Electric field intensity (E) in a region with z > 0 is given as E₁ = 2ax + 3ay + 5az V/m.
The region is a perfect dielectric with a permittivity (ε) of 2.


To calculate the total electric field intensity (Et) in the region z, we need to consider the electric field intensity due to the free charges (Efree) and the electric field intensity due to the polarization charges (Epolarization).


The electric field intensity due to free charges is given by the equation Efree = E₁.

Therefore, Efree = 2ax + 3ay + 5az V/m.


The electric field intensity due to polarization charges can be calculated using the equation Epolarization = -∇(P/ε).

Here, ∇ represents the del operator, P is the polarization vector, and ε is the permittivity.

Since the region is a perfect dielectric, the polarization vector (P) is related to the electric field intensity (E) as P = ε₀(ε - 1)E, where ε₀ is the permittivity of free space.

Substituting the given values, we have P = ε₀(2 - 1)(2ax + 3ay + 5az) = ε₀ax + 3ε₀ay + 5ε₀az.

Taking the gradient of P/ε, we get ∇(P/ε) = (∂(P/ε)/∂x)ax + (∂(P/ε)/∂y)ay + (∂(P/ε)/∂z)az.

Since P/ε = (ε₀ax + 3ε₀ay + 5ε₀az)/2, we have ∇(P/ε) = (ε₀/2)ax + (3ε₀/2)ay + (5ε₀/2)az.

Therefore, Epolarization = -∇(P/ε) = -(ε₀/2)ax - (3ε₀/2)ay - (5ε₀/2)az.


The total electric field intensity (Et) in the region z is given by the equation Et = Efree + Epolarization.

Substituting the calculated values, we have Et = (2ax + 3ay + 5az) + (-(ε₀/2)ax - (3ε₀/2)ay - (5ε₀/2)az).

Simplifying the equation, we get Et = (2 - ε₀/2)ax + (3 - 3ε₀/2)ay + (5 - 5ε₀/2)az.

Therefore, Et = (2 - ε₀/2)ax + (3 - 3ε₀/2)ay + (5 - 5ε₀/2)az V/m.


The electric field intensity (