowledge of units of force (pounds) and energy
  • Familiarity with atomic measurements, such as femtometers and picometers
NEXT STEPS
  • Explore advanced applications of Coulomb's Law in particle physics
  • Research the implications of electrostatic forces in nuclear fusion processes
  • Investigate the energy requirements for separating atomic particles in various elements
  • Learn about the Large Hadron Collider's role in studying fundamental forces
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Physicists, chemistry students, and anyone interested in the fundamental forces governing atomic interactions and their implications in nuclear physics.

Front Office
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I was watching a show the other night about the Large Hadron Collider, and I got to wondering how much force it would take to push two uranium nuclei right up against each other. I figured it would be on the order of a thousandth, maybe a millionth of a pound, which seems like a lot, considering how small an atomic nucleus is.

Boy, was I surprised. IT'S NEARLY 2,000 LBS!

Can that be correct?

I assumed 92 protons per nucleus, each having a positive charge of 1.602x10^-19 coulombs. I assumed a distance between nuclear centers of 15 femtometers (15x10^-15 m). And I used Coulomb's Law for the force between charged objects.

Since I was on a role, I then calculated how much force it would take to separate all the electrons in a copper penny from all the protons, assuming all the positive and negative charges are arranged as penny-sized disks that are 1.55 millimeters apart, which is the thickness of a penny according to Wikipedia. Of course it would be impossible to do this experimentally. But assuming two penny-sized disks separated by 1.55 mm, THE FORCE OF ATTRACTION EXCEEDS THE WEIGHT OF THE EARTH!

That much force to separate the charges would, as I calculate it, would require an amount of energy on the order of two centuries of human energy use (at our current rate of ~13-trillion watts).

I must be making an error.

I assumed 3.5 grams for a copper penny, 100% copper, atomic number 29, and atomic weight 63.5. Used Coulomb's Law here also.
 
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Front Office said:
...
Since I was on a role, I then calculated how much force it would take to separate all the electrons in a copper penny from all the protons, assuming all the positive and negative charges are arranged as penny-sized disks that are 1.55 millimeters apart, which is the thickness of a penny according to Wikipedia. Of course it would be impossible to do this experimentally. But assuming two penny-sized disks separated by 1.55 mm, THE FORCE OF ATTRACTION EXCEEDS THE WEIGHT OF THE EARTH! ...

These 1.55 mm would not be the most difficult part of separating electrons from copper.
The atomic radius of copper is 130 pm, with is about 10-7 smaller, and therefore the force must be 10+14 larger.
 

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