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Capacitor with 2 floating gates

  1. Oct 21, 2013 #1
    Hi,

    I have the following theoretic setup:

    A cap of size C Farad is connected between nodes A and B and loaded with a specific charge (e.g. +10 on side A, -10 on side B).

    In the next step, i disconnect side B and before doing that, "magically" (see footnote) a charge of -3 gets injected on plate B. Since node A is not (yet) floating, it will supply a charge of +3 and the cap is loaded with +/-13 charges now. Node B is floating now (not connected to anywhere).

    Now I so the same with with node A: -3 charges get injected and then the node is disconnected (floating).

    At node A, there are now +10 charges (+13 and -3) and on side B -13 chages. Since both nodes are floating, no node can supply charges to get the thing in equilibrium.

    First of all, at node A, will the three electrons (-3) "recombine" with 3 of the +10 positive charges or not?

    What is the voltage across the cap?

    I assume that on one side, excess charge will move as far away as they can (to a hypothetic other cap plate) because the cap requires same absolute charges on both sides.

    Thanks
    divB







    footnote: The setup are ideal NMOS switches which have a certain charge (-6 in the example above) in the channel. When the MOS switches infinitely fast from on to off, about -3 charges will go to one side and -3 charges will be injected at the cap plates.
     
  2. jcsd
  3. Oct 24, 2013 #2

    rude man

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    By running a Gaussian box between the two plates such that the box ends are inside both plates, that tells you that the surface charge density on the two facing sides must be equal. I think you concluded that already.

    There will therefore be extra negative charge on the outside surface of the negatively charged plate.

    The voltage is the integral of the E field over the plate separation distance. What is the formula for the E (or D) field near a charged conducting surface? (Since there is no free charge in the air space between the plates, so div E = 0, the E field is constant all the way from the A to the B plate.)
     
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