Metal Sphere and Charged Rod

  • #1
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Homework Statement:
A thin, light, spherical metal shell is suspended from a thread. The sphere hangs straight down when a charged rod is held at 3 cm to the left of the metal sphere. It can be concluded that:

A. the charge of the sphere is the opposite sign of the charge of the rod.
B. the charge of the sphere is the same sign as the charge of the rod.
C. the sphere is charged, but the sign of the charge is unknown.
D. the sphere is uncharged.

The answer is (b) according to answer key.
Relevant Equations:
None
(A) incorrect, because opposite signs attract, and the sphere would've been drawn to the charged rod.
(B) correct, according to the answer key, but if the charge of the sphere and the charge of the rod are the same, then wouldn't they repel each other? I'm confused as to why this is the correct choice.
(C) incorrect, because if the sphere is charged, it will either move away or toward the charged rod, but it "hangs straight down" instead.
(D) incorrect, because even if the sphere is uncharged, the presence of a charged rod would induced a charge on the sphere.

Honestly I am baffled by this question. I'm not sure if there's something I'm not understanding. This is not conductive charging since no contact was implied. I'm assuming it has something to do with induction and polarization. Any hints or help is appreciated.
 

Answers and Replies

  • #2
Orodruin
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There are two effects at work here. First there is the force between the rod and the sphere due to the total charges on each. Then there is the force between the rod and the sphere due to the change in the charge distribution on the sphere that is induced by the charge of the rod. What do you know about both of these effects?
 
  • #3
haruspex
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(C) incorrect, because if the sphere is charged, it will either move away or toward the charged rod, but it "hangs straight down" instead.
(D) incorrect, because even if the sphere is uncharged, the presence of a charged rod would induced a charge on the sphere.
@Orodruin has already given you the hint, but note the implication of your two correct statements above. With no charge they would attract, and with a sufficient like charge they would repel. So in between....
 
  • #4
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There are two effects at work here. First there is the force between the rod and the sphere due to the total charges on each. Then there is the force between the rod and the sphere due to the change in the charge distribution on the sphere that is induced by the charge of the rod. What do you know about both of these effects?
The force between the charges of the rod and the sphere due to the total charges on each is determined by Coulomb's law? We know that the sphere is a conductive object, which can stay electrostatically neutral in the presence of a charged rod, since it is able to become polarized.

The force between the rod and the sphere due to change in the charge distribution on the sphere after induction by rod is also determined by Coulomb's law? Can we assume that the "opposite" charges on the sphere have moved closer to the charged rod and "like" charges on the sphere have moved farther away, to the extent that the force of repulsion is equal to the force of attraction, which explains why there is no movement horizontally? Since the "opposite" charges on the sphere is closer to the charged rod, the force of attraction is greater; however since there is no net movement, we can assume that there are more "same" charges on the metal sphere to repel the rod, since it is at a greater distance from the rod?
 
  • #5
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@Orodruin has already given you the hint, but note the implication of your two correct statements above. With no charge they would attract, and with a sufficient like charge they would repel. So in between....
I guess I was tripped up by the wording of the question. It's asking for the sign of the charge, not the magnitude. In my head, I thought that there was no scenario in which the sphere was motionless, since same signs attract, opposite signs repel, and uncharged sphere + charged rod will result in an attraction.

In order for the sphere to remain motionless, the electrostatic forces have to be such in that they cancel each other out, which means the sphere cannot be neutral. It has to possess more "like" charges in order to repel the attractive forces between the charged rod and the "opposite" charges that have redistributed closer to the rod.
 
  • #6
haruspex
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I guess I was tripped up by the wording of the question. It's asking for the sign of the charge, not the magnitude. In my head, I thought that there was no scenario in which the sphere was motionless, since same signs attract, opposite signs repel, and uncharged sphere + charged rod will result in an attraction.

In order for the sphere to remain motionless, the electrostatic forces have to be such in that they cancel each other out, which means the sphere cannot be neutral. It has to possess more "like" charges in order to repel the attractive forces between the charged rod and the "opposite" charges that have redistributed closer to the rod.
Yes.
 
  • #7
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