Coulomb's Law and silver atom

  • #1
notmetalenough
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I'm having trouble even beginning this problem. I'm not quite sure how to convert from the charge, which I believe I can find, to the number of electrons shared. If someone could help me understand what I'm being asked to find, and how to find it, I would really appreciate it. The problem is as follows

Two small silver spheres, each with a mass of 9.0 g, are separated by 0.60 m . Calculate the fraction of the electrons in one sphere that must be transferred to the other to produce an attractive force of 2*10^4 (about 2 tons) between the spheres. (The number of electrons per atom of silver is 47, and the number of atoms per gram is Avogadro's number divided by the molar mass of silver, 107.87 g/mol.)

I have a few questions about this, am I to assume that the charges are equal? If they are not, how would I find them?

for coulomb's law, I figure that if F = (8.99*10^9) * (q_1 * q_2)/r^2, that q_1 * q_2 = 1.802 * 10 ^-6 or a total charge of about 80 microCoulombs. So the number of electrons being shared (since the charge of 1 e = 1.60*10^-19), would be about 5.0 * 10 ^ 12.

So if there are 47 electrons per silver atom, and 9 grams at 107.87 g/mol, then there should be about (6.02 * 10^ 23) / 107.87 atoms in a gram, * 9 grams, * 47 electrons. so about 2.36*10 ^ 24 electrons in one sphere.

if I take the number of electrons being shared and divide it by the number of electrons total, I get about 2.12 * 10 ^ -12 as the fraction of electrons being shared.

Which apparently is wrong. Where do I need to start over?
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
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notmetalenough said:
I have a few questions about this, am I to assume that the charges are equal? If they are not, how would I find them?
Sure the charges are equal (and opposite). You are taking electrons from one sphere and putting them on the other. The total charge (the sum of both spheres) must still be zero.

for coulomb's law, I figure that if F = (8.99*10^9) * (q_1 * q_2)/r^2, that q_1 * q_2 = 1.802 * 10 ^-6 or a total charge of about 80 microCoulombs. So the number of electrons being shared (since the charge of 1 e = 1.60*10^-19), would be about 5.0 * 10 ^ 12.
Recheck your answer for the charge per sphere. You need to solve for q.

So if there are 47 electrons per silver atom, and 9 grams at 107.87 g/mol, then there should be about (6.02 * 10^ 23) / 107.87 atoms in a gram, * 9 grams, * 47 electrons. so about 2.36*10 ^ 24 electrons in one sphere.
This part looks OK.
 
  • #3
notmetalenough
11
0
Doc Al said:
Recheck your answer for the charge per sphere. You need to solve for q.


That was the problem. I just took the square root of the number I got (which was typoed in what you read) and continued with the problem to get the correct answer. At least I found that my concept of the problem wasn't what was flawed. Makes me feel better about my understanding of coulomb's law.
 

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