Correct Answer - Option 3 : D - P = ε
0E
Electric displacement:
Electric displacement, denoted by D, is the charge per unit area that would be displaced across a layer of conductor placed across an electric field. It is also known as electric flux density.
Electric displacement is used in the dielectric material to find the response of the materials on the application of an electric field E.
In Maxwell’s equation, it appears as a vector field.
The SI unit of electric displacement is Coulomb per meter square (C m-2).
Mathematically relation is given by
D = ε0E + P.
where,
ϵ0 = Vacuum permittivity,
P = Polarization vector,
E = Electric field,
D = Electric displacement field
Derivation:
We know that the effect on dielectric placed in an external electric field E0 and an electric field due to polarized charges, this field is called electric field due to polarization (Ep) in relation with electric field vector E.
E = E0 – Ep ...............(1)
Polarization vector, P is equal to the bound charge per unit area or equal to the surface density of bound charges (because surface charge density is a charge per unit area),
Thus \(P = {q_b\over A} = σ_p\) .............(2)
Where qb is bound to charge and σp is surface density of bound charges.
P is also defined as the electric dipole moment of material per unit volume.
P = np
where n is the number of molecules per unit volume.
Displacement vector, D is equal to the free charge per unit area or equal to the surface density of free charges,
Thus \(D = {q\over A} = σ \) ...........................(3)
where q is free to charge and σ is surface density of free charges.
As for the parallel plate capacitor:
\(E = {σ \over ε_0} \)................(4)
\(E_p = {σ_p \over ε_0}\) ..............(5)
By substituting equations 4 and 5 in equation 1, we get
\(E = {σ \over ε_0} – {σ_p \over ε_0}\) or ε0E = σ – σ0
By putting equations 2 and 3 in the above equation, we get
ε0E = D – P
