Medium 1 has the electrical permittivity ∈1 = 1.5 ∈0

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Medium 1 has the electrical permittivity ∈₁ = 1.5 ∈₀ farad/m and occupies the region to the left of the x = 0 plane. Medium 2 has the electrical permittivity ∈₂ = 2.5 ∈0 and occupies the region to the right of x = 0 plane. If E₁ in medium 1 is E₁ = 2 ux – 3 uy + uz V/m, find E2 in medium 2 is—

1. (2 ux – 7.5 uy + 2.5 uz) V/m
2. (2 ux – 2 uy – 0.6 uz) V/m
3. (1.2 ux – 3 uy + 1 uz) V/m
4. (1.2 ux – 2 uy + 0.6 uz) V/m

Correct Option: C

We know that
D_n1 = D_n2
∈_r1 ∈_n1 = ∈_r2 · ∈_n2

E_x2 =	∈_r1	· E_x1
	∈_r2

E_x2 =	1.5 ∈₀	× 2 = 1.2
	2.5 ∈₀

∴ E₂ = 1.2^–u_x – 3^– u_y + 1^– u_z
Note: The field in second medium does not change w.r.t. y and z.
Since the interface plane is x = 0. Which is y, z plane and normal to interface is x-axis when E_t1 and E_t2 resolved the component of E_t1 in y and z direction = resolved component of E_t2 in y and z direction.
Therefore, E₂ component (y and z) is same as that of E₁ i.e. (–3u_y + 1. u_z)