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Capacitors, Electric fields and dielectrics

  1. Apr 18, 2005 #1
    I have a nasty problem here which I have been working at for a while but keep getting rubbish answers.

    I have tried using: E = surface charge density/(dielectric constant * Epsilon0 (8.85*10^-12) but I get a dielectric material which has a lower dielectric constant than air.
    I really need a formula to relate capacitance to electric field.
    Any help would be much appreciated.
  2. jcsd
  3. Apr 18, 2005 #2
    Hint #1: Use Gauss's Law.

    Hint #2: Do you know how to relate the dielectric field (the induced field) to the free field? The dielectric constant enters this expression.

    Hint #3: What is the field between the plates when the dielectric is (a) absent and (b) present.

    Okay now I don't want to give away more than I have...think about it. :smile:
  4. Apr 18, 2005 #3
    First of all--- may i ask if you have converted 100 cm^2 propertly to m^2 ?

    This would greatly increase the value of your charge density--- making kappa much greater.

    If that doesn't work you should take a look at the following:




    Notice something here... we could rewrite the Capacitance equation as:



    [tex]C^{-1}=\frac{dQ}{\kappa_e\epsilon_0AQ}[/tex] but Q/A is sigma!

    So you can write:


    GASP! we know


    so we write




    I think you know what to do now!

    Also, the prudent engineer always does a dimensional analysis to see if the units match up... in this case they should indeed equal inverse Farads. (Please note the use of d here is not a differential).
    Last edited: Apr 18, 2005
  5. Apr 18, 2005 #4
    Although the aforementioned analysis is useless if they do not give you d--- which I just realize they do not... Hmmmmm.
  6. Apr 18, 2005 #5
    [tex]E_{net} = \frac{1}{k}E_{free}[/tex]
    I'm not sure if this helps...
  7. Apr 18, 2005 #6
    Thanks guys, but I have already tried using E = sigma/kEo and solving this equation gives me a dielectric constant of 0.07 something. d is the separation between plates right? well where am I getting this from, I assume I can just put a random value in here and it should work if i keep it the same.
    But I think the problem I am having is the fact that they have equal and opposite charges of 8.9x10^-7, in terms of C=q/v do I just use this as the value of q? because if so shouldnt I beable to put a constant value for v and find out the differences in capacitance and therefore the dielectric constant? and to maverrick, yes I get all that but I do not know how to relate the electric fields, I know that when there is a dielectric inthere that the electric field is increased by a factor of k.

    Im sure this problem is a lot easier than I am making it out to be but I just dont see it.

    (by the way, how do you make the formulae look cool like that)
  8. Apr 18, 2005 #7
    Sorry yeah you must have posted while i typed my reply.
  9. Apr 18, 2005 #8
    where is Daniel we need his help!!

    "(by the way, how do you make the formulae look cool like that)"

    Hehe, its called LaTex... if you know the coding you can make your equations pretty...

    There should be a link somewhere on this site that shows you how to use it.
  10. Apr 18, 2005 #9
    I have Enet ofcourse but not Efree (or it would just be too easy). I'll look into calculating it using 1 as dielectric constant
  11. Apr 18, 2005 #10
    whenever I try to calulate the electric field of the capacitor without a dielectric, I get a value which is less than the electric field of the capacitor with a dielectric and this is impossible.
  12. Apr 18, 2005 #11
    and this is using E=q(enc)/Eo (gauss's law) or E=q(enc)/EoA (which concidently ends up the same equation as gauss's law) and using these values ofcourse gives me an unfeasible dielectric constant of 0.0718
    Last edited: Apr 18, 2005
  13. Apr 18, 2005 #12
    right--- there's no case where [tex]\kappa_e > 1 [/tex]

    Hmm this is quite puzzling.

    I have been looking thru my Resnick and Halliday physics text and am not able to find a single example problem where they do NOT give you the distance between the air gap capacitor.
    Last edited: Apr 18, 2005
  14. Apr 18, 2005 #13
    should I be doubling the charge since there are two of them with opposite signs? I really dont know
  15. Apr 18, 2005 #14
    You wouldnt have to double it... that much I know.

    Thats like saying the previous cases you suggested would mean only one of the plates is charged and the other has a net charge of zero. But we know that this wouldnt be the case.
  16. Apr 18, 2005 #15
    I have the same textbook as you (6th edition). The problem is definatly correct and there is no extra information. This problem is a repeat from the last assignment I had, I guess too many people got it wrong.
  17. Apr 18, 2005 #16
    I could use any random constant value for d right? But then I still have the problem of the Efree being smaller than the Enet
  18. Apr 18, 2005 #17


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    Homework Helper

    [tex]E= q/ \epsilon_0 [/tex] use this to find the electric field of the capacitor, subtract from this the electric field given in the problem to find the electric field due to the induced charges on the dielectric which opposes the capacitor electric field. you'll need to use the proper formula to find solve the rest, if you can't find it...well that would be pathetic.

    this electric field due to the induce charges is opposite in sign to the capacitor field. Nevertheless we can still reuse gauss' law (the one in latex) to find the induced charge.

    hope this helps
  19. Apr 18, 2005 #18
    Thanks, that makes sense. But its stupid that there are no examples of these types of problems anywhere
  20. Apr 19, 2005 #19
    Finally I ended up with this equation:
    where [tex]E_D[/tex] denotes the electric field within the diaeletric. Am I right? Btw, what's the answer?
  21. Apr 19, 2005 #20
    I dont know the correct answer but I got 2.55 using GCT's method. How about you?
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