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.

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).

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)

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.

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

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.

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.

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.

[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.

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?