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Why is the voltage the same?

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data

    C1 = 6.00 uF
    C2 = 3.00 uF
    ##\Delta V = 20.0## V

    Capacitor c1 is first charged by the closing of s_1. Switch S_1 is then opened and the carved capacitor is connected to the uncharged capacitor b the closing of S_2. Calculate the initial charge acquired by C_1 and the final charge on each capacitor.


    2. Relevant equations

    C=Q/V

    3. The attempt at a solution

    See attached.

    ===============


    I get everything in this solution besides when they set Q_1/C_1 = Q_2/C_2.
    Is it ALWAYS true that the voltage across two points are the same?
     

    Attached Files:

  2. jcsd
  3. Mar 12, 2014 #2
    If you connected 2 capacitors, remember that their plates are connected by a conductor.

    Now what happens when you connect two points with a conductor?
     
  4. Mar 12, 2014 #3
    Current is formed, so there is a voltage?
    And since there is no loss of electrons (charge) and the intensity (current) is the same, the voltage is the same?
    Is that correct?
     
  5. Mar 12, 2014 #4
    Well, as long as there is current flowing, you can't really make a problem out of it, at least not here, because it's not in a steady state.

    But yes, when current does stop flowing in the circuit, the voltages are the same on both capacitors.

    Now, there are formulae for the final charge/voltage in such an arrangement. I'd suggest you go through your textbook or look through the forums. I've forgotten the formulae myself, but I do remember that they exist.
     
  6. Mar 12, 2014 #5

    NascentOxygen

    User Avatar

    Staff: Mentor

    You could include the resistance of the wire as a resistor in your crcuit, making it 3 elements. Current will flow through the resistor until the voltage across the resistor falls to zero. At the start, there are different voltages on each side.
     
  7. Mar 13, 2014 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

    When S2 is closed the two capacitors are in parallel so by definition their voltage drops must be the same.
     
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