- #1
Amaelle
- 310
- 54
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:
σ is the surface density of the free charges =Q/S,
doing the same with the second surface I got that
and because my two dielectric are in parallel so E1=E2
this gives me
BUT this not true! so where did my logic fails??
Many thanks in advance!
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:
σ is the surface density of the free charges =Q/S,
doing the same with the second surface I got that
and because my two dielectric are in parallel so E1=E2
this gives me
BUT this not true! so where did my logic fails??
Many thanks in advance!