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A capacitor-like system, is this really a cap?

  1. Apr 20, 2007 #1
    1. The problem statement, all variables and given/known data
    A is an earthed metal plate while B is a positively charged metal plate parallel to A. The separation between the plates is much less than the length of each plate. If A is moved closer to B, what would happen to the electric potential of B and the electric field strength between the plates?

    2. Relevant equations
    I haven't pick the quick capacitor equations as i think there are differences and the equations:
    1.E= V/d, are not practical here as both V and d decreases.
    2.V = Q/C, i doubt if Q change

    At last, I use E= Q/(AK), where K is the permitivity

    3. The attempt at a solution
    1. As A which is negatively charged moves closer to B, it solely establish a more negative potential at B and the resultant of the potential decreases.

    2.On the other hand, the voltage of A should increase as it is closer to B, but as A is earthed, there are some negative charges flow into A to make it 0V, i.e. unchanged.
    Applying E = Q/(AK) we know the E field produced by A increases and that of B is constant as charge on B is constant.
    i.e. E resultant increases

    My teacher told me to treat it like a capacitor and as the capacitor is on a open circuit, there must not be changes on Q, i.e. (Q1-Q2)/2. As a result of V=Q/C decreases and E=Q/(AK) remains unchanged.

    Here comes my actual questions:
    1. Is there really no charge flow even when one plate is earthed?
    2. what is wrong with my approach?

    1.the model answer: V decreases, E constant.
    2. My teacher also told me to think in another way, the earthed A plate is just to make the potential reference constant. and we can see both plates isolated, in this case V can be seem as the potential difference of the plates.
    Last edited: Apr 20, 2007
  2. jcsd
  3. Apr 21, 2007 #2
    i really need help. Please say something about the approaches, is it valid?
  4. Apr 21, 2007 #3


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    1. The charged plate is isolated and so keeps constant Q. The charge on the grounded plate will exactly match that on the charged plate, as soon as the transient current stops.

    2. the subtle thing is that the capacitance increases as the plates are brought together, so the voltage falls - and the field stays constant ( as your teacher as said).
  5. Apr 21, 2007 #4
    Thank you.
    But i am still curious about what is wrong with my arguments that negative charge on A will increase. Can you explain to me?
  6. Apr 22, 2007 #5


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    Usually equilibrium states have the least potential energy and I expect that's what happens here.
    Because the capacitance increases and the voltage drops, there no need for the charge on A to change to maintain equilibrium.
  7. Apr 22, 2007 #6
    Won't B makes A's potential higher than 0v if there are no charge flowing into A?
    Even though it is not a complete circuit, is it possible for some charge flowing into A?
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