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Homework Help: Simple capacitor problem

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A capacitor connected to a battery initially holds a charge of +q on its positive plate and -q on its negative plate. The electric field between the plates is initially E. A dielectric material is then inserted that polarizes in such a way as to produce an electric field of -0.30E (where the - sign indicates that this field opposites that of E). Determine the new charge stored on the positive plate.

    2. Relevant equations
    Eo-E=Ef
    E=kq^2/r^2

    3. The attempt at a solution
    I did my work on Word using the equation tool just to make the notation format easier to read.
    Physics.png
    Is my logic correct here?

    Any help would be greatly appreciated! :)
     
  2. jcsd
  3. Sep 27, 2015 #2

    ehild

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    The equation in red is the Coulomb force between two equal point charges. It is not electric field, and it is irrelevant to this problem.
     
  4. Sep 27, 2015 #3
    Oh duh, it's just q from the source, not q^2

    Could we solve it using the equation for capacitance?
     
  5. Sep 27, 2015 #4

    ehild

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    What i
    What do you mean?
     
  6. Sep 27, 2015 #5
    Well since it's capacitor maybe we can use C=Q/V=Q/Ed?
     
  7. Sep 27, 2015 #6

    ehild

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    That is correct. But the problem is not clear. Is the capacitor connected to the battery during the whole experiment?
     
  8. Sep 27, 2015 #7
    Yes.
     
  9. Sep 27, 2015 #8

    ehild

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    Does the electric field change between the plates then when the dielectric material is inserted?
     
  10. Sep 27, 2015 #9
    The field is the same between the empty space, but is different inside the dielectric, correct?
     
  11. Sep 27, 2015 #10

    ehild

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    No, the field is voltage over the distance between the plates, neither of them changes.
     
  12. Sep 27, 2015 #11
    Okay, so where can I go from there?
     
  13. Sep 27, 2015 #12

    ehild

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  14. Sep 27, 2015 #13
  15. Sep 27, 2015 #14
    Can anyone else help me with this? =/
     
  16. Sep 27, 2015 #15

    ehild

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    Yes, the new charge is Q'=EεA. But you do not know ε.
    The field polarizes the dielectric and dipole chains appear. The ends of the chains neutralize some of the surface charges. How much do they neutralize in this case? It was given, that the dipoles would make a field of 0.3 E, opposite to the original.
     
  17. Sep 27, 2015 #16
    Snapshot.jpg

    Does this image depict what you're describing correctly?
     
  18. Sep 27, 2015 #17
    deleted
     
  19. Sep 27, 2015 #18

    ehild

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    Yes.
     
  20. Sep 27, 2015 #19
    Okay, so then would the new value for ε be the dielectric constant?
    So κ=Eo/E = E/0.7E=1.4286?
     
  21. Sep 27, 2015 #20

    ehild

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    The new electric field is the same as the initial one was, U/d. See Post #10.
     
  22. Sep 27, 2015 #21
    Okay, got it. So voltage is the same because of the presence of the battery and obviously separation does not change, thus field is constant. So how can I find the new charge if we don't know ε?
     
  23. Sep 27, 2015 #22

    NascentOxygen

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    Your sketch in #16 suggests you are assuming that the added material occupies less than the full gap between the plates? However, this fraction doesn't seem to be specified.
     
  24. Sep 27, 2015 #23

    ehild

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    You know that E=E1-E2, as your picture shows. E2=0.3E. The new surface charge corresponds to the field E1, σ'=ε0E1. So what is σ' in terms of σ?
     
  25. Sep 28, 2015 #24
    That's true. I made the assumption because all of the examples that our professor has mentioned has been when there is some separation between the two. But the question never suggests that there is any.

    σ=q/A
     
  26. Sep 28, 2015 #25

    ehild

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    I asked the relation between the new and the initial surface charges.
     
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