For a plane isolated capacitor with a dielectric slab partially introduced inside it, what's indpendent of the medium? $\mathbf E$ or $\mathbf D$?

Question

Say we have an isolated capacitor with charge $Q$ and a dielectric slab, $\varepsilon_r$, partially introduced between the capacitor's plates, then, I'd say the displacement vector should be the same regardless whether it is in $\varepsilon_0$ or $\varepsilon_r$ since according to Coulomb's law, $\mathrm D=\sigma$, and charge is conserved in the plates. Therefore, only the electric field varies depending on the medium: $$\mathrm E=\dfrac{\mathrm D}{\varepsilon}= \left\{\begin{aligned} &\dfrac{\mathrm D}{\varepsilon_0}&\text{ in vacuum}\\ &\dfrac{\mathrm D}{\varepsilon_0\varepsilon_r}&\text{ inside slab} \end{aligned}\right. $$ One could also counter-argue that maybe the surface density of the plates actually changes depending on the medium and that $\mathrm E$ stays constant since $V$, the potential between plates, is the same for parallel capacitors and $d$, the distance between them, remains constant and so $E=V/d$ is independent of the medium too.

What argument is correct then?

As suggested I'll add a picture to illustrate the question:

Answer 1

If we are talking about a typical capacitor with conductive plates, the potential difference between the plates is fixed, so $E = V/d$ is what is constant in both regions of the capacitor, not $D$. Note that the charge density $\sigma$ will be different in the two regions (larger where the dielectric is present), so you cannot treat $D=\sigma$ as constant throughout.

If you had two non-conductive plates uniformly charged with equal and opposite fixed charges, and made a "capacitor" out of them, then $D$ would be uniform inside (besides edge effects), not $E$.

Answer 2

So I'm just going to work in terms of an infinite plane capacitor, because I don't want any edge effect concerns causing a bunch of annoying caveats.

Also: I would be nice if you included a picture, so I could reference it, but you didn't, so I'm going to plow ahead without it.

The field strength is all about divergence, and since $\vec D$ only sees free charge, and there is no free charge between the plates, it remains constant. (Note, with the symmetry:

$$\vec \nabla \rightarrow \frac{\partial}{\partial x} $$

so we can just imagine step functions at boundaries.

Obviously, $\vec E$ has a step change at the dielectric boundary.

Now the concern about image charges in the conductive plates is valid, but here's the out:

For the infinite sheet geometry, distance no longer matters. An infinite sheet of charge produces a uniform field everywhere (or at least on each side, which another step function at the sheet).

Anyway, what capacitor's positive conducting plate (wish I had a picture to reference) see is the near side of the dielectric, which has a negative charge from $d\vec P/dx$, so yes, it you have an image charge (sheet) from that...

However, it also "sees" the far side of the dielectric with it's positive surface charge.

The two charges are equal and opposite, and since the field strength does not depend on distance, boom: their would-be image charges cancel, and the surface charge on the Cap is unaffected.