Are those two capacitors in parallel?

I'm trying to analyze the following situation: we have a parallel plate capacitor where the plates are both squares with side length $$l$$ and on the positively charged plate we have a charge $$Q$$, moreover they're a distance $$d$$ apart one from the other. At some point we introduce between the plates a rectangular conductor whose height is $$h$$ for a section of length $$x$$ and we want to establish the capacitance of the system.

Now we have the following situation:

1. There's a section of capacitor whose surface is $$l(l-x)$$ where there's no conductor between the plates; let's call it $$C_1$$.
2. There's a section of capacitor whose surface is $$lx$$ where there's a conductor between the plates, of height $$h$$, therefore we have a section of height $$d-h$$ where there's no conductor between the plates; let's call it $$C_2$$.

I plan to consider the two sections as different capacitors and I assert that they're in parallel.

Now of course the two plates of the initial capacitor are conductors therefore their surface is equipotential, and corresponding plates of the new conceptual capacitors are part of the same plate of the starting one so they're at the same potential. This holds for both the plates and then there must be the same potential difference between the plates of both the new capacitors, i.e. they're in parallel.

However I wanted to support my thesis in a more formal way so I tried to compute the potential difference between the plates of both the conceptual capacitors. If I take the capacitor $$C_1$$ then to compute the potential difference I can just use the definition: I take as a starting point one on the negatively charged plate, $$A$$ and as ending point one on the opposite plate, $$B$$. Selecting as line one that goes normally from $$A$$ to $$B$$ I find that $$V_{1} = \dfrac{\sigma}{\epsilon_0}d$$ I establish the exact same conventions for $$C_2$$ but this time when I select a line $$\Gamma$$ where to compute the integral, selecting again one that goes perpendicularly to the plate from $$A$$ to $$B$$, that one will cross a conductor and in every point of it (length $$h$$) there's no electric field and therefore no contribution to the potential difference, so $$V_{2} = \dfrac{\sigma}{\epsilon_0}(d-h)$$ So it looks like they're not the same as I expected them to be and this means they're not in parallel.

Of course one of the two must be incorrect, but which one and why?

• Please re-read your question. Capacitors don't have capacitors or conductors between their plates, but they do have dielectrics. Your wording is making the question ambiguous. Also, a drawing would help. – David White Nov 7 at 23:48
• Your assumption that sigma is the same in both parts (V1 and V2) is certainly not correct – lalala Nov 8 at 2:41
• Baffo, your assumptiong that each capacitor plate is an equipotential surface is certainly correct. But you also seem to be assuming that this implies the surface charge density is constant along each plate. Why? – Alfred Centauri Nov 8 at 2:54
• @AlfredCentauri I will support my thesis in a more formal way when I'm back home, but for now this is why: consider a surface $S$ with uniform charge density $\sigma$, then every surface $S'$ that completely lays on $S$ has the same density; in fact the charge $Q'$ on $S'$ is $\sigma S'$ and the density on the new surface (assuming it will be uniformely charged) is $\sigma' = \frac{Q'}{S'} = \sigma$ – Baffo rasta Nov 8 at 7:48
• @DavidWhite Edited some wording, not a native speaker here, if you have edits to suggest you're welcome. The problem I'm facing explicitly requires to have a conductor instead of an insulant between the plates. – Baffo rasta Nov 8 at 7:49