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I General questions with currents

  1. May 19, 2017 #1

    Reviewing for the test and there are still some things I'm not sure about so I thought I'd ask here and hopefully come across someone that can help.

    I swear I knew this, but I have not been diligently reviewing as much as I should so I think I forgot and became confused.

    When a battery is connected to the circuit with a capacitor, what exactly is going on? From what I remember, the conductor connecting the negative side (for simplicity and less confusion refer to this as side one) of the battery with one side of the capacitor will gain electrons from the battery and accumulate electrons on that side of the capacitor. Because it has accumulated electrons on one side of the capacitor, and the other side (side two) of the capacitor being connected to the positive end of the battery, the other side of the capacitor will transfer it's electrons from the capacitor to the positive end of the battery as electrons are attracted towards positive charges and of the repulsive force of the excess electrons accumulated on the opposite capacitor. Is this correct?

    What I'm confused on is HOW the electrons on the negative charges accumulate on the plate that already has an equal amount of electrons and protons, in that wouldn't it the electrons coming onto side one of the capacitor have to experience a force to overcome the repulsive force the electrons already on the plate emitted? Or am I thinking of it backwards in that first the positive end of the battery would attract the electrons from the side of the capacitor it is connected to, which then would allow the accumulation and attracted of other electrons onto the other plate?

    I like to think I have a good conceptual grasp on everything else related to circuits, but I'm not quite convinced of HOW the charges are moving and where the force is coming from that is being applied to the charges.
  2. jcsd
  3. May 19, 2017 #2


    Staff: Mentor

    A small current flows as the capacitor charges. The current represents charges migrating from one plate to the other. The amount of current is proportional to the time rate of change of voltage on the capacitor. The capacitor voltage can not jump instantaneously to the battery voltage.

    Does that help?
  4. May 19, 2017 #3
    I understand that the current is charge migrating per time, usually in seconds, and that charges will continue to flow to the capacitor to store energy so to speak until the potential difference between the capacitor is equal to the battery due to the fact that once it is equal the battery can no longer supply enough force to overcome the repulsive force of the excess electrons on the capacitor.

    But what inside the battery is causing the force exactly? How does the electric field of the battery come into play with all this? Where is it? I think I'm confused in general with the battery and how it's applying the force on the charges in order for it to migrate.
  5. May 19, 2017 #4


    Staff: Mentor

  6. May 19, 2017 #5
    In regards to circuits then, should I just for now with current level of physics I'm at, know that a battery simply is what drives the current? And that's it? I'm in physics 2 electricity and magnetism, from what I can recall is that the battery has a potential difference inside and it drives the electrons to move through the current. The only relevant information I can think of from the top of my head is the potential difference or voltage of the battery and how that pertains to the circuit. This is what I was most confused about as I don't think there was enough in-depth discussion or dissection of the battery.

    Just as a side note if anyone else is reading this, being brief what are the most common mistakes made in the conceptualization of circuits? I don't expect long drawn out answer, but perhaps a brief statement such as what to look out for or what to further investigate or study.
  7. May 19, 2017 #6


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    Homework Helper
    Gold Member

    This much knowledge, along with some basic circuit laws like Ohm's law and Kirchhoff's laws, is sufficient as long as you are studying "circuit theory".
    If you want to know deeper than that, "Maxwell's equations" is what you should study.
    One of the common (at least during my school days) mistakes is that the "heating effect of electric current is because of loss of kinetic energy of electrons flowing through a resistor. "
  8. May 19, 2017 #7


    Staff: Mentor

    If you want to keep it simple, just think of an ideal voltage source and ignore what physics is behind it.

    We are already using ideal wires, ideal resistors, ideal inductor, ideal capacitors, and ideal transformers for circuit analysis, and ignoring Maxwell's equations, why not ideal voltage and ideal current sources too?
  9. May 19, 2017 #8


    Staff: Mentor

    I would recommend that approach. It is just a black box whose behavior is to supply a fixed voltage and will produce exactly that voltage at any current. How it does it is irrelevant. It could have some complicated internal computer controlled circuitry or it could just have a chemical reaction. The rest of the circuit only cares what voltage and current it produces, not how it does so.

    The actual "how" probably won't be covered in a physics or a circuits class. It is more of a chemistry class topic.
  10. May 19, 2017 #9
    Thanks everybody.

    One last time for review for what I simply need to know:

    The electrons on a capacitor are attracted to the positive end to the battery which creates a current, or flow of charge, through the circuit which causes the electrons to migrate from one side capacitor (capacitor/conductor one) to the other. The electrons being attracted to the positive end of the battery come out of the negative end of the battery which then travel through the conductor that is connected to the other capacitor where it can accumulate as those electrons and are able to stay there due to their attraction to the protons on the other capacitor(capacitor/conductor one) which is across from the capacitor it has accumulated on. The flow of charges stops once the potential difference between the capacitors is equal to the potential difference that the battery provides. If the battery is removed and there is simply a circuit of conducting wire from one capacitor plate to the other then the excess electrons will flow back to capacitor/conductor one until there is an even amount of charges on both capacitors and they retain their net charge of zero and where there is no electric field on the capacitors.

    I'm frustrated because I not only knew this before, but I understood in a much clearer way and now am having trouble with it for some reason. I think because it clicked for me so clearly the first time I never took the time to review it, and simply forgot it.

    Another quick thing since on the topic, the "conventional method" is using positive charges to show the flow of charges, as it was easier to show that charges move from high potential to low potential. Why is this? Because can't this also be demonstrated with a negative charge acting on a positive charge? The negative charge traveling to the positive charge is essentially traveling from high potential to low potential as well, or am I missing something?
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