Answer to Question #104213 in Electricity and Magnetism for Akshay

Question

Define electric displacement vector
D

and deduce Gauss’s law for dielectrics

Answer 1

In a dielectric material, the presence of an electric field E causes the bound charges in the material (atomic nuclei and their electrons) to slightly separate, inducing a local electric dipole moment. The electric displacement field "D" is defined as

D≡$\epsilon$₀E+P

where ϵ₀ is the vacuum permittivity (also called permittivity of free space), and P is the (macroscopic) density of the permanent and induced electric dipole moments in the material, called the polarization density.

The displacement field satisfies Gauss's law in a dielectric:

∇ ⋅ D = ρ - ρ_b = ρ_f

Consider a dielectric material kept in an electric field is polarized. It has bound or polarization charge density ρb due to accumulation of bound charges 𝜌𝑏 = −∇𝑃. The applied electric field itself is created by transferring electric charges.They are called as free charges (charges brought from outside e.g. conduction electrons in metals).They give rise to ρf the charge density due to free charges i.e. which is not due to polarization. The total charge density ρ consists of two parts as follows 𝜌 = 𝜌𝑓 + 𝜌𝑏. According to Gauss' law in differential form ∇𝐸 = 𝜌 𝜀0 = 𝜌𝑓 + 𝜌𝑏. 𝜀0 E is the total electric field due to both types (bound and free) charges. and Rearranging the expression and substituting 𝐷 = 𝜀𝐸 + 𝑃 . This is the differential form of Gauss Law in dielectrics.