Finding Electric Field and Displacement in Two Regions

[shevchenko12]

Homework Statement

Now consider a specific example where this boundary lies in the plane y = 0 and we have ϵ1 = 4, ϵ2 = 5 and σ = 5C/m2. The electric field in region 1 (with y > 0) is given by ⃗ E1 = (20,−10, 15) V/m. Find ⃗D1. [1]
Find ⃗D 2 and ⃗ E2 in region 2 with y < 0. [4]

Homework Equations

The Attempt at a Solution

(D_1 ) ⃗=ε_0 (80,-40,60)c/m^2 for the first part don't know if it right
For the second part i would assume i would use

E_1t=E_2t
and
D_{1perpendicular}-D_{2perpendicular}=σ

However I am not quite sure how to apply these...

 

[mark726]

Firstly, in order to find ⃗D1, we can use the formula ⃗D1 = ϵ1⃗E1, where ϵ1 is the permittivity of region 1 and ⃗E1 is the electric field in region 1. Plugging in the values given in the problem, we get ⃗D1 = (80, -40, 60) C/m^2. This is the correct answer for ⃗D1.

Next, to find ⃗D2 and ⃗E2 in region 2, we can use the boundary conditions for electric fields and electric displacement vectors.

Since the boundary lies at y = 0, the electric field and electric displacement vectors must be continuous across the boundary. This means that ⃗E1t = ⃗E2t, where the subscript "t" indicates the component of the vector parallel to the boundary.

Using this condition, we can find ⃗E2 by setting ⃗E1t = ⃗E2t = (20, 0, 15) V/m. This is because the y component of ⃗E1 is 0, since y > 0 in region 1. Therefore, ⃗E2 = (20, 0, 15) V/m.

To find ⃗D2, we can use the formula ⃗D1perpendicular - ⃗D2perpendicular = σ, where the subscript "perpendicular" indicates the component of the vector perpendicular to the boundary. Plugging in the values given in the problem, we get ⃗D1perpendicular = (80, -40, 0) C/m^2 and ⃗D2perpendicular = (0, 0, 0) C/m^2. Therefore, ⃗D2 = (80, -40, -60) C/m^2.

So, the final answers are ⃗D1 = (80, -40, 60) C/m^2, ⃗E2 = (20, 0, 15) V/m, and ⃗D2 = (80, -40, -60) C/m^2.