Asked by ~christina~

Posted by ~christina~ on Sunday, March 30, 2008 at 9:14am.

Two small silver spheres with a mass of 10.0g, are separated by 1.00m. Calcuate the fraction of the electrons in one sphere that must be transfered to the other to produce an attractive force of 1.00x10^4 N (about 1 ton) between spheres (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, 107.87g/mol.



Fe=ke(|q1|q2|)/r^2 =1.00x10^4 N
then
q1/(-1.6021765x 10^-19C)= ans/47= the fraction of e- transfered to the other sphere

<b>I'm not sure how to find the q2 and I think I need to find q1 from the equation below (what would I put in for q2)?</b>

Please help.

Thanks
Answered by ~christina~
I'm not <b>what to put for q2 in the equation above to find q1 from the equation above</b> (what would I put in for q2)?
Answered by ~christina~
I'm asking nicely. =D

(I can almost solve the problem if only someone could tell me what do I put into the force equation for q2 -pretty frustrating that I can "almost" solve it but I need just one part)
Answered by drwls
Since you are transfering charge from one sphere to the other, q1 = -q2. Let the absolute value of both be q and just solve for q. The force will be attraction since the charges are opposite in sign. Once you have q, divide it by e for the numnber of electrons. Compare that to the number of electrons in a 10 g silver sphere. Each sphere would have this number of elctrons: (10g/107.87g/mole)*47 elecrons/atom*6.02*10^23atoms/mole
Answered by ~christina~
do you mean that it would be
q1=-q2
Fe=ke|q1||-q1|/r^2= ke(2q)/r^2
Answered by drwls
No. The product of the two charge absolute values is q^2, not 2q.
Answered by Anonymous
okay. but why do you say to compare that to the number of electrons in a 10g silver sphere?
Answered by drwls
Because that is what the problems asks for -- a fraction. Read it again:

"Calculate the fraction of the electrons in one sphere that must be transfered to the other"

I was just telling you how to do it. It will probably be a very low fraction.
Answered by ~christina~
I got it drwls.

Thank you.
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