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Charge reconfiguration on wires connnected to capacitor

  1. Dec 1, 2014 #1
    Consider a capacitor which is charged to a certain voltage 'V' and having a total charge 'q' in it.The leads of the capacitor are connected to ideal conducting wire strips.Now if I introduce one conductor slab between plates of the capacitor, without touching the plates,as we know from basic electrostatics that the resultant capacitance increases.This means that ,for the same charge, potential difference between plates of the capacitor is reduced.This is easily explainable if we consider the the electric field 'E' between plates and integrate it between plate separation.

    However, please consider the following
    1) Since the wire strips are connected to plates, each wire strip will have the same potential as that of plate to which it is connected

    2)Potential difference between plates must be equal to the potential difference between wire strips

    3)Potential difference is independent of path.Hence if we integrate the electric field 'E' between the wire strips, by following a path without going through the capacitor, I should get the same potential difference.

    But ,when we introduce the conductor slab, to get a reduced potential difference(by integrating ' E' without going through capacitor) between wire strips, the surface charge configuration of wire strips must be altered.In this case surface charges should be reduced,I guess

    I am not getting, which mechanism does this re-configuration of charges.And also, where is the excess charge going to

    If my assumption that the surface charge on the wires are reduced is incorrect,how we will get a reduced potential difference when 'E' is integrated over a path(not through capacitor) between wire strips.
     
  2. jcsd
  3. Dec 1, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    Right. This is completely negligible for most realistic setups. The capacitor plates will get a tiny increase in surface charges and electric field strength.

    You can model this system with two capacitors in parallel, where you increase the capacitance of the significantly larger one.
     
  4. Dec 2, 2014 #3
    thank you for the reply
     
  5. Jan 18, 2015 #4

    Svein

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    Science Advisor

    What you do is equivalent to changing the circuit to having two capacitors in series, with the same resulting capacitance as the original capacitor. I cannot quite see what difference it should make to the rest of the circuit.
     
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