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How does charge pass through a capacitor?

  1. Jan 5, 2014 #1
    Hello. Firstly, I'm new to these forums, so if this topic belongs elsewhere, I apologize.

    I've been struggling with certain concepts pertaining to electrostatics, and would like some feedback regarding as to whether or not I have the right idea in understanding how charge passes through a capacitor.

    For example, let's say we have a simple electrical circuit (i.e. our circuit only has a battery source and a parallel plate capacitor). When we turn on our battery, we create current (I) that flows from the + battery source to the capacitor plate (C) through a copper wire; because our copper wire is composed of atoms, when we turn on the battery, creating an electric field (E) through the wire, the electrons within the wire will shift towards the + battery source. Thus, due to a deficiency in electrons in the wire near (C), our capacitor plate nearest to the + battery will gain a + charge (q+). The capacitor plate will continue to gain q+ until the positive charges build up on the plate, and eventually are repulsed by each other. When this happens, (I) will stop flowing and (C) will be fully charged. The same thing happens with the - battery source as well, though opposite (i.e. the electrons will flow to the capacitor plate, instead of away from it.)

    (At this point, if you see anything that's wrong please offer some input.)

    Now, onto my main question: How does charge pass through a capacitor?

    Let's say between our capacitor plates, we have air, which is a dielectric. When our capacitor plate is fully charged, the atoms within the dielectric will polarize to align their appropriate charge with that of the proper capacitor plate; the electrons in the atoms of the dielectric will shift towards the + plate, and vice versa. I understand that it is because of this effect that our capacitor can now contain even more charge on it from the relationship C = q/Voltage. However, my question arises as to how the charges propagate from one plate to the other when the capacitor discharges. Does the charge simply pass through the dielectric material onto the opposite plate?

    Any help will be appreciated. Thank you.
     
  2. jcsd
  3. Jan 5, 2014 #2

    Pythagorean

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    Charges don't cross the capacitor, the redistribute themselves around the rest of the circuit which gives the final steady state "as if" charges had crossed right over (but now with a transient based on the time constant of the capacitive system).
     
  4. Jan 5, 2014 #3

    sophiecentaur

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    Until it is fully charged, you get an excess of positive charges on one plate and an excess of negative charges on the other plate. Within an 'instant snapshot' during the process, there is no distinction between that and a current flowing.
     
  5. Jan 6, 2014 #4

    sophiecentaur

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    I mean to imply a progressive change / build-up of charge imbalance, here.
     
  6. Jan 9, 2014 #5
    Once a capacitor is charged the power supply can be disconnected. The capacitor can then be discharged through a load. In the case of air dielelectric the electric field dissipates and the voltage gradient tends to zero.

    A better way is the electrolytic capacitors. They are smaller, have small electric resistance and can stay charged longer. The conductance is ionic in the electrolyte.
     
  7. Jan 9, 2014 #6

    sophiecentaur

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    That is not the only purposes of having Capacitors in a circuit. Mostly, they do not operate in a situation that they 'hold' a voltage. They are normally used in conjunction with a Resistor or Inductor to introduce some time dependent function in a circuit.

    The only respect in which an electrolytic capacitor is 'better' is that they can be made with higher capacitance then other forms of construction. Otherwise they tend to be leaky, short lived and require a bias voltage to maintain their value. The voltage limit is often a problem, too. But, when they are useful, they are very useful.
     
  8. Jan 9, 2014 #7
    I know that. I was focusing on large power capacitors used for percussion welding. In this case the capacitor is charged by a DC power supply and a resistor, and then discharged as quickly as possible through a minimal impedance circuit. The spark created goes bang.
    These welders tend to be robust, durable and reliable. I noticed that a lot of people have phobias about capacitors.

    I do know that small capacitors are used in electronic circuits in the way you described.
     
  9. Jan 9, 2014 #8
    When I was learning capacitor theory (way back in the early 80's), one of my instructors gave a great analogy...

    It still helps me to think of the dielectric as a rubber membrane that deforms (stretches) as the cap charges up. Then the potential energy of the rubber provides the energy that discharges the cap.

    It's obviously just a simple analogy, but it helped me a lot. I hope this might be useful to anyone else (if it's not just repeating another post somewhere!)
     
  10. Jan 10, 2014 #9

    Andrew Mason

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    The charges go the other way around: through the conductor connecting the plates. The dielectric material cannot conduct charge - if it breaks down and does conduct, the device is no longer a capacitor.

    AM
     
    Last edited: Jan 10, 2014
  11. Jan 10, 2014 #10

    sophiecentaur

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    I usually hate the water / electricity analogy but that is a good explanation of the way a capacitor allows charge without a direct flow.
     
  12. Jan 12, 2014 #11
    I think of it as the voltage from the battery pushing all of the electrons to one plate and the buildup of negative charge here scares the electrons away on the other side making it positive. The charge on the plates is kept because the positive and negative sides, you could say, tantalize each other with their electric field through the dielectric but cannot pass through because the dielectric doesn't conduct. The greater the capacitance, the less strongly the electric field lines pass through the dielectric to push away the electrons on the other side, and attract to the positive side, and thus the more that have to collect before the capacitor builds up a voltage equal to the battery at which point it stops building up charge and if you disconnect the battery and attach something else it will let loose the charge that had built up like a battery.

    I had to figure that out on my own. Instructors were unclear.
     
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