Final answer:
The electric flux density D2 in region 2 can be found using the continuity condition of electric displacement fields at the boundary between two dielectrics. Assuming no free surface charge and given values for εR1 and E1, the resulting D2 for r = 9 cm is 49.38 ar C/m².
Step-by-step explanation:
To find the electric flux density in region 2 (D2), we need to consider the continuity condition at the boundary between the two dielectric media. According to the boundary conditions for electric fields at the interface of two dielectrics, the normal component of the electric displacement field (D) must be continuous across the boundary unless there is a surface charge present. Since we're given that the dielectric shells are perfect (assuming no free surface charge), we can write:
εR1 ⋅ E1 = εR2 ⋅ E2
Given εR1 = 2 and E1 = (2000/r²) ar V/m for region 1 for r ≥ r1 (outside the sphere of radius 4 cm) and assuming there's no free charge present at the interface, we have:
εR1 ⋅ E1 = εR2 ⋅ D2/εR2
We can then solve for D2 as follows:
D2 = εR1 ⋅ E1
Therefore, D2 = 2 ⋅ (2000/r²) ar V/m
For r = 9 cm:
D2 = 2 ⋅ (2000/9²) ar V/m
D2 = 49.38 ar C/m²
The electric flux density D2 in region 2 is 49.38 ar C/m² when r = 9 cm.