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Understanding the capacitor deeply

  1. Jun 10, 2008 #1
    hello every body,
    i am trying to understand the capacitor behavior as deeply as i can ...
    consider a simple capacitor with two plates A and B

    ---| |---

    1- why the charge on both plates are equal in magnitude ?
    when we apply a voltage to a capacitor,charge will accumulate on side A.this charge will build an E field which will pass thru the insulator and exert force on the plate B.So charges on plate B move and accumulate together.Ok but how do we know that this E field caused by Charges on plate A will exactly attract the same amount of charge that was on plate A on plate B??
    why these two has to be equal ? Conservation of Charge Law ?

    2 - why is that the higher the frequency , the more is the current ? is the following intuition wrong ?
    when we apply a ac voltage to plate A of a capacitor , charges density on plate A will change with voltage.
    when charge density changes , E field caused by them in the Capacitor changes too.
    This changing E filed will cause the plate B charge density changing .
    Ok but I am thinking that when the E filed is changing very fast(high freq.) the plate B charges may not find enough time to respond to it and move to the new configuration.so the charge density(voltage) on plate B may lag the charge density(voltage) of plate A. So why is that at high freq. the capacitor acts like a short curcuit or the impedance becomes lower?:confused:
     
  2. jcsd
  3. Jun 10, 2008 #2
    simple answer:afaik the current through a capacitor depends only on the capacitance and the rate of change of voltage. (i assume that this is because the total charge stored in the capacitor depends only on the voltage).

    more complex answer:i think the capacitance may depend on the frequency though. i dont know anything about that.
     
  4. Jun 10, 2008 #3
    capacitance depends only on geometrical aspects + the electrical field permitivity of the insulator between the plates.The insulator characteristics may generally change with frequency but that is not our discussion.
     
  5. Jun 10, 2008 #4
    1.When voltage is applied to the capacitor, positive and negative charges go onto different plates. Say positive charge on plate A and negative one on plate B.

    Allright, let's imagine you somehow inject electrons into plate A of the capacitor, these electrons will repel the electrons hidden within the other uncharged plate, right? And these other electrons will be forced out of the B plate of the capacitator. However, if the other capacitor plate isn't connected to anything, then those electrons cannot leave the other plate! They will continue to repel the electrons in plate A, thus very few electrons can be forced into plate A. This means that if charge is injected into capacitor's plate A, then an equal amount of charge is pushed out of the capacitor's plate B.

    Anyway, that's kind of the way I understand this, although a little childish it might as well be correct :)

    2. " Ok but I am thinking that when the E filed is changing very fast(high freq.) the plate B charges may not find enough time to respond to it and move to the new configuration.so the charge density(voltage) on plate B may lag the charge density(voltage) of plate A. So why is that at high freq. the capacitor acts like a short curcuit or the impedance becomes lower? "
    I agree with you. The higher the frequence the smaller is the resistance of the capasitor X=-j/wC (j - complex number) and the explanation you give seems reasonable, I can't really think of anything better.

    cheers
     
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