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Changing Dielectric parts

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data

    This was done on my Physics II class, ans the Professor has'nt want to explain it to us

    A 2 Circular Metallic Plate Capacitor, maybe of a diameter of 15cm, was connected to a constant sourch of DC voltage untill it was charge to 10Volt, after this the source was DISCONNECTED from the Capacitor, and of course the voltage on it remained at 10V

    Then, the professor took a DRY plastic racket with 500ml of water INSIDE, and puts it between the plates, leaving no space between them, it was very precise, the plates where adjusted to the specific width of the racket. By doing this the voltage went down to 4V

    And then, he takes the racket out the Capacitor, bringing the voltage back to 10V, and empties all the water and closes the little overture. After this he put the racket back between the plates and the voltage comes down to 4volts AGAIN.

    2. Relevant equations

    Why does the change of the Dielectric makes no difference?

    3. The attempt at a solution

    The water leaving the racket gives up a large quantity of ions, making the air so conductive to the field as it was before
  2. jcsd
  3. Nov 25, 2011 #2

    rude man

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    Homework Helper
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    What is the formula for capacitance with dielectric constant ε?

    What is ε for water and for air?
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