Polarization density in a dielectric?

In summary: So the formula P = σi = the induced charge on the dielectric? is incorrect.In summary, polarization density is defined as the response of bound charges in a material to an electric field, which is represented by the equation P = D - ε0E. This can also be written as P = ε0χeE, but is only true for simple materials. It is related to the dielectric constant κ, which affects how well a material can polarize. The electric displacement field D and the dielectric constant κ are more commonly used to describe this phenomenon, and the polarisation density is considered a response function. The formula P = σi is incorrect, as P is not equal to the induced charge on
  • #36
Guys, OP here. Can somebody please clear this up for me?

I made this post:
Electric dipole moment it measures the strength of the dipole, right?

But what practical conclusion, in this case, can you draw from multiplying the dipole moment with the particle density, other than measuring how polarized the dielectric is?

In some cases the polarization of the dielectric equals its induced charge density. Can you please explain to me why this is so, and in what situations it is so?

As an answer to Janu's post:
One important definition of polarization which was not mentioned so far is that it is the total electric dipole moment of neutral set of molecules divided by the volume they occupy, or, which is the same thing, number density of molecules times their average dipole moment:

P=N⟨μ⟩This applies well to dielectric media (the definition is unambiguous).

In case the medium is conducting, like metals, this concept of polarization does not apply, because there are no neutral molecules.
 
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  • #37
As Jano L has written in #29, ##\nabla \cdot P=-\rho##, which is the relation between the polarisation and the induced charge density.
 
  • #38
Ah, thanks. I appreciate your help, but what exactly does the formula mean?

I learned from Calculus 2 recently that the gradient of a function F(x,y,z) is the direction in a 3D-area in which F(x,y,z) grows the most. How does this tie with P? Geometrically, when is ∇P = ρ? Note: I haven't learned about vector-fields yet.

Could you please explain this to me?
 
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  • #39
It's not the gradient of P but its divergence ##\nabla\cdot P=\partial P_x/\partial x+\partial P_y/\partial y +\partial P_z/\partial z##. I. e. the difference of the polarization flowing in some volume and the one flowing out is minus the bound charge density.
 
  • #40
So the difference between the polarization flowing out and in per volume is the bound charge density? Is there an intuitive explanation for this?

Sorry for being a bit incompetent in the math department - we have just started on gradients, and have not learned about vector fields yet.

PS: Since you're quite skilled in physics, would you mind assisting me here, too? https://www.physicsforums.com/showthread.php?t=676590
 
  • #41
Nikitin said:
So the difference between the polarization flowing out and in per volume is the bound charge density? Is there an intuitive explanation for this?

the difference between the total electric field (E) flowing out and in per volume is the total charge density: divE = ρtotal

for convenience, we can (and do) arbitrarily split the total charge into the free charge plus the bound charge: ρtotal = ρfree + ρbound

we then define the free (D) field as being that produced by the free charge, and the bound (-P) field as being that produced by the bound charge: divD = ρfree, -divP = ρbound

(i've left some constants out)
 

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