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Capacitor PE

  1. Jan 23, 2009 #1
    The following 3 step process refers to a simple RC circuit with a battery and an initially open switch:

    1- the switch is closed, allowing the capacitor to charge;
    2- after the capacitor has charged, a slab of dielectric material is inserted between the plates of the capacitor and time passes;
    3- the switch is opened, and the dielectric removed.

    PE(capacitor)=1/2 CV^2

    I said that: delta(PE)1>0; delta(PE)2>0; delta(PE)3<0

  2. jcsd
  3. Jan 23, 2009 #2


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    When you are putting the dielectric in the capacitor, you are effectively increasing the capacitance by the relationship that C/Co = ε the dielectric constant for the material, where ε > εo

    As to your answers, I don't really know what the question is.

    I would point out in step 3 that since the charge remains the same, but the capacitance is now influenced by εo instead of the higher ε of the removed dielectric, that the voltage will drop across the cap.
  4. Jan 23, 2009 #3
    My question is whether or not the change in PE of the capacitor would reflect my "answer:" delta(PE)1>0; delta(PE)2>0; delta(PE)3<0 (the numerals are respective of the steps)
  5. Jan 23, 2009 #4


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    I guess I don't really understand your notation.

    I just described what happens.
  6. Jan 23, 2009 #5


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    :confused: You decrease ε so you increase V according to pion's equation and physical intuition - you would need a higher voltage to establish that charge without the dielectric!?
  7. Jan 23, 2009 #6
    yeah, that is what my answer shows right? a drop in V means a drop in PE. so PE final - PE initial < 0 ???
  8. Jan 23, 2009 #7


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    That's correct. I misstated it in typing it.

    Thanks for catching that.
  9. Jan 23, 2009 #8
    soooooo, i am correct?
  10. Jan 23, 2009 #9


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    No. I misstated my original description.

    The equations were right. My English description was wrong.
  11. Jan 23, 2009 #10
    ay.... i don't understand what happens then. this is my logic:

    1- capacitor is charging (PE final is greater than PE initial therefore deltaPE>0)
    2- capacitor has dielectric inserted (PE final is greater than PE initial because dielectric allows more capacitance therefore deltaPE>0)
    3- capacitor no longer charging and dielectric is removed (PE final is less that PE initial because capacitor is now able to discharge therefore deltaPE<0)
  12. Jan 24, 2009 #11


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    3 does not correspond to 3 of your first post. Switch stays open, the charge has to stay there ideally. Yes, if the switch is then closed the capacitor will largely discharge.
  13. Jan 24, 2009 #12
    yes 3 is the same on both posts:

    3 (first post): the switch is opened, and the dielectric removed.
    3 (last post): capacitor no longer charging [the switch is opened] and dielectric is removed

    I am restating my logic. Is this correct? Anyone!

    delta=final-initial; numerals correspond to the steps.

    delta(PE)1>0; delta(PE)2>0; delta(PE)3<0
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