Let us derive the relation between polarization vector (P), displacement (D) and electric field (E):

In the last article of polarization, we have discussed about the effect on dielectric placed in an external electric field E₀ and there will be electric field due to polarized charges, this field is called electric field due to polarization (E_p). (You can see the figure in that article).

Rewrite equation (1) of that article, that is:

E = E₀ – E_p (1)

Polarization vector, P = P is equal to the bound charge per unit area or equal to the surface density of bound charges (because surface charge density is charge per unit area),

Thus P = q_b/A = σ_p (2)

Where q_b is bound charge and σ_p 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 number of molecules per unit volume.

Displacement vector, D= D is equal to the free charge per unit area or equal to the surface density of free charges,

Thus D = q/A = σ (3)

where q is free charge and σ is surface density of free charges.

As for parallel plate capacitor (already derived in earlier articles):

E = σ /ε₀

Let us derive the relation between polarization vector (P), displacement (D) and electric field (E):

In the last article of polarization, we have discussed about the effect on dielectric placed in an external electric field E₀ and there will be electric field due to polarized charges, this field is called electric field due to polarization (E_p). (You can see the figure in that article).

Rewrite equation (1) of that article, that is:

E = E₀ – E_p (1)

Polarization vector, P = P is equal to the bound charge per unit area or equal to the surface density of bound charges (because surface charge density is charge per unit area),

Thus P = q_b/A = σ_p (2)

Where q_b is bound charge and σ_p 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 number of molecules per unit volume.

Displacement vector, D= D is equal to the free charge per unit area or equal to the surface density of free charges,

Thus D = q/A = σ                                                                                 (3)

where q is free charge and σ  is surface density of free charges.

As for parallel plate capacitor (already derived in earlier articles):

E = σ /ε₀ (4)

E_p = σ_p /ε₀ (5)

By substituting equations 4 and 5 in equation 1, we get

E = σ /ε₀ –  σ_p /ε₀

Or ε₀E =  σ  – σ₀

By putting equations 2 and 3 in above equation, we get

ε₀E  = D – P

or D = ε₀E  + P

This is the relation between D, E and P.

(4)

E_p = σ_p /ε₀ (5)

By substituting equations 4 and 5 in equation 1, we get

E = σ /ε₀ – σ_p /ε₀

Or ε₀E = σ – σ₀

By putting equations 2 and 3 in above equation, we get

ε₀E = D – P

or D = ε₀E + P

This is the relation between D, E and P.