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Capacitor and dielectric

  1. Jun 25, 2009 #1
    A capacitor is constructed commercial wrap two sheets of aluminum separated by two sheets of paper coated with paraffin. Each aluminum foil and paper has width of 7cm. A sheet of aluminum has 4x10^{-6}m and thickness of paper is thick 25x10^{-6}m and a dielectric constant of 3,70. What is the length that the leaves should be desired is the capacitance of 0.95 pF? (Capacitor as the parallel plates)
     
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  3. Jun 25, 2009 #2

    negitron

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    Seems like you might need an equation which relates plate area, separation and dielectric constant to the total capacitance, yes? And, of course, some super-basic geometry.
     
  4. Jun 25, 2009 #3
    Yes. But what equations ?
     
  5. Jun 25, 2009 #4

    negitron

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    Presumably for your coursework, you have covered these. There is only one which applies here. See if you can't figure it out and post it here; I'll tell you if you have the correct equation. Then, you can take the next step.
     
  6. Jun 25, 2009 #5
    C = (E. * A) / (d) With dielectric C = (E. * A * k) / (d)

    How do I find the area?
     
  7. Jun 25, 2009 #6

    negitron

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    In your equation above, A is the area, specifically, the area of overlap between the two plates (which in your problem is the area of either plate since it's implicitly assumed they overlap completely) and it is stated is square meters. C, of course, is the capacitance in Farads, E is the permitivitty of free space (8.854x10^-12 F/m ), k is your dielectric constant and d is the separation between the two plates (in meters). Now, apply your algebra to isolate A on one side of the equation.
     
  8. Jun 25, 2009 #7
    A is length of the leaves X width ? What is the value of d?
     
  9. Jun 25, 2009 #8

    negitron

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    Yes, exactly. And since the width is known, the rest is easy

    What do you think the value of d is? What must determine that value in the problem setup?
     
  10. Jun 25, 2009 #9
    d is thickness of the paper X thickness of aluminum foil ?
     
  11. Jun 25, 2009 #10

    negitron

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    You're getting there. Since charge collection is a surface phenomenon only, the separation distance need only concern itself with the inner surfaces of the plates. It's how far apart these surfaces are that matters.

    And remember to keep in mind how many layers of paper are being called for here.
     
  12. Jun 25, 2009 #11
    Yes. d is is (thickness of the paper X thickness of aluminum foil) / 4. Why have 2 thickness of the paper and 2 thickness of aluminum. Correct ?
     
  13. Jun 25, 2009 #12

    negitron

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    No, there is a reason the problem does not specify the thickness of the foil.
     
  14. Jun 25, 2009 #13
    I do not understand what is d in formula.
     
  15. Jun 25, 2009 #14
    I find:

    C = (E * x * y * k)/d
    C * d = E * x * y * k
    x = (C * d) / (E * y * k)
    x = 1,036x10^{-5}m

    It is correct ?
     
  16. Jun 25, 2009 #15

    negitron

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    Almost, but you're forgetting to take into account the number of layers of paper. Your answer is correct for a single layer of the stated paper.
     
  17. Jun 25, 2009 #16
    How so? Why do I have two sheets the distance is 50x10^{-6} and not 25x10^{-6} ?
     
  18. Jun 25, 2009 #17

    negitron

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    I'm pretty sure the problem as stated gives the thickness of one sheet of paper and specifies that two such sheets are used. Is that not the case? You're in a better position than I to definitively answer that, as I don't have (and probably couldn't read) the problem in its original language.

    To be clear, if the problem gives the total thickness of the two sheets of paper, your answer is correct; if it gives only the thickness of one you obviously must multiply by 2.
     
  19. Jun 25, 2009 #18
    Yes thank you. If possible, I am very grateful if you help me in other problems. Thanks
     
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