Final answer:
To find the electric field E2 and electric displacement D2 in region 2 with dielectric constant εr2=3, given E1 in region 1 with εr1=2, we use the continuity of tangential electric fields and normal components of dielectric displacement across the boundary. The relationship D=εoεrE is employed.
Step-by-step explanation:
The question asks about the electric field E2 and electric displacement D2 in region 2 of a space divided by a plane with different dielectric constants on either side. Given that the electric field E1 in region 1 with dielectric constant εr1=2 is x^2y - y^3x + z(5+z), we can determine E2 and D2 for region 2 with dielectric constant εr2=3 using the boundary conditions of the electric fields and dielectric displacement at the interface.
At the boundary between two different dielectrics, the tangential components of the electric field are continuous (they have the same values on both sides of the boundary), and the normal components of the dielectric displacement are also continuous. Thus, knowing the electric field in region 1 allows the direct calculation of these components in region 2. The electric displacement D is related to the electric field E by the equation D = ε0εrE, where ε0 is the vacuum permittivity and εr is the relative permittivity or dielectric constant of the medium.