Electric Charges: Forces on A and C Explained

AI Thread Summary
Two equally charged conducting spheres, A and B, repel each other with a force of 2.0 x 10^5 Newtons. When an uncharged sphere, C, touches A, it acquires half of A's charge, resulting in both A and C having a charge of q/2. The electric force on A remains unchanged, while the force on C, positioned halfway between A and B, can be calculated using the formula F = k(3/2)q*(1/2)q/r^2. This leads to a new force expression that combines the charges of A and B with the charge of C. The discussion emphasizes the importance of charge distribution upon contact between conductors.
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Homework Statement


Two equally charged identical conducting spheres, A and B, repel each other with a force of 2.0 x 10^5 Newton. Another identical uncharged sphere, C, is touched to A and then moved over right next to B.
What is the electric force on A now?
What is the electric force on C after it has touched A and is halfway between A and B?


Homework Equations


F=K q^2/r^2



The Attempt at a Solution


I let q arbitrarily equal 1.0x 10^-6 Nm^2/C^2
I determined the distance to be 21.2 m
If A is negatively charged at 1.0 x 10^-6 does C touching it take half of that charge?
If so, does A and C now have a charge of 5.0 x 10^-7?
I'm not sure what happens to the charge when A and C touch.
If they are halved, then the electric force on A remains the same and
the force on C remains the same when it is at half the distance.
I would love some help with this, please.
 
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Don't assume arbitrary numerical values and make things messy. Let's suppose the charge on A and B is q each. If their dist is r, F = k*q^2/r^2.

After touching, the charge q will distribute itself equally on A and C, since they are conductors. So, A and C each has q/2.

If you now place C very close to B, then there is a net charge of 3q/2 there, and q/2 at A. So, force F1 = k(3/2)q*(1/2)q/r^2 = k*3q^2/(4r^2).

We can now find how many times is F1 of F.

Can you do the 2nd part now?
 
Yes, I've got it now. Thank you very much!
 
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