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Charing a capacitor hard time visualizing what's happening.

  1. Apr 5, 2010 #1
    Okay here goes, I've read different things online about charging a capacitor so I just want to clear this up once and for all.

    When you are charging a capacitor, I always understood it as the electrons shifting from the wire connected to the negative end of the battery onto the plate, and the electrons on the positive end of the plate shifting away, to ultimately give a positive and negative charge on the plates (which equals the voltage of the battery).

    However, I've read several sources where it says that electrons actually physically move from the positive plate to the negative plate to create this electric potential.

    Could it be that both are happening? Or just one? And if so, which one? Thanks in advance.
  2. jcsd
  3. Apr 5, 2010 #2
    Ultimately it doesn't matter. The end result is the same.
    Those electrons don't have little id badges, so how would anyone know if any of those that left the one plate had turned up on the other? :)
    What's important is that a charged capacitor has managed to lose a certain amount of charge (electrons) from one plate, and gain the same amount on the other.
    Exactly how it does this will depend on the circuit. You can charge a capacitor, for example, by connecting it to another capacitor which is already charged.
    In this case it's common to imagine the electrons flowing from one capacitor to the other.
  4. Apr 5, 2010 #3
    Thanks Stonebridge.

    I do understand that the end result is the same, however I just wanted to be able to visualize it conceptually since I've read different things online and just wanted to see which one is right. Not that it's SUPER important, but it just bothers me if I can't understand it.

    From what I've been reading in my book, there are two different kind of situations. A steady-state situation (in which no current is present in any branch of the circuit containing a capacitor) and a non steady-state situation (in which charges are already moving and a current exists in the wires connected to the capacitor.

    It seems as if the steady-state situations, have just the electrons moving from the wire to the plate, and from the plate to the wire, while in non steady-state situations, charge is actually transferred between the plates.

    Don't really know in what situations you would have a steady-state situation and a non-steady state situation, but maybe someone can clarify? Book is so unclear.
  5. Apr 5, 2010 #4
    "However, I've read several sources where it says that electrons actually physically move from the positive plate to the negative plate to create this electric potential."

    This is plain wrong. Electrons move out of the positive plate toward the positive termanal of the battery. The negative charges move out of the negative terminal of the battery to the wire. The charges in the wire move into the opposite plate.

    Perhaps this is what was implied and you didn't read carefully enough. Do you think that was it? It happens to me all the time.
  6. Apr 5, 2010 #5
    You are right to question this and it's always a good idea to try to visualise something if it helps you to understand what's going on.
    Maybe what causes confusion here is the difference between the "traditional" idea of "charge" moving or being transferred; and the idea of electrons as the actual mechanism for what is happening.
    In the traditional study of electrostatics, we always talk about "charge" on the plates, and "charge" moving from place to place; without really worrying too much about what the charge actually is.
    Traditionally, charging a capacitor involved transferring an amount of charge from one plate to the other. The end result was that one plate had a charge of +Q and the other plate a charge of -Q. We say the capacitor now has a charge Q.
    If this was done by connecting to a battery, for example, it was traditional (originally) to imagine a charge +Q flowing from the positive terminal of the battery onto the + plate and a charge +Q flowing away from the other plate into the negative terminal.
    Later on it was realised that it's negative electrons that are the mechanism, and the picture changed to electrons flowing from the -ive terminal onto the negative plate and flowing off the other (+ive) plate into the cell. The end result is of course the same. Some charge has been "transferred" from one plate to another. This results in a charged capacitor.
    If you look more closely and think of the movement of electrons in the circuit, the first thing to note is that there are billions of electrons on the plates of the capacitor and in the wires connected to it. The actual number of extra electrons that move onto a plate to charge it is relatively small. [Work it out from Q=CV and use a couple of volts and a typical capacitance of a picofarad; charge on one electron = 1.6 x 10^-19C]
    When you connect up the charging circuit, the current flows for a very short time. The actual electrons in the conducting wires have a drift velocity of around about a millimeter per second. The number of free electrons in the connecting wires is of the order of 10^29 per cubic meter.
    So what actually happens is that a small number of electrons in the connecting wire, near to the capacitor, get pushed onto the plate by the applied electric field. They don't move very far at all.Does this help you to picture what's going on. Have you studied drift velocity of electrons in a conductor, and about the number of "free" electrons in a typical wire?
    Keep on questioning if you want more answers. There are plenty on here willing to help.
    Last edited: Apr 5, 2010
  7. Apr 5, 2010 #6


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    Short version, electrons are shifting as opposed to being tranported over the entire path. The ones that left the positive plate into the wire are not the same ones that moved from the other wire onto the negative plate.
  8. Apr 5, 2010 #7
    @Phrak Your answer is what I thought happened until I read this in my book. Maybe I read this wrong, but from what it seems like, it sounds like electrons are actually moving across the capacitor.

    "When charging a capacitor, to increase the charge on the capacitor, work is required to remove extra electrons from the positive plate and move them to the negative plate. This work against the electric field is stored as electric potential energy." (They show a diagram of an electron jumping from the positive plate to the negative plate).


    Thanks for your answer again. I think I am getting a better understanding of it. So basically, electrons that are in the wire already are induced to move to the negative plate because of the applied electric field, while electrons on the positive plate are induced to move away. Is this figure at the bottom of the webpage what you're trying to explain?


    @jeff reid

    Thanks for your answer! Just trying to see if there are any scenarios where an electron would actually move from the positive plate to the negative one, instead of just shifting.
    Last edited by a moderator: Apr 25, 2017
  9. Apr 5, 2010 #8
    The figure is exactly what I was trying to say. It just about sums it up.
    Last edited by a moderator: Apr 25, 2017
  10. Apr 5, 2010 #9
    Alright that's what I thought Stonebridge. Thanks!

    I've attached a picture of what my book said, which really confused me.

    Is this example something completely different? It's talking charging a capacitor which is why I got confused.

    Attached Files:

  11. Apr 5, 2010 #10


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    It's the same thing .. I guess the arrows on the left side of the diagram are intended to represent the shifting of the electrons through the wires connecting the power supply to the system.

    One interesting point about capacitors is that the charging process is rather inefficient .. you actually lose half of the energy put into the system. In order to create a charge difference of Q between the two plates, you need to put in energy equal to QxV, where V is the potential difference between the plates. On the other hand, the formula for the energy stored by a parallel plate capacitor is 1/2(QxV), or half of that amount.
  12. Apr 5, 2010 #11
    Yes, that is confusing! The diagram shows you the net effect of charging a capacitor, but not the mechanism which includes wires and battery.
  13. Apr 5, 2010 #12
    Yes, I think the others have answered for me.
    You just need to be aware that when talking about charging up a capacitor, you will often hear it described as "transferring charge from one plate to another".
    Just remember what actually happens to electrons in the charging circuit.

    Imagine I had two boxes A and B each with 100 identical billiard balls in them.
    The next moment, there's 99 in A and 101 in B.
    It doesn't really matter whether I took one out of A and put it in B; or took one out of A and put it in my pocket, then from my other pocket, found another ball and put it in B.
    The end result is the same.
  14. Apr 6, 2010 #13
    @spectracat Thanks for clarifying! That diagram was what threw me off and why I thought an electron actually physically moved from the positive plate to the negative one.

    @phrak Yes it is confusing! haha, but yes I will just have to look at it as the net effect of charging a capacitor and not the actual mechanism.

    @stonebridge Thanks again for all the help! much appreciated. now I can finally sleep at night.. hehe
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