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Varied Electrical Potential Difference with Dielectric.

  1. May 19, 2009 #1
    Hello All,
    I am wondering why the presence of a dielectric reduces the potential difference of a capacitor (after having been separated from its energy supply). I understand how it works, based around maintaining the same charge, I just don't see why.
    Thank you for the Help.
  2. jcsd
  3. May 19, 2009 #2
    My apologies,
    I figured its due to a reduction in Electric of the initial capacitor system due to an induced field from the dielectric and its atomic components.
    Sorry again.

    A question I do still have after reading a bit more is what would happen if the capacitors initial Electric field and Potential difference were to be ridiculously huge almost theoretical, and then it polarized the dielectric between it. What would happen to the molecular components and object itself. Would it just heat up due to the intermolecular movement and possibly liquefy. or even better vaporize?
    Last edited: May 19, 2009
  4. May 19, 2009 #3
    increasing the voltage on a capacitor by changing the capacitance (at constant charge) is a technique used to generate high voltages. This can be done either by mechanical means (changing the plate spacing) or by changing the dielectric. The vibrating reed electrometers use mechanical vibration to generate a small ac voltage. Variants of the Wimhust (rotating electrostatic) machines used this principle. A parallel plate capacitor can be constructed with ultrapure water (dielectric constant e = 80) as a dielectric. C = e e0A/d, where A=plate area, and d=spacing. After charging the plates, removing the water (or other dielectric) will increase the voltage, because V = Q/C = Qd/(e e0A).
  5. May 19, 2009 #4


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    Science Advisor
    Gold Member

    Just to clarify, the reduction occurs because the molecules/atoms of a dielectric will polarize within the material. This polarization will create local electric fields that will counteract the applied field. Over the bulk of the material, this results in a reduction of the overall applied field. However, it does not result in a reduction in the stored energy because the energy lost in the applied field is now stored in the polarizations in the dielectric.

    The production of heat would require continual agitation of the molecules. A capacitor, under DC currents, would orient and polarize the dielectric's molecules once (and in an ideal capacitor we do not consider it consuming power under AC or DC currents either) and thus not allow for a continual production of heat. However, there is the chance of dielectric breakdown if you have a large enough applied field. A dielectric experiences polarization of its molecules and under a large enough electric field this will result in the electrons being stripped from the molecule/atom, making the material locally conductive. Once a path of conductive material is created in the dielectric, a current can arise which will discharge the plates. Once the plates are discharged, there is no longer an applied electric field (or a large one) and the dielectric can sometimes resume it's normal properties.

    A classic example is lightning. The electric field between a cloud and the ground becomes large enough to make the air conductive, turning it into a plasma, along the path of the discharge. This can often permanently damage the dielectric in a capacitor (even in air the oxygen often turns into ozone, but ozone is fairly unstable though).
  6. May 20, 2009 #5
    Lightning is now thought to follow ionization tracks of cosmic rays in its tortuous path to (from?) ground. Ions from cosmic rays provide a low resistance path, just like ions in the old spark chambers once used in high energy physics. See
    and others on the web.
  7. May 20, 2009 #6


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    Staff: Mentor

    The magic phrase is "dielectric breakdown." Exactly what happens depends on the nature of the dielectric. It might be ionization (e.g. in a gas), or rupturing of molecular bonds. It's why capacitors usually have a maximum voltage value stamped on them, along with the capacitance.
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