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Question about charge induction

  1. Aug 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Untitled 2.jpg

    2. Relevant equations
    There are no equations here, only theoretical knowledge

    3. The attempt at a solution
    Hi,

    My approach was this:
    I know that with the positive charge on the sphere being lowered into the cup I will have the inner area of the cup have a negative charge and the outer area will have a positive charge as the positive charge of the sphere attracts the negative electrons.

    Because the outer area is going to become positively charged by induction, then I will have electrons come up from the Earth to neutralize it, so the inside surface will become negatively charged, and the outer surface will be neutral.

    However, the correct answer is B.

    Could anyone please offer any insight into why the answer is B?

    Thanks in advance!
     
  2. jcsd
  3. Aug 26, 2016 #2

    Merlin3189

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    You are correct while the charged ball is still in the cup. But what happens when the ball is then removed? Where is the charge then?
    Remember Faraday's "ice pail" experiment.
     
  4. Aug 26, 2016 #3
    The ball is removed without touching the pail, so the negative charge could not transfer to it....

    I honestly have no idea apart from that.

    I do know that we got a net negative charge on the pail.

    Maybe when we remove the sphere, we have the whole net negative charge redistribute itself evenly on the surface? However, that is not the right answer "B".
     
  5. Aug 26, 2016 #4

    Merlin3189

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    Which is correct - negative charge on the inside and neutral (grounded) on the outside. So what is the total overall charge on the jar?

    If the ground remained connected, then when you remove the ball, all the electrons which were attracted onto the cup by the ball, now flow back to earth.
    BUT we first disconnect the grounding wire, while the ball is still in the cup. Removing the wire does nothing, because the electrons have long since gone where they wanted to and now are in happy equilibrium.
    Then when the cup is isolated again, we remove the ball. The electrons may want to go back to earth (now that the ball is no longer attracting them), but have no way of doing so.
     
  6. Aug 26, 2016 #5
    We have a net negative charge.

    We first disconnect the ground wire, yes. The excess electrons are on the cup as usual.

    When the cup is isolated, we remove the ball, and the electrons would want to go into equilibrium. That would mean that they spread out easily over the whole cup, not just an inside or outside side, right? So, we would have 100 percent net negative charge evenly distributed along the whole cup, which is obviously not the case.

    What is my logical fallacy here?
     
  7. Aug 26, 2016 #6

    gneill

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    Like charges mutually repel (and want to end up as far from each other as possible given the constraints of the system). So where should the charges migrate to?
     
  8. Aug 27, 2016 #7

    Merlin3189

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    No *logical* fallacy, but as gneill says, they behave differently than you are thinking. Think about his question.

    But more generally you are making an untrue assumption when you say "evenly distributed". It is not just hollow conductors that can have a surprising charge distribution. The shape of a conductor affects the charge distribution. But you may not have gotten that far just yet.

    When you've sorted this question, it might be worth looking at some of the many descriptions of Faraday's experiments with his ice pail (ice bucket.) Some of the details - quantity of charge, detecting polarity, using multiple balls, touching the ball to the inside of the cup, the hole(s), etc. - illustrate a lot of electrostatic ideas.
    The "Faraday cage", based on the principle from this expt is still of practical use.
     
  9. Aug 29, 2016 #8
    Wouldn't the charges migrate all over the cup evenly as the inside and outside spheres surfaces contact each other?

    OK. Let's step back. The inside is negatively charged, but the outside is neutral after drawing up the electrons from the ground.
    The answer is B. How can negative charge then reside on the outside surface and no charge reside on the inner surface, which is the opposite of this situation that we have here?

    I would understand if the electrons from the net negative electric charge distributed themselves all over the inside and outside surfaces so that both surfaces would be negatively charged, regardless of how evenly things are distributed.

    I am selfing E&M via MIT Opencourseware, and I have not run into anything of this sort yet.

    Thanks for the help!
     
  10. Aug 30, 2016 #9

    gneill

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    Any system that is free to move will "seek" to move to the lowest energy state that it can find. For mutually repellent mobile charges that means them ending up located as far from each other as possible and with the least overall density. The charges "flee" from the center to the periphery, trying to escape each other and reduce the mutual potential energy by maximizing the overall spatial extent that they fill.

    Objects with the most geometric symmetry are the simplest to work out how the charge will distribute, with the sphere or spherical shell being the ideal cases. The cup shape has nearly as much symmetry, but every degree of symmetry that's taken away makes it that much harder to just "feel" the solution intuitively.

    Sometimes it can be helpful to conduct a thought experiment. Suppose you imagine starting with an uncharged cup and you are able to add point charges one at a time (you may think of them as individual, indivisible electrons if you want). You may add them anywhere you like, but they are all free to move after they are introduced to the cup material. A single charge on the cup might roam around freely anywhere in the cup material, there being no other charges to tell it what to do. So what happens the moment a second charge is deposited anywhere on the cup? How will the pair of charges move to try to lower the potential energy? What if then a third charge is introduced, and so on?

    It soon gets tricky to mentally rearrange the charges in the given geometry in order to find a minimum potential energy, especially when there are multiple possible arrangements. But I hope you can see that in all cases where there's more than one charge the arrangements with the lowest potential energy will have the charges on the outside surface of the cup.

    Note that the real experiment has been done for this shape, too, and the conclusions confirmed. That's the so-called "Faraday's Ice Pail" series of experiments that has been mentioned by others. It is worth reading about it.
     
  11. Aug 30, 2016 #10

    Merlin3189

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    Agree with gneill. Just to add another idea.
    How about drawing in lines of electric field, which join charges of opposite potential. Where do the lines go from charges on the inside of the cup? (Especially if you imagine adding a lid, as Faraday actually tried.)

    In case you're not so familiar with them, In the absence of anywhere obvious to go nearby, lines go off to earth or go off indefinitely in search of earth! Field lines try to be as short as they can, without crossing., but "prefer" to be spread out rather than close together (a bit of a contradictory balancing act of course. But then you can't take these lines too seriously, they just illustrate what happens in a lot of situations.)
     
  12. Aug 30, 2016 #11
    Thank you for your help in trying to decipher my question, everyone. I might have come up with a good logical set of steps that may explain why the right answer is B. Please let me know if I mess up.

    here is where I am at now: I have a negatively charged inner surface and an electrically neutral outer surface.
    I have electric field lines going from the outside surface to the inside surface, so the electrons will naturally want to move in the other direction (and to minimize their surface density).

    The excess electrons will then move from the inner surface to the outer surface completely, leaving the inner surface electrically neutral and the outer surface having a net negative charge, hence the answer is B. Right?
     
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