Question

Two homogeneous isotropic dielectrics meet in the plane y =0. For y < 0, Eᵣ₁ = 8 and for y >0, Eᵣ₂ = 2. A uniform electric field E₂ = -Aₓâₓ + Aᵧâᵧ - Aâ kV/m exists for y > 0. Take the last 3 digits of your student ID. If any of these digits is 0, replace it with 1. Use these digits for the values of Aₓ, Aᵧ, and A respectively and find:

a). E₁ for y ≤ 0.

Answer 1

E₁ within the first dielectric (y ≤ 0) can be found using the continuity of the electric field across the dielectric boundary and the relative permittivities of the materials. With Eᵣ₁ = 8 and Eᵣ₂ = 2, E₁ is scaled by Eᵣ₁ for parallel components and Eᵣ₂ for the perpendicular component.

Step-by-step explanation:

To answer how to find E₁ for y ≤ 0 when two homogeneous isotropic dielectrics meet in the y=0 plane, we need to consider the boundary conditions for electric fields at the interface. The given electric field in the dielectric for y > 0 is E₂ = -Aₓâₓ + Aₖâₖ - Aₗ kV/m. Using the given relative permittivities Eᵣ₁ = 8 and Eᵣ₂ = 2, along with the assumption that the electric field components perpendicular to the boundary must be continuous, we use the fact that the normal component of the electric field E scales with the relative permittivity.

Therefore, if Eₖ is the perpendicular component, we have Eₖ₁/Eᵣ₁ = Eₖ₂/Eᵣ₂. Assuming the normal component Eₖ remains unchanged across the boundary and knowing Eᵣ₁ is 4 times Eᵣ₂, we find that the electric field within the first dielectric (y ≤ 0) is E₁ = -Aₓâₓ/Eᵣ₁ + Aₖâₖ/Eᵣ₂ - Aₗk/Eᵣ₁ kV/m. Note that because Aₓ and Aₗ are parallel to the boundary, they scale with Eᵣ₁, while Aₖ (perpendicular) scales with Eᵣ₂.