The plane π§=0 separates two linear dielectric media with relative permittivities ππ1=4 and ππ2=3, respectively. There is no free charge at the interface. If the electric field in the medium 1 is πΈβ 1=3π₯Μ+2π¦Μ +4π§Μ, then the displacement vector D2ββ in medium 2 is: (π0 is the permittivity of free space)
At the boundary between two dielectric media, the following conditions hold:
The tangential components of the electric field (Etββ) are continuous:
The normal components of the displacement field (Dnββ) are continuous:
Step 2: Separate tangential and normal components of E1ββ
The electric field in medium 1 is: E1ββ=3x^+2y^β+4z^. - Tangential components (E1tββ): These are the components parallel to the interface (z=0): E1tββ=3x^+2y^β. - Normal component (E1nββ): This is the component perpendicular to the interface (z-direction): E1nββ=4z^.
Step 3: Determine E2ββ in medium 2
Using the continuity conditions:
Tangential components remain unchanged:
Normal components scale with the ratio of relative permittivities:
Step 4: Calculate D2ββ in medium 2
The displacement vector D is related to E by: D=Ο΅0βΟ΅rβE. For medium 2 (Ο΅rβ=3): D2ββ=Ο΅0ββ 3β (3x^+2y^β+316βz^). Distribute Ο΅0ββ 3: D2ββ=Ο΅0β(9x^+6y^β+16z^).
Final Answer:
The displacement vector in medium 2 is: D2ββ=(9x^+6y^β+16z^)Ο΅0β.