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Disclaimer: This is not a homework question. This came from an online resource that I've been using to tutor a student. After much thought, I found myself going in circles on this one.

  • Two neutral conducting spheres, A and B, which are initially in contact, are brought close to a rod (positively charged). Sphere B is taken away. The investigation is repeated with a negatively charged rod, same procedure of steps. Which of the following is true?

(It's not clear to me, by the way, if they mean that we bring the rod close, remove sphere B, and then repeat the procedure with sphere B absent. Or just repeat the entire procedure, with sphere B present again.)

A. The rod induces charge separation, so sphere A gains a charge opposite that of the rod. Removing B, sphere A retains that charge and is therefore attracted to the rod in both cases.

I understand that the rod should induce a charge separation. But it's not clear to me how that relates to the fact that there are two spheres in contact. The problem makes no distinction regarding where, relative to the spheres, the rod is placed "close" by. So if we factor out relative location, then it seems like, due to lack of contact, all we should expect is a temporary charge separation. After which the sphere, being removed, wouldn't affect anything.

B. Charges from the rod transfer to sphere A, making them the same charge. There is repulsion between these two objects in both cases since they are like charges.

This answer doesn't make sense to me due to the fact that the rod doesn't come into direct contact with either sphere.

C. Because sphere A remains neutral throughout the experiment, it will attract any nearby charged object because of induced charge separation. So attraction in both cases

I ruled this out as close to nonsensical for awhile until I started to regard it as the most likely answer due to problems I had with the others. Currently it's my top contender.

D. The gravitational force between objects is always attractive, and at this scale the gravitational forces are more dominant than electric forces in both cases so attraction wins

Nonsensical answer of course.

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Because the spheres are made of metal, electrons are free to flow between them. So when the rod containing positive charge is brought close to the spheres, the electrons within the two-sphere system will move toward the rod.

This means there will be a migration of electrons from sphere B to sphere A (assuming sphere A is closest to the rod). This will cause the two spheres to be polarized as they are in contact (but the whole two-sphere system is still electrically neutral). But each individual sphere, will have a net charge. Sphere B will have a net positive charge while sphere A will have a net negative charge. When sphere B is moved away from sphere A the positive charge will redistribute itself uniformly about sphere B, and therefore there will be net negative charge uniformly distributed around sphere A. Sphere A attracts the rod.

It's not clear to me, by the way, if they mean that we bring the rod close, remove sphere B, and then repeat the procedure with sphere B absent. Or just repeat the entire procedure, with sphere B present again

It means the process is repeated with a negatively charged rod. The same will happen but with opposite net charge on each sphere. That is, there will be net positive charge on sphere A and negative charge on B. Sphere A attracts the rod.

all we should expect is a temporary charge separation. After which the sphere, being removed, wouldn't affect anything.

As written above, since the charged rod polarizes both spheres, removing one sphere results in net opposite signed charge on either sphere.

This answer doesn't make sense to me due to the fact that the rod doesn't come into direct contact with either sphere.

As written above, it makes sense that the rod is brought near to the spheres and not in contact, since charge is free to move on the spheres, all we need is polarization of charge. In fact, if the rod did make contact, there would be charge flow, which will result in equilibrium (equal charge on each sphere).

A. The rod induces charge separation, so sphere A gains a charge opposite that of the rod. Removing B, sphere A retains that charge and is therefore attracted to the rod in both cases.

This is obviously the correct answer, as we have seen in the analysis above.

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  • $\begingroup$ This makes sense, thank you $\endgroup$ Oct 7, 2021 at 5:58
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    $\begingroup$ You're quite welcome David. Good luck with your tutoring. Cheers. $\endgroup$
    – joseph h
    Oct 7, 2021 at 5:59

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