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A question of charged spheres and voltages

  1. Jan 26, 2005 #1
    I have a question regarding charged spheres and voltage measurements. I've talked to my physics professor about it, and he wasn't quite sure of the answer (although to be fair, I asked briefly at the end of class.)

    Imagine that you had a spherical shell, charged to a high potential (for the sake of argument, we'll say 100 kV). Now, in that shell, there's a small hole, where a small, insulated copper wire is piped in. At the other end of the wire, which is very far away, there is a constant 12V DC source, and the positive terminal is connected to the wire. Here's my question. If one was to measure the voltage of the wire with respect to the negative terminal inside the spherical shell, what would one read? Furthermore, if a capacitor was placed inside the sphere and attached to the wire's end, would it charge up (i.e., would current flow through the wire)? I've attached a picture for clarification.

    It seems to me that there are two conflicting ideas here. First, I know that the E-field inside the shell is zero. However, just outside the sphere, it is relatively high. If I was to draw a voltage vs distance from center graph, at infinity, the voltage would equal 12V. At r, the radius of the sphere, it would equal 100kV, having slowly increased, and inside the sphere, it would drop to zero. So, I guess my question is: would positive holes be able to overcome the large increase in potential in order to decrease their overall potential (from 12V to 0V)? That is, can holes (or electrons, for that matter), act like water in a siphon?

    Attached Files:

  2. jcsd
  3. Jan 26, 2005 #2
    I din't get your question properly.
    Anyways, the thing I noticed was this -
    The elctric field inside the shell is zero and so the potential is CONSTANT and not zero. And the potential inside is equal to the potential at the surface i.e. 100 kV.

    That may or may not clarify anything, but it is present. Sorry if it wasn't relevant.
  4. Jan 27, 2005 #3

    Andrew Mason

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    The answer has to be 12 volts. I don't see how you can get a potential difference anywhere along the wire. (I gather that it is a uniformly charged sphere, and that the sphere is a conductor as well)

    If you use a gaussian cylinder spanning the opening, just around the conducting wire, the charge density contained in the cylinder is [itex]E\epsilon_0 = \sigma[/itex] which has to equal 0 assuming that the wire was uncharged to begin with: so the field E = 0. The field lines are radial from the sphere, so no flux from the sphere would be intersected by the wire.

  5. Jan 27, 2005 #4
    I agree that at most points along the wire, the flux passing through it is zero. However, I think that the problem arises at the opening, where there are some field lines pointing inward, due to the fact that some extra charge will gather at the edge. In this case, if one were to imagine a Gaussian cylinder around the wire, there would be flux pointing inward, indicating negative charge buildup at the opening. Wouldn't this electron buildup decrease the voltage measured at the end?
  6. Jan 27, 2005 #5
    I'm sorry but I can't make sense out of the question. You asked
    What is the "negative terminal" that you speak of? I see a red square inside the shell and that red square is connected to the "+" terminal of a 12V battery.

    I take it that you want to know the difference in potential between two points/between two equipotential surfaces, right? If so then what are those points/surfaces?

    Note that when you poked a hole in the shell and placed that wire in it then its not the perfect spherically symmetric charged shell that we all know and love. The difference here is that another conductor is placed inside the shell and that conductor (red wire) has charges on it which are induced by the battery. The shell and battery will fight it out of course. :smile: By that I mean that the arrangement of charges on the wire are altered by the presence of the charged sphere.

    Also, the E-field inside the shell is not exactly zero now since you've introduced a wire with charges on it into the shell and there is a corresponding electric field from those charges. The situation is extremely complicated to give a precise answer though.

    One thing to note is that there is no reason to assume that the potential at infinity is 12V. If there are no sources at infinity then we always choose the potential at infinity to be zero.

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