Metal sphere drawn and then repelled by charged rod

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When a positively charged glass rod is brought near an uncharged metal sphere, the sphere is attracted due to the rod's positive charge attracting the sphere's negative charges while repelling its positive charges. Upon contact, some electrons may transfer from the sphere to the rod, slightly neutralizing the rod's charge and making the sphere positively charged. Once both objects are positively charged, the sphere is repelled by the rod due to like charges repelling each other. The discussion highlights the importance of charge transfer, even in nonconducting materials, and the role of electric fields in attraction and repulsion. This phenomenon illustrates fundamental principles of electrostatics.
MonkeyDLuffy
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Homework Statement



An uncharged metal sphere hangs from a nylon thread. When a positively charged, nonconducting glass rod is brought close to the metal sphere, the sphere is drawn toward the rod. But if the sphere touches the rod, the sphere suddenly flies away from the rod. Explain why the sphere is first attracted and then repelled.

Homework Equations



Assuming there are only horizontal components making a contribution,
a = the distance from the rod to a point on the sphere
L = the length of the rod
Q = total charge
k = constant
Ex = (kQ) / (a2 + aL)

The Attempt at a Solution


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At first the sphere is attracted by the rod because the positive charge on the rod attracts the negative charge on the sphere while repelling the positive charge on the sphere. This part I understand.

Initially I thought that when sphere touched the rod the excess positive charge would flow into the sphere and the rod would repel the sphere. However, because the rod is nonconducting there should be no flow of charge.

In another attempt I thought that perhaps when the sphere made contact with the rod, then it was close enough for the electric field which repels the sphere to overcome the electric field which attracts the sphere. But if the repulsive field increases, I'm assuming the attractive field is increasing at the same rate.

From here I'm not sure what other angles I can look at this from. Could it be the case the rod is not perfectly non conducting so some positive charge does flow into the sphere and thus the repulsion occurs?
 
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MonkeyDLuffy said:
At first the sphere is attracted by the rod because the positive charge on the rod attracts the negative charge on the sphere while repelling the positive charge on the sphere. This part I understand.
That alone is not sufficient. The positive and negative charge have the same charge (in magnitude). The effect you are looking for needs one step more.
MonkeyDLuffy said:
Initially I thought that when sphere touched the rod the excess positive charge would flow into the sphere and the rod would repel the sphere. However, because the rod is nonconducting there should be no flow of charge.
Positive charges are nuclei of atoms, those don't move in solids. The rod might not be a good conductor, but there are no perfect insulators. Some charge will be transferred.
 
mfb said:
Positive charges are nuclei of atoms, those don't move in solids. The rod might not be a good conductor, but there are no perfect insulators. Some charge will be transferred.
mfb is absolutely correct, what moves around are electrons (negatively charged). I can conceive that upon contact, some electrons move from the sphere to the non-conducting rod, but not much though, and the potential of the non-conducting rod remains positive, even though it is now a little less than it started. That will be enough though to make the sphere positively charged.
Now that both objects are positively charged, you can guess what happens next!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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