What is the resulting compressional force on Earth?

[dekoi]

1.) If 1.00g of hydrogen is separated into electrons and protons, and the protos are placed on Earth's north pole while the electrons on the south pole, what is the resulting compressional force on Earth?
So i would need to get the attractive force of the electrons and protons at the north and south poles. I could use the formula
F_e = k_e \frac{q_e q_p}{r^2}
Where q_e is the charge of the electrons and q_p that of the protons. The r (i suppose) would be half of the circumference of earth. Correct? The biggest problem is getting the number of electrons and protons from the mass of the hydrogen. How do i do this?! I am sure I am overlooking somethign.
2.) This is a general theory question.
Suppose i have 2 stationary charges x-distance apart. Another charge is on the same axis and is movable. What does it mean for the system of charges to be in stable equilibrium? I would assume it means for the net force on charge three to be 0.
If i am told to solve for the position of charge three in respect to the other two stationary charges, i equate the two forces acting on it and solve for "x". But i could get two values for x (its a second-degree equation). I always thought that it makes more sense to take the positive value, since that would mean the third charge lies in between the other two, but the answer sheet tells me otherwise. One of the answers says that the distance of the third charge is (e.g.) -1.54cm. How would stable equilibrium be reached if the charge is not even in the middle of the other two?

 

[Astronuc]

In #1, r is the separation between the charges. So with protons at one pole and the electrons at the other, one must use the polar diameter for r. This is a classic physics problem.

1 gram of H = 1 gram-mole of H = Avogadro's number of H atoms which gives the same number of protons and electrons.

 

[dekoi]

Yes, that worked. Thank you.
Can anyone answer the second question, please?

 

[mezarashi]

In my definition, stable equilibrium means that if the position is perturbed any delta x from the equilibrium position, the restoring forces will act to return it to the original equilibrium. Consider a configuration with 3 positive charges. Place a fourth in the middle. It will remain there if it is perfectly in the center. If you touch it just a bit, it will most definitely speed off in that direction. This is an unstable equililbrium.

As for the negative sign. How is the book defining where the two charges are placed? -x/2, +x/2 or etc. For the case of stable equilibrium being outside, however, it is possible if the two charges are of opposite and also different magnitudes. In this case, being between the two charges will achieve an unstable equilibrium. The opposite is true for like sign charges.