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Calculate the maximum charge that can be stored in the capacitor

  1. Apr 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Each plate of a parallel plate capacitor has an area of 3.00 cm2. A dielectric material with
    dielectric constant κ = 2.40 and a dielectric strength of 5.00 kV/mm is placed between the plates.
    Calculate the maximum charge that can be stored in the capacitor.


    Answer: 31.9 nC

    2. Relevant equations



    3. The attempt at a solution

    C=(K[tex]\epsilon[/tex]A)/d

    and I know that E=V/d


    K=2.4
    [tex]\epsilon[/tex]=8.85*10^-12
    A=0.3
    E=5.00

    but i cant find d, which is what i probably need
     
  2. jcsd
  3. Apr 12, 2008 #2

    Hootenanny

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    Note that the effective electric field, [itex]E_f[/itex] can be written as,

    [tex]E_f = E_0-E_p[/tex]

    Where [itex]E_p[/itex] is the electric field due to the polarisation charges in the dielectric and [itex]E_0[/itex] is the electric field in the absence of a dielectric material.
     
    Last edited: Apr 12, 2008
  4. Apr 12, 2008 #3
    is there another method of approaching this question? I dont believe we've learned this.

    Could you expand your equatiion please
     
  5. Apr 12, 2008 #4

    Hootenanny

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    How much physics have you done? Have you worked with Gaussian surfaces and Gauss' law before?
     
  6. Apr 12, 2008 #5
    i missed a week of school when I tore a ligament in my ankle so I missed those lectures, but I do have the notes and I know about gauss' law

    E=q/E0A

    Could you please just start me off, I have exams comming up
     
  7. Apr 12, 2008 #6

    Hootenanny

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    It can be shown that the electric field of a dielectric capacitor may be written thus,

    [tex]E = \frac{\sigma}{\kappa \varepsilon_0}[/tex]

    Where [itex]\sigma[/itex] is the surface charge density and [itex]\kappa[/itex] is the dielectric constant. If I have a little more time this evening I'll post a write up of how it is derived (or you could just look at your notes).
     
  8. Apr 12, 2008 #7
    that formula is nowhere in our notes...hmm. I'm going to ask a buddy of mine for the solution, but if you dont mind could you post you method too.

    Surface charge density...is that a constant?
     
  9. Apr 13, 2008 #8

    Hootenanny

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    I'm afraid until you've answered the question, forum guidelines prohibit me from posting the derivation.

    Yes, once the capacitor is charged and the dielectric is inserted, the charge density is constant.
     
  10. Apr 13, 2008 #9
    so what is the charge density?
     
  11. Apr 13, 2008 #10

    Hootenanny

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    The [surface] charge density on a plate is the total charge on the plate divided by the plate's area.
     
  12. Apr 13, 2008 #11
    Is this right

    Total charge on plate is q= E*E0A -- gauss' law
    Plates area is, of course, 3.00
     
    Last edited: Apr 13, 2008
  13. Apr 13, 2008 #12

    Hootenanny

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    That is only true in the absence of a dielectric material.
     
  14. Apr 13, 2008 #13
    first off, id like to thank you for your help, im learning a lot.

    the only formulas i have for a dielectric material are:

    E = 1/2(qV) = 1/2(q0)(V0/k) = E0/k

    C=q/v = qk/V0 = kC0
     
  15. Apr 13, 2008 #14

    Hootenanny

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    I'm sorry I think I've just dragged you the long way round. Going back to the two equations in your OP,
    Can you think of a way of eliminating d from the first equation using the second?

    And this equation,
    may also come in handy at some point :wink:
     
    Last edited: Apr 13, 2008
  16. Apr 13, 2008 #15


    haha yes, yes i can.
    I had a feeling there was another, easier method. Thanks for you help
     
  17. Apr 13, 2008 #16

    Hootenanny

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    No problem :smile:
     
  18. Apr 13, 2008 #17
    I got the answer, but I have a question. Why do we use 3.00cm^2 instead of m^2
     
  19. Apr 14, 2008 #18

    Hootenanny

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    I'm not sure what you mean, it doesn't matter which units one uses provided one converts them all to the same type. However, I noticed in your OP that,
    You should note that [itex]3cm^2 \neq 0.3m^2[/itex].
     
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