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Electrostatic voltage and current

  1. Aug 31, 2014 #1
    If we had a van de graaff generator, near the sphere there is a metal plate ( not touching) and it's wired to the ground (a resistor is connected to the wire) , as the voltage on the surface of the sphere increases, the voltage at the plate does too, creating a potential difference between the plate and the ground and a current will flow, will this current depend on the resistance of the wire? ( no electrical discharges occurs )
     
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  3. Aug 31, 2014 #2

    Simon Bridge

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    Yes - the current through any resistor depends on the potential difference across it and it's resistance.
    A wire is another resistor - usually with a very small value of resistance.
    If you have an ammeter in there, this will also affect the current.

    Note: you are not describing a static situation.
     
  4. Sep 1, 2014 #3
    But when the voltage reaches its maximum the current will stop, right??
     
  5. Sep 1, 2014 #4

    Simon Bridge

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    Voltage where?
     
  6. Sep 1, 2014 #5
    There is a potential at the surface of the sphere, as we increase the distance ( from the center of the sphere) increases in radial direction the potential decreases, but the potential reaches a maximum after that the generator cant build up more charge
     
    Last edited: Sep 1, 2014
  7. Sep 1, 2014 #6
    If the case i am describing is not clear, i can draw a sketch and upload if you want
     
  8. Sep 1, 2014 #7

    Simon Bridge

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    Oh OK - some notes:
    A single location cannot have a voltage by itself - you'll notice you need two leads to make a voltmeter work right?
    So the VdG ("Van der Graaf" - saves typing) sphere may reach a max voltage wrt the ground (we can now read that all voltages wrt the ground unless otherwise specified ... I needed to be sure).

    You are using the sphere to induce a voltage on a plate, and the plate is connected to the ground by a wire.

    In a secondary school level model:
    What happens is that the sphere separates charges in the plate - repelling like charge and attracting the opposite.
    This means that charges move about the plate until the electric field due to the charge separation is equal and opposite the electric field due to the VdG sphere. That's a current, but it does not last long.

    Basically you end up with like charges on the opposite side of the plate to the VdG.

    Attach a wire from the opposite side to the ground and charges from the ground can travel up the wire to cancel the charges on the far side of the plate. You can think of the ground as a large source for any charges that may be attracted or a sink for a any charges that may be repelled. Note though: this is a very qualitative description. The process is called "charging by induction" - you may have heard of it already?

    Re your question then:
    ... no. As you saw from the above, the voltage on the VdG sphere can be at it's maximum and there is still a current. The current flows as long as there are charges available to flow.
     
  9. Sep 1, 2014 #8
    Okay, thanks
     
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