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Electrical work done

  1. Jan 25, 2010 #1
    A capacitor C1 is connected to a supply. When a potential difference of 4.0 V is applied across the capacitor, it stores a charge of 0.80 nC

    a) Calculate the electrical work done by the supply as it transfers this charge

    The equaton used is W=QV

    Why is it QV and not 0.5QV, which is usually used?


    b) Mark on the diagram above the magnitudes and polarities of the charges stored on the plates of the capacitor

    ________
    l..............l
    =C.........4V
    l_______l

    This is what the diagram looks like. Ignore the dots. Its basically a emf on the right and a capacitor in series.

    The answer is +0.8 (nC) on top plate and –0.8 (nC) on bottom plate .

    How do we know which plate is +/-?

    c) With capacitor C1 charged to 4.0 V, the supply is removed and a second, uncharged capacitor C2 is connected in its place.

    Capacitor C1 transfers 0.2nC to the plates of capacitor C2. As a result the potential difference across C1 falls to 3.0 V

    Draw a graph of charge stored on C2 (y axis) against pd (x axis).

    The graph is a straight line from (0,0) to (3 , 0.2)

    However, shouldnt the voltage be 1, as there is 3 volts aross C1, and since the capictor are in series, C2 should be 1V to make 4V total?
     
    Last edited: Jan 25, 2010
  2. jcsd
  3. Jan 25, 2010 #2

    rl.bhat

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    The work done by the supply is not equal to the energy stored in the capacitor.
    b) The plate which is connected to the positive terminal of the power supply will have the positive charge.
     
  4. Jan 25, 2010 #3
    a) So for supply W=QV always?

    b) is a positive terminal the one where the side touching the wire has a line longer than the one next to it?
     
  5. Jan 25, 2010 #4

    rl.bhat

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  6. Jan 25, 2010 #5
    Thanks.

    What about for part (C)?
     
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