Capacitors, Electric fields and dielectrics

  • Thread starter coldturkey
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  • #1
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I have a nasty problem here which I have been working at for a while but keep getting rubbish answers.

Two parallel plates of area 100cm^2 are given charges of equal magnitudes 8.9x10^-7 but opposite signs. The electric field within the dieletric material filling the space between the plates is 1.4x10^6 V/m.
a) Calculate the dielectric constant of the material. b) Determine the magnitude of the charge induced on each dielectric surface.

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.
 

Answers and Replies

  • #2
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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:
 
  • #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:

[tex]C=\frac{\kappa_e\epsilon_0A}{d}[/tex]

and

[tex]E=\frac{\sigma}{\kappa_e\epsilon_0}[/tex]

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

[tex]C=\frac{\kappa_e\epsilon_0AQ}{dQ}[/tex]

then

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

So you can write:

[tex]C^{-1}=\frac{d\sigma}{\kappa_e\epsilon_0Q}[/tex]

GASP! we know

[tex]E=\frac{\sigma}{\kappa_e\epsilon_0}[/tex]

so we write

[tex]C^{-1}=E\frac{d}{Q}[/tex]

thus

[tex]C=\frac{Q}{Ed}[/tex]

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).
 
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  • #4
Although the aforementioned analysis is useless if they do not give you d--- which I just realize they do not... Hmmmmm.
 
  • #5
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[tex]E_{net} = \frac{1}{k}E_{free}[/tex]
I'm not sure if this helps...
 
  • #6
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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)
 
  • #7
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Theelectricchild said:
Although the aforementioned analysis is useless if they do not give you d--- which I just realize they do not... Hmmmmm.
Sorry yeah you must have posted while i typed my reply.
 
  • #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.
 
  • #9
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futb0l said:
[tex]E_{net} = \frac{1}{k}E_{free}[/tex]
I'm not sure if this helps...
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
 
  • #10
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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.
 
  • #11
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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
 
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  • #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.
 
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  • #13
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should I be doubling the charge since there are two of them with opposite signs? I really dont know
 
  • #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.
 
  • #15
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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.
 
  • #16
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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
 
  • #17
GCT
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Two parallel plates of area 100cm^2 are given charges of equal magnitudes 8.9x10^-7 but opposite signs. The electric field within the dieletric material filling the space between the plates is 1.4x10^6 V/m.
a) Calculate the dielectric constant of the material. b) Determine the magnitude of the charge induced on each dielectric surface.


[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
 
  • #18
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Thanks, that makes sense. But its stupid that there are no examples of these types of problems anywhere
 
  • #19
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Finally I ended up with this equation:
[tex]\kappa=\frac{Q}{Q-\epsilon_{0}AE_D}[/tex]
where [tex]E_D[/tex] denotes the electric field within the diaeletric. Am I right? Btw, what's the answer?
 
  • #20
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I dont know the correct answer but I got 2.55 using GCT's method. How about you?
 
  • #21
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[tex]C=\frac{Q}{(E-E_{D})d}[/tex]
[tex]C=\kappa\frac{\epsilon_{0}A}{d}[/tex]
[tex]E=\frac{Q}{\epsilon_{0}A}[/tex]
then I equal both of them, and I got 1.16
 
  • #22
25
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There is nothing even similar to having a dielectric constant of 1.16
2.55 is very similar to polystyrene (2.6)
 
  • #23
Yeah hes right--- i have the same text with that table too :D
 
  • #24
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its ok


Using [tex]\displaystyle{E =\frac{\sigma}{\epsilon_0 k_e}}[/tex] I get the result [tex]\displaystyle{k_e = 7.1832...[/tex], where
[tex]\displaystyle{\sigma =\frac{Q}{A}}}[/tex] .
Be sure you have converted the [itex]
\displaystyle{100 cm^2}[/itex] to [itex]0.01m^2[/itex].
 
  • #25
yeah he got it!
 

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