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Can an open circuit voltage field charge a capacitor?

  1. Jun 13, 2008 #1
    What happens when you sandwich an uncharged capacitor between two charged capacitors?

    1. Just as a piece of Styrofoam can have a static electric charge, a capacitor's dielectric material can be charged, without the requirement of capacitor plates.

    2. Given 3 equal pieces of dielectric material, with two of the pieces fully charged with static electricity. If the uncharged dielectric material is sandwiched between the two charged pieces, without creating a closed circuit. Will the uncharged dielectric material gain a static electric charge?

    3. Does the charged versus uncharged state of the center piece of dielectric material depend on the polarity of the charge of its neighboring charged pieces of dielectric material?

    4. If the center piece of dielectric material indeed becomes charged, will shorting this capacitor discharge all three pieces of dielectric material?

    5. Does the charging of the center piece of dielectric material depend on its distance from neighboring charged capacitors? In other words, does charging the center capacitor within an open circuit require exposure of the electrostatic field of its neighboring charged dielectric material.

    I claim that w/o a closed circuit, the center piece of dielectric material will not become charged. Can you prove me wrong?
  2. jcsd
  3. Jun 13, 2008 #2


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    The dielectric material is strictly insulative. It would not be charged so long as dielectric breakdown does not occur. However the presence of an electric field would polarise the dielectric medium.

    If they are all strictly insulative materials, then the centre item would not be charged. It would, however be "stuck" due to electrostatic attraction with the other 2 materials.

    I don't get this one. If it's uncharged it'll stay uncharged until some of the electrons from its neighbouring insulators slowly "leak" to/from the item. If it's charged, it wil stay as charged until again leakage occurs.

    Charges can't flow freely because they are all insulative materials. How do you propose to "short" them?

    Again, how do you charge a dielectric? The dielectric for a capacitor is just to ensure the charges on each plate "stay" where they are, thereby creating a potential difference via electrostatic attraction. And in a capacitor, it is the conductive plates which are charged, not the dielectric.
  4. Jun 13, 2008 #3


    Staff: Mentor

    Assuming the voltage applied is lower than the dielectric breakdown voltage then it won't become charged, but it will become polarized. Do you understand the difference between "charged" and "polarized" for non-conductive materials?
    Last edited: Jun 13, 2008
  5. Jun 13, 2008 #4
    Perhaps the following URL provides an overview of methods to charge insulative materials.

    Perhaps I should provide some detail on the necessity of plates in the Wimshurst generator example. The following is Wimshurst’s original publication:
    The following is a quote from the Wimshurst’s publication:

    “Another feature is that the neutralizing current may also be broken without reducing the excitement, but then the charges alternate from positive to negative with each half revolution of the disks.”

    That means that without neutralizing bars, sparks occur each half revolution of the disks, regardless of the number of plates on each disk.

    The following URL mentions some causes of triboelectric effects.
    Last edited: Jun 13, 2008
  6. Jun 13, 2008 #5


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    I don't see how that is relevant. Your original setup consisted of a dielectric material sandwich in between a capacitor. There was no mention of friction. I don't doubt that you can charge an insulator with static electricity; examples of this phenomenon are common in everyday lives as well as introductory physics courses. I answered your questions based on the setup of a dielectric material in a capacitor or sandwiched between 2 charged insulating materials.

    Secondly you did not mention in your description of the setup that a conductor would be brought in contact briefly with the dielectric material sandwiched in between the 2 charged insulators. This step is necessary in order for the sandwiched dielectric to gain a net charge by electrostatic induction. I interpret the term "charged" to mean a "net charge", not mere polarisation itself, which is why I replied as above. I did mention that the dielectric material would become polarised in the capacitor setup, but not that it would be charged.

    As DaleSpam said, you have to distinguish between "polarised" and "charged" for insulative materials.
  7. Jun 13, 2008 #6
    There is a difference between charged and polarized. Charged material contains voltage's energy. Polarized material contains polar molecules.

    Some texts claim that dielectric material becomes polarized when exposed to an electromagnetic field.

    Consider that when a vacuum capacitor becomes charged, the vacuum contains a voltage gradient, but the vacuum is not polarized. We know that the vacuum within vacuum tube diodes and triodes contained a voltage gradient, because the voltage gradient accelerated thermionic electrons. We know that the tube's vacuum was not polarized because the vacuum itself contained no atoms or molecules that could be polarized.

    Anything (including conductors, resistors and capacitors) that contains a voltage gradient contains voltage’s energy. If polarization occurs w/o a voltage gradient, then that polarized material is w/o a voltage’s energy. The dielectric material within a charged capacitor indeed contains a voltage gradient.
    Last edited: Jun 13, 2008
  8. Jun 13, 2008 #7


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    Where did I say a vacuum can be polarised? I was referring to dielectric materials,not vacuum. Furthermore, where did I deny that the insulating dielectric cannot contain a potential gradient? The textbooks you cite are referring to dielectric insulative materials, and clearly not vacuum in particular. It is apparent that in the case of a vacuum, there is nothing to be polarised, so what point are you trying to make here?

    I still don't see how this connects with your original questions.
  9. Jun 13, 2008 #8
    Perhaps Defennder should re-read his 1st response:
    In my responses to Defennder’s 1st response, I (tried to) put forth the idea that:
    1. Dielectric material can be charged w/o the requirement of dielectric breakdown, albeit one of the charging methods is the triboelectric effect.

    2. The charge within dielectric material is not because of polarization of dielectric material, because a vacuum is dielectric material. A vacuum is dielectric material, because a vacuum is a poor conductor of electricity and a vacuum can store voltage’s energy.

    Perhaps dielectric material simply contains voltage's energy, instead being a converter of voltage's energy to polarization energy.
  10. Jun 13, 2008 #9


    Staff: Mentor

    Almost, a charged material has a different number of positive and negative charges, there is a net charge. A polarized material has an uneven distribution of charges, but no net charge, e.g. a surface layer of polar molecules with all of the + ends pointed out and the - ends pointed in.
  11. Jun 13, 2008 #10
    Thank you DaleSpam for identifying which characteristics differ in a charged versus polar molecule. I really did not know that charged material has a different number of positive versus negative charges. I am having some trouble verifying your net charge hypothesis. Perhaps you could either provide some more detail, or provide references that support your hypothesis.

    Your incite into the nature of polarized material is fantastic. Many people think that most polarized molecules are bipolar, thus result from displacement of a single charge. Do I understand you correctly; the surface layer of a polar molecule has multiple + ends pointed out and multiple – ends pointed in? I did even not know that the + ends of a polar molecule point out, or that the – ends point in.
  12. Jun 13, 2008 #11


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    Perhaps pzlded should re-read his original post:

    My response which you convenient quoted completely out of context was written with the bolded part of your original post in mind. You provided the setup, neglected to mention friction or any way in which it could be used to charge the insulator, nor provided means by which charging by electrostatic induction can be used and then you baselessly accused me of claiming that insulators cannot be charged at all.

    This is trivial. Nowhere have I denied that insulators may be charged. Again I reiterate that your setup says nothing about friction or partially discharging the polarised material, which can be done easily if it is conductor, but not if they are dielectrics. Quoting out of context is disingenous. I have already said this was very common and self-evident.

    Dear pzlded, can you distinguish the difference between "polarised" and "charged" ? Please look at DaleSpam's last post more carefully for the difference. There is no "charge" within a dielectric material in a capacitor setup. If you want to charge it, it is by means of as you said tribo-electric effect and electrostatic induction. As for your point of vacuum being dielectric material, you are splitting semantical hairs. You want to fault textbooks for leaving out the trival possibility that vacuum, which is literally empty space cannot be polarised, when it is readily apparent.

    And it turns out, if I'm not mistaken, that there is actually something known "vacuum polarization" in quantum physics, thought I don't think that is relevant here.
  13. Jun 14, 2008 #12


    Staff: Mentor

    It isn't really a hypothesis, it is just the definition of polarized and charged. For reference, what I am describing is straight out of my 2nd semester freshman physics text book: Serway, "Physics for Scientists and Engineers" 3rd edition. Look at Chapter 26 with a special emphasis on section 26.7 and the figures on page 727.
  14. Jun 14, 2008 #13
    Hi DaleSpam,

    The force between the charges is said to be the source of voltage’s energy. Near the + plate, there is a net excess of + charges; near the – plate, there is an excess of – charges. Kirchoff’s current law specifies that “current in equals current out”, therefore + charges in equal + charges out. In other words, if the + plate gets more charge than the – plate, the charges got there in violation of Kirchoff’s law.

    In the diagram of a polarized molecule, the little +’s and – ‘s represent the field from the + charge at one end of the molecule and the – charge at the other end.
  15. Jun 14, 2008 #14


    Staff: Mentor

    That is OK, Kirchoff's law is not a fundamental law and gets broken all the time. For example, Kirchoff's law doesn't work in antennas either.

    You use Kirchoff's law for analyzing small circuits and you usually only apply it at nodes of your circuit diagram.
    Last edited: Jun 14, 2008
  16. Jun 15, 2008 #15
    Hi Defennder,

    Perhaps I did misinterpret your response. Let us start over. I would like you (or others) to provide whatever assistance you can in clarifying a problem. I would like to know more about the nature of the field beyond a capacitor’s plates.

    A vacuum separates two charged insulators. The charged insulators attract each other.
    Does the portion of electrostatic field within the vacuum contain a voltage gradient?

    If you claimed that electrostatic field between the insulators has a voltage gradient:
    Why am I not been able to detect the voltage gradient with a high impedance voltmeter?

    If you claimed that, the electrostatic field within the vacuum has no voltage gradient:
    How do point charges produce both a field within an insulator with a voltage gradient, and the attractive field within a vacuum that does not have a voltage gradient?
  17. Jun 15, 2008 #16


    Staff: Mentor

    Are you sure about that? I would think you would be able to detect it. Perhaps not with a standard voltmeter, but certainly with a device designed for the purpose.
  18. Jun 15, 2008 #17


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    You probably have to use a sensitive electrometer to detect it. I'm not familiar with experimental devices, so I won't be able to help you here, sorry.
  19. Jun 16, 2008 #18
    Placing a conductor between two charges does not change the electrostatic force between the charges. Adding capacitor plates to each of the two isolated charged insulators can reduce the voltage gradient between the two insulators to zero, w/o changing the force between the two charged insulators.

    Point charges do not produce a force w/o a voltage gradient. Is there a theoretical explanation of how a conductor can reduce a voltage gradient w/o reducing force between two charges?
  20. Jun 17, 2008 #19


    Staff: Mentor

    Your first sentence is correct, but not the second sentence. The definition of the voltage field is that it is the gradient of the voltage is the e-field, and the definition of the e-field is that it is the force on a test charge. So if the voltage gradient is reduced at the insulator then the force is necessarily reduced also, by definition.

    Note that a conductor in the middle will have a zero e-field inside the conductor, but will not change the e-field outside the conductor (for an infinite sheet charge and conductor at least, I am not sure about more complicated geometries).
  21. Jun 17, 2008 #20
    Hi DaleSpam,

    Thank you for your thoughtful opinion.

    The conductors in an electronic vacuum tube are tube elements that control (not short) the vacuum voltage gradients between elements. There is no current in the conductor inserted between to charged insulators within a vacuum, because there are no point charges in the vacuum. W/O current, the conductor does not convert vacuum volts to thermal energy or magnetism.

    Perhaps you could more precisely explain the interaction between the conductor and the volts between the two-isolated (open circuit) charged insulators.
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