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Homework Help: Plate capacitor

  1. Mar 20, 2010 #1
    1. The problem statement, all variables and given/known data
    charge a plate capacitor, disconnect the battery, increase the distance between the plates: capacitance decreases (C=keA/d, as d increases, C must decrease),
    V increases (q=CV, q is constant, C decreases, V must increase),
    electric field stays constant (E=V/d, both V and d increase proportionally),
    energy stored increases (energy= 1/2 C V*V, C decreases, but V increases and squares).


    2. Relevant equations
    see above.


    3. The attempt at a solution
    does it make sense that both the voltage and the energy increase as the distance between the plates increases? conceptually, I would think they would both decrease. can you help clarify? Thanks...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 20, 2010 #2

    kuruman

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    If the charge is fixed, the electric field (volts per meter) between the plates is constant and uniform regardless of the separation. So if you increase the distance, you would have fewer volts per meter unless the voltage increases.

    The plates attract each other. To separate them some more, you need to do work on them by pulling them farther apart against the electrostatic attraction. That work appears as additional potential energy stored in the capacitor. It's like stretching an already stretched spring even more.
     
  4. Mar 20, 2010 #3
    Thanks and I have some follow on question.

    for the voltage increase: I see your point and it make sense mathematically and logistically. It just runs a little against the grain to think the voltage would increase when the plates are pulled apart.
    for the energy stored increase: so you mean the energy increase is kinetic instead of electric?
     
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