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Capacitors in series -- microscopic view

  1. Jan 1, 2016 #1

    cnh1995

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    I learned about surface charge feedback theory in electrical circuits a few months ago and it has been extremely helpful for me to intuitively understand many concepts in electrical circuit analysis including conservative and non-conservative fields. I initially referred the paper written by Chabay and Sherwood and later, their famous book "Matter and Interactions" along with some helpful videos on youtube. Here I've attached an image(screenshot) from the book's pdf. As far as surface charges are concerned, I don't understand how the field reaches the light bulb since there are two "breaks" in the circuit. How do the surface charges arrange themselves in order to get the field past the two dielectric barriers and reach the bulb? I can imagine the scenario in case of single capacitor since both of its plates are connected to the source terminals. What happens when there are multiple capacitors is series? Please help me understand this intuitively. Screenshot_2016-01-01-19-17-46.png
     
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  3. Jan 1, 2016 #2

    mfb

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    Fields are not limited to conductors. The light bulb has to be at some potential, and no matter what exactly this is, if you feed the circuit with AC current will flow through it.
     
  4. Jan 1, 2016 #3

    cnh1995

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    Thanks for your reply! I understand that there is a voltage across the bulb. But I want to know how exactly the field travels from source to the bulb "through" the dielectric by means of surface charges. The circuit seems to be discontinuous but it isn't as the source voltage reaches the bulb. What is the mechanism here? Would you please elaborate?
     
  5. Jan 1, 2016 #4
    When a emf pushes one additional electron to one plate of a capacitor, another electron is pushed off the other plate. So the first plate is now negative, the second, positive. If there is another capacitor in series, it happens again.
     
  6. Jan 1, 2016 #5

    davenn

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    the bulb ISNT going to light up and stay lit .... not with a DC supply

    depending of the value of the capacitances, ( as that will affect the size of the charge flow) the bulb may light for a "micro second" ie a VERY BRIEF time till
    the charge on the capacitor equalises then it will "go out"

    the bulb will ONLY stay light IF there is an AC supply
     
  7. Jan 1, 2016 #6

    cnh1995

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    Ok. But but even for brief time(say 2s), the bulb will be lit and there will be voltage across it. How will the field travel to the bulb through the dielectric,even for a short time? Suppose the voltage here is V dc. Initially, all the voltage i.e.V will be across the bulb. That means the voltage has reached the bulb "through" the dielectric. That's what I want to understand in terms of surface charges.
     
  8. Jan 1, 2016 #7

    davenn

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    :
    the capacitor dielectric doesn't stop the electric field. on the contrary, when a capacitor is charged, an electric field is developed within the dielectric between the capacitor plates. regardless of if the dielectric is a physical solid, say ceramic or if it is air or a vacuum. The dielectric will primarily just affect the overall capacitance value of the capacitor.

    alw34 hinted at what is happening ... lets look at that again ......

    lets look at the basics ...

    The electric field travels/propagates around the circuit at near the speed of light. And it is the EMF imparted by the electric field that causes the electron flow.
    This is why we don't have to wait for electrons from the battery to reach the bulb and cause it to light up. The electric field causes that to happen almost instantly ( in a small circuit aka short conductors).

    so ... when we connect the capacitor and bulb to the battery and complete the circuit, there is a movement of electrons from the negative battery terminal towards the capacitor ( in your diag.). For EVERY electron that moves onto the first plat of the cap, one moves off the opposite plate and towards the bulb.
    This results in the first plate becoming negatively charged and the other plate becoming positively charged. This continues till the voltage across the capacitor equals the battery voltage.
    NOTE: because of the one electron on = one off, the net charge on the capacitor is always zero.

    OK so for this very brief time, there are electrons being forced off the second plate and through the bulb and back to the positive terminal of the battery.
    Again, for every electron that leaves the second plate, one electron enters the positive battery terminal


    the voltage doesn't reach anything through anything .... ie. the voltage doesn't/isn't something that moves

    how's that ?


    Dave
     
  9. Jan 1, 2016 #8

    cnh1995

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    What you explained is indeed true and I understand it. Consider a resistor connected across a battery. Here also, initially voltage across the resistor is 0. Within nanoseconds, the surface charges build up across the resistor till the field everywhere on the circuit is same, to ensure the same current. Now, during this initial nanosecond period, the elctrons travel through the wire and sufrace charges are built up across the resistor. After that, the actual steady current starts which is same throughout the circuit as alw43 said. What you explained is this current that starts after the surface charge builds up across the bulb.
    Screenshot_2016-01-02-10-31-31.png
     
    Last edited: Jan 2, 2016
  10. Jan 1, 2016 #9

    ehild

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    upload_2016-1-2_5-57-26.png

    You have the arrangement above. The capacitor is charged, then the switch is closed. There is a potential difference across the capacitor, but the same potential difference is across the open switch. So there is some electric field between the ends of it, When the switch gets closed, that electric field effects the electrons in the piece of conductor and they start to move, and an electric field builds up in the whole wire, including the lamp. The potential drops from the + terminal of the capacitor to the negative one also along the wire, and the current flows through the wire, and the capacitor discharges, without any charge carrier crossing the barrier between the capacitor plates.
     
  11. Jan 2, 2016 #10

    cnh1995

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    Thanks for the reply..But here the capacitor is already charged. Discharging mechanism is easier to understand in terms of surface charges.In the original diagram, I want to know how surface charges carry the "information" to the bulb through the dielectrics while charging. The bulb is not conductively connected to the source and still there is a voltage across it (and there should be, I agree). But how do the surface charges carry the signal to the bulb? What happens when they encounter the dielectrics?
     
  12. Jan 2, 2016 #11

    ehild

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    I do not follow you. A piece of wire connected to a charged metal, becomes part of a metallic system, and equipotential. The ends of the wire at the switch are at the same potentials as the connected capacitor plates. If you close these ends, the surface electrons start to move, changing electron densities and potential along the wire. The non-equilibrium potential distribution leads to transport of charge all along the wire. The information propagates in the wires, and current flows through the wires instead of through the gap.
     
  13. Jan 2, 2016 #12

    cnh1995

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    Yes..I understand that. I think you are not getting my confusion. Consider the following diagram containing two series capacitors.

    screenshot_2016-01-01-19-17-46-png.93868.png
    As the title says, the bulb is isolated. Initially, the field will propagate from the source to the bottom plates. Now there is dielectric barrier on each capacitor. How does the field get past the barriers by means of surface charges and establish potential difference across the bulb? The current starts "after" the p.d. has been established and is same throughout the circuit as explained by you and others earlier. My question is how is this voltage developed across the bulb despite there are two breaks in the circuit? How do surface charges arrange themselves to send the field past the dielectric?
     
  14. Jan 2, 2016 #13

    davenn

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    OK, forget surface charges, you are dwelling on it needlessly .... WHY ?
    in this situation I think it is leading to confusion
    it is JUST a movement of charge WITHIN the cross-section of the conductor and I have already explained how those charges move. Yes, and according to Gauss's Law there will be a movement of charge away from the centre of the conductor
    Skin effect of charges/electrons moving on/near the surface of a conductor only becomes prevalent when dealing with an increasing AC system

    Because of the bulbs resistance, there is a brief moment when there is a voltage difference across it
    The EM field propagates around the circuit before the charge. It is the propagation of the EM field that generates a force on the electrons causing the movement of charge.
    The EM field DOESNT need the charge to move, other than the initial motion when the EM field is first generated, as said, it then becomes the other way around.
    The EM field propagates around the circuit in the space outside the conductors. The conductors just act as a waveguide

    Any deeper than that and some one else will have to explain ( I have my limitations :wink::wink: )


    Dave
     
    Last edited: Jan 2, 2016
  15. Jan 2, 2016 #14

    ehild

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    An isolated bulb would not light unless it is put into an electric field very-very strong. You can get such a strong electric field by static electricity. Rub a plastic ruler and hit the lamp with it: you may get a flash of light for a very short time, caused by the ions of the gas in the bulb accelerated by the field.

    Otherwise, I do not understand your set-up and question. Is the bulb really isolated and not connected to anything?
     
  16. Jan 2, 2016 #15

    davenn

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    no it's not ..... the "isolation" is DC isolation caused by the capacitor(s)
     
  17. Jan 2, 2016 #16

    cnh1995

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    I don't know much about the waveguide approach. But if I place a 9V battery on a table and a bulb adjacent to it without any wires connecting it to the source, will there be 9V p.d across the bulb through the EM field propagated through space, even for nanoseconds?
     
  18. Jan 2, 2016 #17

    davenn

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    no, cuz there is no flow of charge from the battery to the bulb ... no physical circuit
    BUT this is VERY different to a situation where you have an AC source of an EM field ... say a radio transmitter
    it's quite easy to light a bulb that isn't physically connected to the source ... IT IS lit by the propagating EM field
    I have done this in years gone by using a LED to detect the RF/EM field

    I have also done this in yrs gone by .....

    tubes_and_pylons_at_dusk.jpg

    The fluoro tubes are lit only by the radiated EM field from the HV power lines


    Dave
     
  19. Jan 2, 2016 #18

    cnh1995

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    That is very cool! Now, in my original diagram, if I put two open switches in the place of capacitors, would there be 9V across the bulb, even for nanoseconds? To make the charges flow in the circuit, there must be some voltage present. Voltage is first developed across the bulb and then the charges start flowing. How does the battery voltage reach the bulb? Through conductors or through the space? Where exactly am I screwing this up?
     
  20. Jan 2, 2016 #19

    davenn

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    no, I don't think so, as the contacts in the switch would not have a large enough capacitance to store a big enough charge to produce the same flow of charge like seen with a capacitor
    Keep in mind, the capacitor value would been to be significant to see a measurable effect big enough to momentarily light a bulb. I would expect at minimum, 100's or 1000's of uF

    Do some experimenting and see if you can make a bulb light up :smile:


    no, the flow of charge, as I said earlier, generates a small and brief voltage difference across the resistance of the bulb. If the bulb had zero resistance, then no voltage could be generated across it

    it doesn't except during that very brief time that the capacitor charges to the battery voltage.
    Once the cap reaches battery voltage across it, there will no longer be a charge flowing and you would not measure any voltage difference across the bulb

    D
     
  21. Jan 2, 2016 #20

    cnh1995

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    Yes, I am talking about that very brief time when the capacitors charge to the battery voltage. How is that voltage developed across the bulb when there are two dielectric breaks in the circuit? Is it the EM wave through the space around the conductors or is it through the conductors and the dielectrics?
    And congratulations for the award!:smile:
     
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