Question

If D=4 aₓ-6 aᵧ+8 a nC/m² in the region y<0 where ε=2.4 ε₀, find D in the region y>0 where x=9 ε₀. Assume that a surface charge density of -3 nC/m² exists on the interface.

Answer 1

To find D in the given region, calculate the electric field using the surface charge density and permittivity. Then, use the electric field to find the value of D in the specified region.

Step-by-step explanation:

To find the value of D in the region where y > 0 and x = 9ε₀, we need to consider the relationship between electric displacement (D) and electric field (E). In the region y < 0, we know that D = 4aₓ - 6aᵧ + 8anC/m², and we are given ε = 2.4ε₀. Since the surface charge density on the interface is -3nC/m², we can calculate the electric field using the equation E = σ / ε, where σ is the surface charge density and ε is the permittivity of the medium.

Substituting the given values, we have E = -3nC/m² / (2.4ε₀). Now, in the region where y > 0 and x = 9ε₀, we can use the electric field we just calculated to find the value of D. Since D = εE, we have D = ε(Ex, Ey, Ez). We know the values of Ex and Ez are 0, and Ey is equal to the electric field we calculated. Therefore, D = (0, -3nC/m² / (2.4ε₀), 0).