Suppose also that 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?
The answer should be in Newtons
2007-09-10 07:28:40 · 3 answers · asked by ferrari2388 1 in Science & Mathematics ➔ Physics
It will be the charge that defines the force.
30*10^23 protons or electrons is a charge (each pole) of 480*10^3 Coulombs (1 electron = 1.602*10^-19 C).
Force = (k *Q1*Q2) / r^2
r is the diameter of the earth from pole-to-pole ~12.6*10^6 m
k is 1/ (4*pi*epsilon0). but you have to take into account the relative permittivity of the earth, which will be somewhere on the order of 2-5 times the permittivity of space. So, I would make k = 1 / (10*pi*epsilon0)
epsilon0 = 8.854*10^-12 C^2 / Nm^2
It all works out to be 5.2 MegaNewton
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2007-09-10 07:53:32 · answer #1 · answered by tlbs101 7 · 3⤊ 0⤋
1g of H has 6.10²³ atoms
5g of H has 30.10²³ atoms. So there's 30.10²³ Eletrons and 30.10²³ Protons in 5g of H.
That's just half of the problem. I can't figure out how to solve the second one.
ADDED...
If the awnser of the user bellow me is right.
Just multiply the values for 9,8 (The gravity of earth) to found the weight in Newtons
2007-09-10 14:35:23 · answer #2 · answered by Anonymous · 0⤊ 0⤋
That's about 4.9975193 grams of protons and about 0.0024807 grams of electrons.
Are you trying to calculate electrostatic force of attraction between them or something. lol! I have no idea.
2007-09-10 14:55:11 · answer #3 · answered by Anonymous · 0⤊ 0⤋