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Capacitor Problem!

  1. Nov 12, 2011 #1
    I came across a question in my Physics Paper
    "What is the work done by a battery in charging a capacitor" (Assume circuit has no resistance).

    This is what i thought...
    The Energy in the field between the capacitor plates is 0.5CV^2 where C=capacitance and
    V=Battery EMF.So the ans should be 0.5CV^2.

    BUT the actual solution in paper was...
    Net Work done=Net Charge * Voltage
    W=QV
    W=CV^2 .

    What is the correct ans?
     
  2. jcsd
  3. Nov 12, 2011 #2

    rock.freak667

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    Homework Helper

    The energy supplied by the battery is E = QV and Q=CV to get E=CV2

    the energy stored by the capacitor is E=½ CV2 (if I remember correctly, the other ½ is lost as heat).
     
  4. Nov 12, 2011 #3
  5. Nov 16, 2011 #4
    It cant be lost as heat...without resistance!
     
  6. Nov 16, 2011 #5
    The energy supplied by the battery is Q x V. The energy stored on the Capacitor is 0.5Q x V.
    Energy can be lost by
    1) Resistance in the connecting wires
    2) Sparking when the switch is closed
    3) Electro-magnetic radiation from the connecting wires as the charging current flows. The flowing charge is a changing current and a changing current produces electro-magnetic (radio) radiation
    In the absence of 1) and 2) there is always 3)
     
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