In a dielectric, what is the relation between the displacement field D, the electric field E, and the polarization P?

In a dielectric, what is the relation between the displacement field D, the electric field E, and the polarization P?

• A. D = ε0 E
• B. D = ε0 E + P
• C. D = E/ε0
• D. D = μ0 H

Answer: (B)

The displacement field reflects how the material’s bound charges produced by polarization affect the electric field inside the dielectric. It’s defined so that Gauss’s law looks like ∇·D = ρ_free, separating the free charges from the bound charges associated with polarization. This leads to the relation D = ε0 E + P, where ε0 E is the vacuum contribution and P is the polarization density (dipole moment per unit volume). In a linear dielectric, P is proportional to E (P = ε0 χe E), giving D = ε0 E + ε0 χe E = ε0(1 + χe)E = εE, with ε = ε0 εr. So the displacement includes both the direct field and the material’s response through polarization. The other forms don’t capture this combination: using only ε0 E neglects polarization, D = E/ε0 misstates the units and relationship, and D = μ0 H is a magnetic-field relation, not electric.

The displacement field reflects how the material’s bound charges produced by polarization affect the electric field inside the dielectric. It’s defined so that Gauss’s law looks like ∇·D = ρ_free, separating the free charges from the bound charges associated with polarization. This leads to the relation D = ε0 E + P, where ε0 E is the vacuum contribution and P is the polarization density (dipole moment per unit volume). In a linear dielectric, P is proportional to E (P = ε0 χe E), giving D = ε0 E + ε0 χe E = ε0(1 + χe)E = εE, with ε = ε0 εr. So the displacement includes both the direct field and the material’s response through polarization. The other forms don’t capture this combination: using only ε0 E neglects polarization, D = E/ε0 misstates the units and relationship, and D = μ0 H is a magnetic-field relation, not electric.