Fully filled capacitors with parallel dielectrics problem

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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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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 !