Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitors in Series

  1. Jun 22, 2006 #1
    Can anyone help me out with this?

    I like your word picture of the capacitor circuit. I've added GND and Vsupply to complete the circuit. I've apppended numbers for each plate and wire. I will be referring to each plate/wire/air numbers throughout the message.

    GND - wire1 - plate1 - air1 - plate2 - wire2 - plate3 - air2 - plate4 -wire3 - Vsupply

    where : plate1 and plate2 make capacitor1
    and plate3 and plate4 make capacitor2

    Here is what I understand (Stop me if I'm incorrect):
    1). When Vsupply is initially applied to this circuit. Electrons will build up on plate1 which in turn repels electrons away from plate 2 thus leaving positive charge built up on plate2. The electrons leaving plate2 causes electron to build up on plate3 which in turn repels electrons away from plate4(leaving behind positive charge) and the process is done and the circuit settles into equilibrium when Vcapacitor1+Vcapacitor2 = Vsupply. There is no current flowing thru the capacitors per se, but just displacement currents when Vsupply is first applied. No currents or displacement currents flow thru the circuit after the circuit settles into equilibrium.

    2). Quote from my physics book. "For a series combinations of capacitors, the magnitude of the charge must be the same on all the plates when in equilibrium." That is Q-(plate1) = Q+(plate2) = Q-(plate3) = Q+(plate4). By the way, you can look this up in any physics book that explains two capacitors in series.

    Here are my questions again:
    It's obvious that I'm interested in the region that consists of plate2 - wire2-plate 3

    1). When VSupply is increased then the voltage across each capacitor will increase which in turn will increase all the charge on each capacitor according to the Q = CV equation where Q = Q-(plate1) = Q+(plate2) = Q-(plate3) = Q+(plate4). This must mean that more additional electrons must leave plate2 and migrate over to plate3? My question is: How can there be more electrons to move from plate 2 onto plate3 with the addition of more voltage. Why doesn't "all" the electrons move over to plate3 with any small amount of Vsupply. Since the plate2 - wire2-plate 3region is a conductor (very low resistance). I would think that all the electrons would flow to plate 3 with any small amount of Vsupply voltage and then not change with any additional Vsupply voltage because "all" the electrons would have already migrated over to plate3.

    1a) According to Gauss's Law, there is no "net" charge stored inside a conductor when in equilibrium and there is no electric field inside a conductor when in equilibrium. Also, all "net" charge must reside on the surface of a conductor when in equilibrium. So in the case of plate2 - wire2-plate3 region (which is a conductor), plate3 must borrow more additional electrons when Vsupply is increased. Is it borrowing the additional electrons from the surface of plate2 or from within the wire2? In which case if it is borrowing more electrons from wire2 then the electrons would not be considered "net" charge, but some other type of charge. Is there a difference? You stated earlier that the "net" charge within the plate2 - wire2-plate3 region is zero.

    2). You stated that the net charge is zero in the plate2 - wire2-plate3 region. However, there is negative charge stored on plate3 and positive charge stored on plate2. I am wondering why this does not create potential difference form plate2 to plate 3? I know that it does not because its a conductive wire in the circuit that has very low resistance. However, I was under that assumption that charge imbalance would create an Efield which would create a voltage drop. Obviously I am wrong, but can you explain?

    3). Last but not least. This is a doozy!. What is voltage? I know that voltage is a scalar quantity the equals the amount of work done to move a positive point charge from A to B within an electric field. It the line integral of the efield across some distance ds. This is the physics book definition, but what does it mean from a charge perspective. Say you have a node A in a circuit that is a very long distance away from node B. The nodes are far away from each other as to not be effected by any Efield produced by the charges that resides in the nodes. Say NodeA is 5V and NodeB is 3 Volts. What is the difference between the nodes from a charge standpoint? Is there more electrons in NodeB than NodeA? Or is there a greater charge density in NodeB than NodeA?
  2. jcsd
  3. Jun 23, 2006 #2


    User Avatar

    Staff: Mentor

    -1- Because the voltage only generates enough force to move some of the electrons. Bigger voltage means more force, so more electrons move. The electrons are coming from the bulk metal and being pulled to the surface of the - plate.

    -2- The metal is an equipotential surface. The E fields are between the plates where there is a separation of charge.

    -3- Voltage is a potential. When you do work on a charge against an electric field, the charge is gaining potential energy, and hence moving to a higher voltage. When you let the charge go and it moves in response to the E field, the charge loses potential energy, and goes to a lower voltage. For your NodeA/B question, I need to think about that some. Not enough time to come up with a good answer at the moment....
  4. Jun 23, 2006 #3


    User Avatar

    Staff: Mentor

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook