Wouldn't $D = \epsilon E$ mean that a higher permittivity constant gives a higher electric flux? (And what exactly is the displacement field?)

Question

A quote from the Wikipedia page for Permittivity:

More electric flux exists in a medium with a low permittivity (per unit charge) because of polarization effects.

This makes sense to me.

My understanding is that, exposed to an electric field, atoms will develop correspondingly aligned induced dipoles. The effects of all of these dipoles, except the ones at the very edges of the medium, will cancel each other out (being adjacent in the medium to the opposite pole of the next atom). The uncancelled layer of positive charge at one end and negative at the other result in an electric field within the medium that opposes the surrounding electric field. $D$ denotes electric flux, which takes into account all charges and resulting fields involved, not just the original, which is denoted $E$.

Wouldn't $D = \epsilon E$ mean that the greater the permittivity (the greater the opposing electric field due to polarization in the medium), the greater the electric flux? What happened to the electric flux being diminished with greater opposing field?

(I've been seeing 'electric flux', 'electric flux density', 'electric displacement field' and '$D$' used interchangeably. Are my definitions wrong, perhaps, and I'm trying to merge two different concepts?) <- It seems that this was the case. New question: What exactly is the displacement field ($D$), and what does it represents physically?

Answer 1

Note that the Gauss law comes from the the Maxwell Eq. DivE=constant X charge density. When you place a charge in a dielectric the polarization so developed gives a bound charge opposite to that of that embedded. The total charge therefore decreases. The electric flux will therefore decrease since the total charge will now give lesser E. In other words the E due to original charge and that due to the polarization will be in opposite directions.

Answer 2

Electric flux is the surface integral of the electric field along or over a given surface .

By Gauss's law , it depends upon the charge enclosed inside the surface . However, I guess what you mean by electric flux is not the total flux through the surface but just the flux through one of it's surfaces .

Let us consider a slab made up of a material having a dielectric constant , k, which by definition is the factor by which an existing electric field in vacuum is reduced when that material is placed in that region of space.

Let this slab be placed between the two uniformly charged sheet having opposite charges and having surface charge density ,s.

Now the net electric field between the two plates initially is S/e, where e= permittivity of vacuum .

Now when a slab with dielectric constant K is placed the field in between the plates is S/Ke. Now let us consider a Gaussian surface which in the form of a cylinder whose one circular surface lies on one of the plates and another at the surface plate facing surface of the dielectric slab .

Now the two surfaces of the dielectric slab can also be considered as two sheets with uniform charge density . As the field inside the slab is resultant of field due to the plates and the aligned molecules of the dielectric slab we have by superposition principle : the charge density at one of the surfaces of the slab = s{1-(1/k)}

Now let us consider the flux through the mentioned Gaussian surface. The net flux is equal to {1/e} times the net charge enclosed inside the surface ,which is equal to sA[1-{1-(1/k)}] =sA{1/k} , where A is the surface area of the base circular area of the cylinder. Now the flux through the surface lying on the sheet is = sA/e.

so the flux through the surface lying on the slab = sA/e - sA/ek . As, K is directly proportional to the relative permittivity you can see smaller the permittivity ,greater is the flux through the surface lying on the slab . However, if you consider the total flux through the medium , it will turn out to be zero as the net charge enclosed is zero.