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E Field?

  1. Jul 4, 2011 #1
    I was reading this question out of a book. A small metal sphere hangs by an insulating thread within a larger hollow conducting sphere. A conducting wire extends from the small sphere through, but not touching, a small hole in the hollow sphere. A charged rod is used to transfer positive charge to the protruding wire.
    And it wants to know what objects will be charged.
    This is not a homework question just wondering.
    SO just the inner smaller sphere will have positive charge on it, since the charge can't flow across the insulating thread. But the electric field inside a conductor is zero so the positive charge on the smaller sphere will attract negative charge on the inner shell of the larger sphere and then leave positive charges on the very outer shell of the sphere.
    But what if the smaller sphere was hanging to the shell by a wire. Would the charge then just spread out uniformly to make the E field inside the conductors?
    And then also why would the positive charge leave the rod to the metal sphere. Is the E field pushing the charge onto the sphere. And is the E field of the rod causing the metal sphere to be polarized and this attracts the positive charge from the rod?
  2. jcsd
  3. Jul 4, 2011 #2
    If hanging by a wire, all the charge would go all the way around to the outside of the large sphere. There would be zero electric field in the hollow space.

    When hanging by the insulator, there is a non-zero electric field between the spheres.

    The positive charge leaves because the voltage on the rod is higher. This isn't given in the book but it is the case. The voltage is a function of the shape of the rod and the number of charges already on it. Q=CV as in a capacitor.
  4. Jul 4, 2011 #3


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    Conductors are equipotential surfaces and the potential is dependent upon the surface charge that exists on the conductor. So if you have a charged conducting rod, then it will have a different voltage than your neutrally charged spheres and shells. However, when you bring the charged rod in contact with your small sphere then the charge must redistribute itself across the joined surfaces so that a new equipotential is reached.

    As for a more physical picture, then yes I think you can take the induction of the charge separation on the small sphere to be the culprit. As you bring the positively charged rod into proximity of the sphere, it induce charge separation. The electrons will bunch up on the surface closest to the rod leaving positive ions on the opposite side. Upon touching, the electrons be pulled over by the positive ions on the rod thus depriving the sphere of some of its electrons resulting in a positive charge.

    You can even see this easily enough yourself. Make a pith ball using a foil candy wrapper (gum or Ferrero Rocher) and wrap the foil around a piece of string. Then take a plastic comb and rub it against wool or cotton (sock does nicely). When you bring the comb into close proximity of the pith ball the charge induction can become so severe that the ball is physically attracted to the comb.

    This does not necessarily mean that the charge is evenly distributed. We can make that claim in some cases by virtue of symmetry though. For example, if you connect the inner and outer sphere by a conducting wire as you propose then the charge distribution should be symmetric about the axis of the wire (say along the \phi direction) but it can vary along the \theta direction due to the breaking of symmetry.
  5. Jul 4, 2011 #4
    Ill have to get some gum and a string and make that. thanks for both of your answers.
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