Fully filled capacitors with parallel dielectrics problem

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Homework Help Overview

The discussion revolves around a capacitor filled with two different dielectrics, where one occupies a third of the surface area and the other occupies two-thirds. The original poster is attempting to calculate the electric field within the capacitor and the surface charge density, while encountering confusion regarding the application of Gauss's law and the uniformity of charge density across the dielectrics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster applies Gauss's law to analyze the electric field and charge density in the capacitor but questions the validity of their conclusions regarding the uniformity of the electric field across the dielectrics. Participants discuss the implications of differing charge densities on the plates due to the presence of different dielectrics.

Discussion Status

Participants are actively engaging with the original poster's reasoning, with some providing clarifications regarding the non-uniform charge density on the capacitor plates. The discussion is ongoing, with participants exploring the implications of the differing dielectric constants and their effects on the electric field and charge distribution.

Contextual Notes

There is an indication that the original poster may have made assumptions about the uniformity of charge density, which is being questioned by others in the thread. The specific configuration of the dielectrics and their respective areas are central to the discussion.

Amaelle
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consider that we have two dielectrics inside a capacitor as shown in the picture, let0s consider also that Q is the charge of the capacitor and d the distance between the two plates , the first dielectric occupy a surface of S/3 with a dielectric constant of er1 and the second a surface of 2S/3 with a dielectric constant of er2, the question is calculate the electric field inside the capacitor and the surface density of the induced charge (see the uploaded image)
During the calculation I faced the following doubts: I applied gauss theorem to the flux density in both areas of the dielectrics and I got:
QEAJg.gif


σ is the surface density of the free charges =Q/S,

8YAxA.gif


doing the same with the second surface I got that
y48Zg.gif


and because my two dielectric are in parallel so E1=E2

this gives me
MHZ6c.gif


BUT this not true! so where did my logic fails??

Many thanks in advance!
dielectric.png
dielectric.png
dielectric.png
 

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Amaelle said:
BUT this not true! so where did my logic fails??
Charge density on the plates will not be uniform.
 
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Likes   Reactions: Amaelle
Thanks for your prompt reply, it makes sens but can you elborate more?
 
I just go why the charge density is different from dielectric to dielectric
many thanks !
 

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