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Disturbing the Faraday Cage

  1. Jun 3, 2010 #1
    Hi everyone!

    I was thinking..then I decided to post this thought to ask your opinion about that.

    We have this capacitor made by different layers, starting from the top we have: electrode, dielectric material (let's assume 10um thick, permittivity k=600), conductive layer (good liquid conductor, 2um thick, conductivity sigma=1000-10000uS/cm), dielectric material (again 10um thick, permittivity k=600), electrode.

    Now we apply an electric field to the electrodes of the capacitor. The conductive layer in the center will not allow the field to pass through it (electric field into the conductive layer keeps being zero).

    Question: let's apply now another electric field across the conductive layer, orthogonal to the other one (10-100 times higher), is it possible then to assume that the generated current is "disturbing" the electrons in such a way to alter the shielding effect?

    Thank you very much for your contribution!

    Attached Files:

  2. jcsd
  3. Jun 3, 2010 #2
    What "generated current"? How much voltage (current) is passing through the conductive liquid layer before and after you apply othogonal E field.

    (I assume the orthogonal E field direction on your illustration is out of the plane of the page.?)
    Is this a DC field??

    Is there a measured voltage across the electrodes of the capacitor after the ortho E field is set up?
    Last edited: Jun 3, 2010
  4. Jun 3, 2010 #3
    The "generated current" through the conductive layer is the one indicated by J (a current density) in the picture. It's generated by the DC field in the x-direction (let's say). The electric field applied on the capacitor in the y-direction (across the electrodes) is a DC field (or a low frequency varying field, up to 100Hz).

    The state "after" was just to explain the structure, they are co-exisisting at the same time.

    Easily, we apply 1V in the y-direction across the electrodes, and 100V on the x-direction across the conductive layer.

    Anyway, I was wondering if there is any way to get a finite permittivity from a conductor without employing really high frequencies field.

    Thank you again
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