# Relation between polarization vector (P), displacement (D) and electric field (E)

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 (Ep). (You can see the figure in that article).

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

E = E0 – Ep (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 = qb/A = σp (2)

Where qb is bound 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 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)

Ep = σp0 (5)

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

E = σ /ε0 –  σp0

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.

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 (Ep). (You can see the figure in that article).

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

E = E0 – Ep (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 = qb/A = σp (2)

Where qb is bound 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 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 (Ep). (You can see the figure in that article).

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

E = E0 – Ep (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 = qb/A = σp (2)

Where qb is bound 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 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)

Ep = σp0 (5)

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

E = σ /ε0 –  σp0

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)

Ep = σp0 (5)

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

E = σ /ε0 σp0

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.

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