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Change in energy of a capacitor

  1. Feb 12, 2014 #1
    A certain capacitor has a capacitance of 5.0 μF. After it is charged to 5.0 μC and isolated, the plates are brought closer together so its capacitance becomes 10μF. The work done by the agent is about:

    I assumed Qfinal = Qinitial since I believe there is still some separation, right (it's still a capacitor, isn't it?)? So, I did W = ΔU and let Uf = (1/2C)Q2 and I did Ui = (1/2C)Q2 and I increased the capacitance by 2 since area is doubled, right? But how does the distance of separation change?

    Any help please?
     
  2. jcsd
  3. Feb 12, 2014 #2

    gneill

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    Staff: Mentor

    MathewsMD, Please be sure to use the template provided when starting a new thread in the homework sections. Otherwise you might find yourself garnering infraction points...

    The area of the capacitor plates is not changing. The increase in capacitance is due to the decrease in separation between them (What's the formula for the capacitance of a parallel plate capacitor?).

    You're correct that the charge remains the same. So what's the change in energy stored on the capacitor between the initial configuration and final configuration?
     
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