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Charging and Discharging of two capacitors

  1. Dec 18, 2014 #1
    Two capacitors with different capacitance are charged with two different voltages. When we connect both the capacitors with the similar terminals (that is positive terminal of one capacitor to the positive terminal of the other and vice-versa) the net charge is the sum of the former individual charges.

    Whereas when we connect both the capacitor (without any battery or cell in the circuit) with the opposite terminals (that is positive terminal of one capacitor connected to the negative of the other and vice-versa) the net charge is taken as the difference of the former individual charges of both the capacitors.

    Physically how does this happen?
     
  2. jcsd
  3. Dec 18, 2014 #2

    Doc Al

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    When the terminals are connected together, the net charge is always the sum of the original charges, but you must keep track of the signs. In one case, you are adding positive charges together, Q1+Q2; in the other case, you are adding positive charges to negative charges, so it looks like Q1 + -Q2 = Q1 - Q2.
     
  4. Dec 18, 2014 #3

    CWatters

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    Like charges are cumulative.
    Opposite charges cancel each other.
     
  5. Dec 19, 2014 #4
    Alright, so when we connect the capacitors the segment which connects each plates adds up or cancels out according to the connection?
    But how does this happen in terms of the electrons which are the basic constituents that contributes to the charge flow?
     
  6. Dec 19, 2014 #5
    The segments' charges and capacities are negligible compared to capacitors' charges and capacities.
    If there is a potential difference between segments prior to the series connection of the caps, small equalizing electrical transient takes care of it.
     
  7. Dec 19, 2014 #6

    Doc Al

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    That's right.

    The electrons are free to redistribute themselves until equilibrium is reached.
     
  8. Dec 19, 2014 #7

    CWatters

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    If you have an excess of 2000 electrons (eg a -ve charge) on one bit of metal and an excess of 1000 on another then when you connect them together you have an excess of 2000 + 1000 = 3000 electrons.

    If you have an excess of 2000 electrons on a bit of metal and a deficiency of 1000 (eg a +ve charge) on another then when connected you have of 2000 + (-1000) = 1000.
     
  9. Dec 19, 2014 #8
    I think that gives a sort of 'quantum explanation' for the problem.
    Thanks for your support everyone!
     
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