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Capacitance/Capacitor Conceptual Problem

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Assume we have two neutral plates a small distance away from each other. We place charge Q on one plate, leaving the other plate uncharged. We then measure a potential difference V across the plates. Some bright student notes the ratio of Q and V is not the value of the capacitance. What is the value of the capacitance?

    2. Relevant equations

    C = Q/V

    3. The attempt at a solution

    So I understand that capacitance is a geometric property of the object and what it is made of. But for this question I think that since only one of the plates is charged, therefore the Q in the equation would really only be Q/2 thus the ratio would be C = Q/2V.
     
  2. jcsd
  3. Jan 22, 2014 #2

    ehild

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    Hi wompkins, welcome to PF!


    Q is the whole charge given to the capacitor, but you are right, only Q/2 charge appears on the inner surfaces of the plates. Q/2 on one plate and -Q/2 on the other. So what is the potential difference between them?

    ehild
     
  4. Jan 22, 2014 #3
    Ok so would the overall be a factor of 1/4.

    Another thing is that we gave the charge to just ONE plate, so wouldn't that charge -Q stay on that one plate and polarize the other plate to +Q on the inside of the plate?
     
  5. Jan 22, 2014 #4

    lightgrav

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    yes, so the E-field lines cross the gap but do not pierce the metal.
    Since both outer surfaces are covered with negatives, the potential of the pair is negative with respect to infinity.
    ... it is not obvious that this potential ought to be ignored.
    Parallel-plate capacitors usually hold zero net charge, so their average potential is zero w.r.t. the "V=0 at infinity".
     
  6. Jan 22, 2014 #5

    ehild

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    No, the Q charge on the first plate distributes evenly on both sides. So there is Q/2 charge on the inner surface and Q/2 outside. The electric field of the inner charge induces equal and opposite charge to the inner surface of the other plate, leaving Q/2 positive charge on the outer surface.

    From very far away, the capacitor looks like a single positively charged object, with potential about kQ/R with respect to infinity.

    Between the plates, the electric field corresponds to the surface charge density σ=Q/(2A): E=σ/ε0. From that you get the potential difference U between the plates.

    The students measure that potential difference U, and calculate the capacitance as C=Q/U - which is wrong.

    ehild
     
  7. Jan 24, 2014 #6

    utkarshakash

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    A general rule which will help you:

    If the two plates of a parallel plate capacitor are given charges Q1 and Q2, the inner plates acquire charges (Q1-Q2)/2 and the outer plates get (Q1+Q2)/2.
     
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