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Question about capacitor voltage and inner dielectric effects?

  1. Dec 23, 2011 #1
    A capacitor is connected to a HVDC power supply which provides a minimal current.
    When the volage reaches the capacitor plates the dielectric material in the capacitor polarizes.

    Now, if there existed a barrier between the dielectric and the capacitor plate the capacitor could theoretically also operate like a transistor. If the voltage applied to the plates was high enough to break through the barrier free electrons will enter into the dielectric.

    My quesiton is, if this could occur would the excess electrons entering into the dielectric material cause the capacitors voltage to rise above the power supply voltage?
  2. jcsd
  3. Dec 23, 2011 #2


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    Science Advisor

    No, the capacitor voltage can only rise to the supply voltage. Where would any extra voltage come from?

    If the dielectric breaks down, the voltage across the capacitor drops rapidly.
  4. May 19, 2012 #3
    When the dielectric breaks down, the water conducts electricity. Conductivity allows current, so potential energy is converted into electron movement. There is a scientific effect known as Avalanche Effect which may occur in this situation, but I'm not sure. there may be temporary electron holes causing an avalanche, which again means lots of current, and potential energy is converted into movement of electricity rather than it's previous static built up state.

    It is interesting that dielectric breakdown is not taught as an alternative to faraday electrolysis in schools. As far as I know dielectric breakdown means that no salts (electrolytes) are needed, so this would be a cheaper method of breaking down water, since salts cost money to add to a device, and salts clog up/dirty up the device with contaminants. Theoretically, distilled water could be used without any salts and still produce hydrogen, since distilled water does conduct electricity without any salts at a dielectric break down. WIthout the dielectric breakdown, water is not a good conductor. Faraday did not have this knowledge back in the day when he was doing his tests.
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