# 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 E_{0} 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 = E_{0} – 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 /ε_{0 }(4)

E_{p} = σ_{p} /ε_{0 }(5)

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

E = σ /ε_{0} – σ_{p} /ε_{0}

Or ε_{0}E = σ – σ_{0}

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

ε_{0}E = D – P

or D = ε_{0}E + 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 E_{0} 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 = E_{0} – 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 /ε_{0 }

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 E_{0} 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 = E_{0} – 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 /ε_{0 }(4)

E_{p} = σ_{p} /ε_{0 }(5)

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

E = σ /ε_{0} – σ_{p} /ε_{0}

Or ε_{0}E = σ – σ_{0}

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

ε_{0}E = D – P

or D = ε_{0}E + P

This is the relation between D, E and P.

_{ }(4)

E_{p} = σ_{p} /ε_{0 }(5)

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

E = σ /ε_{0} – σ_{p} /ε_{0}

Or ε_{0}E = σ – σ_{0}

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

ε_{0}E = D – P

or D = ε_{0}E + P

This is the relation between D, E and P.

dear sir,

will u plz explain me about electric field due to polarisation

Ep= free charge/(k€*A)

where k is relative permitivity and € is permitivity of vaccume

or k*€= $

where $ is the permitivity of medium.

now since Ep is field induced within dielectric then why you have used € in the term Ep???

means you have asumed both external and induced field in vaccume? but in vaccume polarisation of wot will happen?

means electric polarization is zero

plz explain me why have you taken € (permitivity of vaccume) in both Ep and external applied field???

my mail id is

p.aaditya@hotmail.com