# Compressional force

1. Sep 1, 2010

### J89

1. The problem statement, all variables and given/known data

Suppose 7.00 g of hydrogen is separated into electrons and protons. Suppose also the protons are placed at the earth's North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on the Earth?

2. Relevant equations

Fe = Ke*|q1||q2|/r^2, radius of earth is 6.38*10^6 * 2 to get diameter is 1.3*10^7...

3. The attempt at a solution
I believe this is the correct way:

7 * 6.02^1023 = 4.21*10^24 (since it is 7 hydrogens, and 1 hydrogen = 1 Avagrado's number)
= [(4.21*10^24)*(1.60*10^-19)]^2/(4^(6.38*10^6)^2) * 8.99*10^9 = 2.51*10^7. I think the reason we multiply by 4 is because we need to get the diameter for both proton and electron, giving 4.

Last edited: Sep 1, 2010
2. Sep 2, 2010

### Andrew Mason

$$F = \frac{Q_1Q_2}{4\pi\epsilon r^2}$$
where $\epsilon = \epsilon_r\epsilon_0$ ($\epsilon_r$ being the relative permittivity of iron).