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Charging of capacitor

  1. Nov 30, 2015 #1
    I was going through charging of capacitor .I did not understand one thing .Plates of capacitors are uncharged initially but when battery is connected to it ,(positive terminal of the battery is connected to the Plate 1 and negative terminal of the battery is connected to plate 2)electrons from one of the plates (plate 1)start coming out from there to the positive terminal of the battery and as a result that plate (plate 1) of the capacitor becomes positively charged .This continues up to the time when potential of plate 1 becomes equal to the potential of the positive terminal of the battery .Similarly from the time we connect battery to the plate2 elcectrons from the negative terminal start going to the plate 2 in order to be as close as possible to plate 1 as it has positive charge which attract electrons.
    Electrons of plate 1 stop because the potential of plate 1 equals potential of the positive terminal .Why electrons of negaitive terminal stop?

    But the thing I don't understand is why electrons stop (going from negative terminal of the battery )at the same time when electrons from plate 1 stop going towards positive potential of the battery?Why ?
     
  2. jcsd
  3. Nov 30, 2015 #2

    phinds

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    Why would there be any further buildup of charge on either plate after equilibrium is reached? Do you think it could even mean anything to have "equilibrium" on only one of the two plates.
     
  4. Nov 30, 2015 #3
    Can you please answer it?
     
  5. Nov 30, 2015 #4

    phinds

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    I DID answer it. Equilibrium is reached.
     
  6. Nov 30, 2015 #5
    In terms of potential of terminals of battery.
     
  7. Nov 30, 2015 #6

    jbriggs444

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    As has been pointed out in your related thread, the individual potentials of the terminals of the battery are irrelevant. Only the potential difference between those terminals matters.
     
  8. Nov 30, 2015 #7

    gneill

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    Gracy, the charges on both plates are always equal in magnitude: charge being pushed onto one plate pushes the same amount of the same polarity charge off of the other. So if one unit of positive charge is pushed onto one plate, one unit of positive charge is pushed off of the other (leaving behind a net negative charge of the same magnitude). The result is an electric field between the plates and thus a potential difference between the plates. When you use the formula V = Q/C for the potential difference across a capacitor it is understood that there is a charge Q on one plate and -Q on the other.

    Charge will move onto the capacitor so long as the potential difference across the capacitor is less than the potential driving the charge (i.e. the battery's potential difference). When this potential difference is the same as that of the battery's potential difference, charge stops moving onto or off of the plates. Charge will only move when there is a potential difference to drive it and a conducting path for it to follow.

    When the potentials match, charges stop moving, the capacitor potential stops changing, and the current is zero. We call this an equilibrium condition, or "steady state" when all the activity is over.

    The thing to remember is that it is potential difference that drives currents. If you connect just one lead of a capacitor to a battery, no charge will flow. Also, a "charged" capacitor is still a net neutral object, it having equal and opposite charges on its plates.
     
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