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This is one of the questions from my textbook with no solutions.

One plate of a capacitor carries a charge of $+0.567\space pC$. The other plate carries a charge of $-0.567\space pC$. If the capacitor plates are parallel discs with diameters of $2\space cm$ calculate:
(i) the charge density on the plates,
(ii) the flux density between the plates.
Calculate the strength of the electric field between the plates if the dielectric between the plates is
(iii) air and
(iv) polythene.
What is the absolute permittivity of polyethylene? (Assuming flux density is constant i.e. ignore field fringing effects)

My answer:

(i)For the first part of the question I worked out the charge density to be $4.512\times 10^{-4} Cm^{-3}$

(ii) In parallel plate the capacitors, we assume that the electric field between the plate is uniform i.e. it doesn't spread out. Given that it is uniform, electric flux at the surface is the same as electric flux between them. Therefore electric flux between plates is $4.512\times 10^{-4} Cm^{-3}$

(iii) The relative permittivity of air is $1.000594$ using the equation for relative permittivity of material I got $\displaystyle{\frac{1.000594}{8.854\times 10^{-12}}= 1.120\times 10^{11}}$

And then using Electrical flux density equation I got the electrical field strength between the plates to be $1.120\times 10^{11}\times 4.512\times 10^{-4} = 50990288.32$

This seems like to big of a number

Could anyone give me any hints how to approach this question or let me know whether I am right?

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Your numbers seem to be completely wrong.

(i) The charge density should be an areal density here in $Cm^{-2}$, not a volume density as you assume. How do you arrive at this number?

(ii) The electric flux density is the electric displacement, which according to Gauss' law should be equal to the areal charge density on the plates.

(iii) The electric field strength you will get by dividing the electric displacement by the absolute dielectric permittivity of air an the material $$E = \frac{D} {\epsilon_0\epsilon_r}$$

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