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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 chargeenter image description here

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:

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

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doing the same with the second surface I got that

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

this gives me

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

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The answer is the density of the free charges σ on the capacitor plates isn't uniform. it varies from region 1 to region 2.

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