# Are these two calculations of electrostatic force correct?

1. Oct 18, 2011

### Front Office

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.

2. Oct 20, 2011

### lalbatros

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.