Clarification required: polarization density and displacement field

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Hi!

I can't wrap my head around the formula [tex]\textbf{D} = \epsilon \textbf{E} + \textbf{P}[/tex].

The electric field, [tex]\textbf{E}[/tex], goes through the dielectric. It polarizes the atoms/molecules, which creates a field going the opposite way, which cancels out at least part of [tex]\textbf{E}[/tex] (right?). Now, for some reason textbooks say that this polarization field [tex]\textbf{P}[/tex] is defined as pointing from negative to positive, thus going the same way (with, not against) the original field.

If [tex]\textbf{D}[/tex] is supposed to be the resulting field in the dielectric, why are they added and not subtracted? Or am I misinterpreting the point of [tex]\textbf{D}[/tex], and it really stands for the original field plus whatever the polarization reduces?

Thanks!
Seb
 

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  • #2
tiny-tim
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Welcome to PF!

Hi Seb1! Welcome to PF! :smile:

(have an epsilon: ε and try using the B tag just above the Reply box :wink:)
If [tex]\textbf{D}[/tex] is supposed to be the resulting field in the dielectric, why are they added and not subtracted? Or am I misinterpreting the point of [tex]\textbf{D}[/tex], and it really stands for the original field plus whatever the polarization reduces?
No, you're right :smile:

there's free charge and bound charge, and together they make the total charge,

corresponding to the free field D the bound field P together making the total field E

… so (apart from factors) E is D plus P, except that for historical reasons P is defined as minus P (if you see what I mean :wink:), so it comes out as εE = D - P :smile:
 
  • #3
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P isn't the same as the depolarization field (i.e. the electric field created by the polarization charges). The unit of D, epsilonE and P is C/m2, which for P can be read as dipole momentum per volume unit (Cm/m3).

The strength of the electric field generated by P's polarization charge will depend not only on P, but also on the geometry. For example, if a sphere and a plane has the same polarization P per unit volume (the polarization of the plane assumed perpendicular to the plane) then the depolarization field, i.e. the electric field generated by the surface charges, will be three times as strong inside the plane as inside the sphere. So the formula [tex]\textbf{D} = \epsilon \textbf{E} + \textbf{P}[/tex] doesn't explicitly include the depolarization field, only the polarization (i.e. charge separation).
 
  • #4
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Thanks Tim!

So D is the field imposed on the dielectric by the free charges in, say, the capacitor; P is the field created by the polarization surface charges; and E is the resulting net field in the dielectric.

Since P is defined as going from negative to positive, the net field is E = D - P, or in other words, D = E + P. Correct?

Shoestring, is this just a consequence of the fact that polarization causes surface charges, and those are closer together on two sides of a plane than on two sides of a sphere?

Thanks again guys :)
 
  • #5
tiny-tim
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hi sk21! :smile:
So D is the field imposed on the dielectric by the free charges in, say, the capacitor; P is the field created by the polarization surface charges; and E is the resulting net field in the dielectric.

Since P is defined as going from negative to positive, the net field is E = D - P, or in other words, D = E + P. Correct?
yup! :biggrin:

(except … i've never like the use of the word "net" in this context … i prefer "total" …

"net", according to my dictionary, and as distinct from "gross" etc, means "left after all deductions", whereas E is a total, with no deductions :wink:)
 
  • #6
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Shoestring, is this just a consequence of the fact that polarization causes surface charges, and those are closer together on two sides of a plane than on two sides of a sphere?
The depolarization field is derived from the surface charges caused by the polarization, but it's not just a question about distance. The scale doesn't matter, only shape.

For example, if you compare a sphere and a plane having the same diameter and P, there'll be more surface charges on the plane compared to on the sphere, both because of the extension of the plane, but also because of the perpendicular angle between P and the surface. And if you compare a huge sphere and a plane, both having the same polarization P, the charges on the sphere will be further apart, and on most of the surface they'll also be smaller than on the plane, because of the cosine factor. (The surface charge is given by the scalar product of P and the normal vector n.)
 
  • #7
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I'm not sure about this, but I think the whole point of making the sum [tex] \epsilon \textbf{E} + \textbf{P}[/tex] has something to do with how it (or rather its time derivative) relates to the curl of the magnetic field. The unit of the time derivative of P is that of current density (C/s)/m^2, and the curl of the magnetic field depends on both the current density and the time derivative of the electric field.
 

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