induced charges inside a capacitor with two dielectrics

Question

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

1. I have no problem in calculating the electric field inside the capacitor, but what pauzzeld me most was the answer of the book

because the interface between the two dielectric is parallel to the electric field E1=E2=E we can say that the electric field is conserved HOW come ???? i could find those results analytically but how could they spot this just by saying such sentence???

2.While calculating the induced charges, they just calculate the induced charge on the left side of the dielectris (the ending point of the electric field) ?? how come because i thought that the induced charge in this case should be on both internal surface of the dielectrics

Many thanks in advance

Answer 1

You can consider the two dielectrics as two capacitors connected in parallel. Since the two are connected across the same potential difference and the distance between both the capacitors(upper and lower one) are same, E= V/d , the electric field in both the dielectrics is the same.

As the dielectric is neutral, the charge appearing on the right side will be exactly opposite of that appearing on left side. So, I guess there's no need to mention charges on both sides of dielectric. Hope it helps!