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Magnetic field and displacement current

  1. Nov 26, 2009 #1

    crx

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    Is there really a magnetic field around a capacitors (parallel plate) in vacuum dielectric fed by a variable voltage? Is there an experiment that can prove that we don't need actually a moving electrical charge to create a magnetic field, but a variable electric field in vacuum its enough?
     
  2. jcsd
  3. Nov 26, 2009 #2
    The relation between charge Q and voltage V on a capacitor C is
    Q = CV
    The derivative is
    dQ/dt = I = C dV/dt

    So it requires a current I to produce a variable voltage dV/dt on a capacitor. So there is a moving electrical charge around the capacitor.
    Bob S
     
  4. Nov 26, 2009 #3
    Here is an example of the displacement current creating a real current. Take an air capacitor C with gap d and area A. Put and maintain a voltage V on it. So the charge
    Q = CV = e0AV/d.
    Now insert a dielectric of relative permittivity e and thickness d and area A in the capacitor. Now
    Q' = C'V
    where C = ee0A/d
    There is a current in the external circuit that increases the charge Q on the plates to maintain the voltage V on the capacitor.
    Bob S
     
  5. Nov 26, 2009 #4

    crx

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    Yes,but this is because of the dielectric molecules are shielding and weakening the electric field, so the capacitor will need more charges to reach the power supply voltage, so there will be a current in the external circuit.
    What i would like to know is that if a plate capacitor with no dielectric in vacuum (with a pretty large gap ), connected to a AC supply, will have a magnetic field exactly in the area between the plates where there are no moving charges, but only variable electric field...
     
  6. Nov 26, 2009 #5
    From Maxwells equations,
    Curl H = sigma E + e e0 dE/dt = e e0/d dV/dt,
    so a varying voltage across the capacitor creates a magnetic field, even when the conductivity sigma = 0.
    Bob S
     
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