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Capacitance, Charges, Voltages

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider a charged parallel-plate capacitor. How can its capacitance be halved?
    Check all that apply.

    Double the charge.
    Double the plate area.
    *Double the plate separation.
    *Halve the charge.
    *Halve the plate area.
    Halve the plate separation

    2. Relevant equations

    C = Q/V


    3. The attempt at a solution

    I know for sure that the capacitance can be reduced if the area decreases and separation increases. The part I am unsure is if the capacitance is decreased if the charge is decreased, although I am leaning more towards it does, since there will be less charge spread around the area of the plate. Thanks in advance. :smile:
     
  2. jcsd
  3. Feb 14, 2008 #2
    The relevant equation you posted answers this
     
  4. Feb 14, 2008 #3
    So capacitance is directly proportional to charge.
     
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