1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polarization density in a dielectric?

  1. Mar 2, 2013 #1
    This is polarization density. acd3b9486ed684a80b38dc7984ef1c1c.png Can somebody explain to me why exactly this is so? My book just says that's how polarization is defined.. Can you guys give me an intuitive understanding of why? Only thing I know is that χ is related to κ, the dielectric constant, and κ is related to how well a dielectric polarizes.
     
    Last edited: Mar 2, 2013
  2. jcsd
  3. Mar 2, 2013 #2
    hmm, isn't the formula above equivalent with P = ε0(E0-E), where E = the electric field inside the dielectric, and E0= the electric field without the dielectric?

    Thus P = σi = the induced charge on the dielectric?

    Wow, why would anyone define something so simple and obvious in such a over-complicated way???
     
  4. Mar 2, 2013 #3

    DrDu

    User Avatar
    Science Advisor

    How do you get E0?
     
  5. Mar 2, 2013 #4
    Isn't E=E0/k?
     
  6. Mar 2, 2013 #5

    BruceW

    User Avatar
    Homework Helper

    I think most people are more familiar with the term 'electric displacement field' [itex]\vec{D}[/itex], for which [itex]\epsilon_0 \vec{E}_0=\vec{D}[/itex]
     
  7. Mar 2, 2013 #6

    BruceW

    User Avatar
    Homework Helper

    And strictly, the polarisation density is defined as:
    [tex]\vec{P} = \vec{D} - \epsilon_0 \vec{E} [/tex]
    (which is the second equation which you wrote). And the first equation you wrote:
    [tex]\vec{P} = \epsilon_0 \chi_e \vec{E} [/tex]
    Is only true in a material which is nice and simple. Since you are just starting to learn about this stuff, you can usually assume that the material is nice and simple. It is only when you've got used to the concepts that you will begin to say "what if the material is not nice and simple", and then the polarisation density will no longer be proportional to the electric field. But for now you can assume it is.
     
  8. Mar 2, 2013 #7

    BruceW

    User Avatar
    Homework Helper

    Anyway, the intuitive explanation is that the bound charges in the material react to an electric field in such a way that will reduce the electric field in the material. The most simple model we can use, is to say that the response from the bound charges is the same as the electric field causing it, multiplied by a constant. (And the constant is positive, since the minus sign is already taken into account in the equation which defines the polarisation density).

    In fact, the word 'response' is very good, because the electric susceptibility is an example of a response function. Response functions are widely used in physics, and can be very useful.
     
  9. Mar 2, 2013 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Nikitin! :smile:
    Yes, but it's usually written εoE = -P + D,

    where εoE is the total field, made up of -P (the bound field) and D (the free field)

    (for historical reasons, P is minus the bound field)

    (the εo is only there as a conversion factor because for historical reasons the three types of electric field are described by two different electric units: E is defined as force per charge, while D and P are defined as charge per area)

    i think it's because D is so much easier to calculate (it's just the surface charge density, in coulombs/m2), while P itself is difficult to calculate or measure directly, so it's easier to calculate εoE and D and subtract
     
  10. Mar 2, 2013 #9

    DrDu

    User Avatar
    Science Advisor

    No, at best E=D/k. But D is usually not the field which would be there if the dielectric were absent.
     
  11. Mar 2, 2013 #10

    BruceW

    User Avatar
    Homework Helper

    if the dielectric is the sole contributor of bound charges (and if the dielectric doesn't do anything else), then the D field is the same as the electric field that would be there if the dielectric were absent (except the constant factor of epsilon-0). Right?

    Edit: or I guess you mean that the time-evolution of the system would be different if we did remove the dielectric?
     
    Last edited: Mar 2, 2013
  12. Mar 2, 2013 #11
    What does polarization, P, equal to? Is it simply equal to induced charges?
     
  13. Mar 2, 2013 #12

    DrDu

    User Avatar
    Science Advisor

    No, that's exactly the point why we introduce all these additional fields like D and H.
    The point is that D fulfills different boundary conditions than does E_0. E.g. compare the D field of a sphere which is brought into a constant electric field with the constant field before the dielectric was entered.
     
  14. Mar 2, 2013 #13

    DrDu

    User Avatar
    Science Advisor


    The definition of P is not unique and different conventions are in use.
    For problems which rapidly changing fields one often uses
    ##P=-\int_{-\infty}^{t}j_\mathrm{int}(t') dt'##, where j is the internal current density (either both bound and unbound current density or only bound current density) in the material.
    At lower frequencies it is convenient to split j into a rotational and an irrotational part. The former is taken care off introducing magnetisation, the latter is used in the definition of polarisation.
     
  15. Mar 2, 2013 #14
    Yes, if the field exists in between two parallel metal plates and the charge on those plates stays constant.
    Without a dielectric (in a vacuum or air) D and E are the same. I'm omitting e0 here since it just serves as a unit conversion factor.
    So since D is equal to the amount of charge on the plates divided by their area, it will not change when the dielectric is removed. And without the dielectric E = D, therefore D is the same as E would be without the dielectric.
    By the way, in the (microscopically small) gap between the plates and the dielectric E is also equal to D. That is unless there is electric breakdown occuring in the gap.
     
  16. Mar 2, 2013 #15

    BruceW

    User Avatar
    Homework Helper

    Ah, yeah of course, you are right, now that I think about it. In the case of electrostatics, the electric field will have zero curl, but the D field will have non-zero curl. So we can only say that the divergence of the D field is the same as the divergence of the E field that would be there if the dielectric were absent (except the constant factor of epsilon-0).
     
  17. Mar 3, 2013 #16
    Why would D have a non zero curl in a static situation?
     
  18. Mar 3, 2013 #17

    BruceW

    User Avatar
    Homework Helper

    As DrDu says, this is the whole reason for introducing the D and P fields. We want a field P that is going to be non-zero inside the material and zero outside. And generally, this means the curl of P is going to be non-zero. But the electric field doesn't curl in time-independent situations, so when we write the equation relating P,E and D, we know that the curl of D must equal the curl of P, to ensure that E doesn't curl. So then we know D must have non-zero curl.

    I'm guessing your next question might be: "Why must P have curl, just because it is non-zero inside the material and zero outside?" Well, because the E field has no curl, this means the tangential component of the E field at the boundary must be continuous. But for the P field, we don't want to make this restriction. We will often want it to be discontinuous at the boundary. For example, for the simple material, if the E field is parallel to the boundary, then we want the P field (inside the material) to be parallel to the boundary too. But outside the material, we want the P field to be zero. So we want the tangential component of the P field at the boundary to be discontinuous. Therefore, we must allow the P field to curl.
     
  19. Mar 3, 2013 #18
    In the case of plate-capacitors, V=V0/k => E=E0/k when q=constant.

    How about for static fields? I've just begun learning about condensers.
     
  20. Mar 3, 2013 #19

    Jano L.

    User Avatar
    Gold Member

    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:

    $$
    \mathbf P = \mathscr{N} \langle \boldsymbol \mu \rangle
    $$

    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.
     
  21. Mar 3, 2013 #20

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    For me this is the most intuitive way to understand this. Hopefuly the OP has learned about simple electic dipoles. Here is the way I think of it:

    First a definition. If you have a charge +q and a charge -q, separated by [itex]\mathbf{d}[/itex] (a vector that starts at -q and ends at +q) then the dipole moment is [itex]\mathbf{p} = q \mathbf{d}[/itex]. Now when we consider a dielectric media, there are molecules that become distorted when an external field is applied, and the distortion can be modeled by a dipole.

    If we look on a microscopic scale, the field and dipoles make a rapidly varying (in the spatial sense) electric field in the vicinity of each molecule, in fact we usually do not care about the fluctuations on these tiny scale sizes. Instead, we are usually interested in macroscopic behavior. So one approach is to average the fields, and the polarization vectors, over volumes that contain many of the elementary dipoles and yield (approximately) continuous vector fields, but are still small enough volumes so that the scale sizes we typically are interested in are preserved. When we do this, we obtain what are usually called the macroscopic Maxwell equations. The terms associated with the dielectric media involve this average polarization density,
    [tex]
    \mathbf{P}(\mathbf{r}) = \frac{1}{\Delta V} \sum_{\mathbf{r}_i \, in\, \mathbf{r}+\Delta V} \mathbf{p}(\mathbf{r}_i)
    [/tex]
    Note that due to the sign convention for electric dipoles, [itex]\nabla \cdot \mathbf{P} = -\rho_b[/itex] where I have let [itex]\rho_b[/itex] represent bound charge density. There is nothing fictitious about these charges, they are really there. To convince yourself of the sign, consider a bunch of elementary dipoles all with their -q charges at the origin (say), and all of the [itex]\mathbf{d}[/itex] vectors pointing radially outward. Clearly this configuration has positive divergence at the the origin, where there is a bunch of negative charges.

    Also, [itex]\frac{\partial \mathbf{P}}{\partial t}[/itex] represents a current. To understand this think about a single dipole. if the time derivative is positive, it means either the charges are increasing (meaning positive charge must be moving from the -q location to the +q location) or the vector [itex]\mathbf{d}[/itex] is increasing, which simply means (for example) the +q charge is moving away from the -q charge, so is a positive current.

    Hence, to include dielectric media into our macroscopic Maxwells equations (again, where we usually think of all quantities as having been averaged) we obtain,
    [tex]
    \begin{eqnarray}
    \nabla \cdot \epsilon_0 \mathbf{E} & = & \rho - \nabla \cdot \mathbf{P}. \\
    \nabla \times \mu_0^{-1} \mathbf{B} & = & \mathbf{J} + \frac{\partial }{\partial t}\epsilon_0 \mathbf{E} + \frac{\partial }{\partial t} \mathbf{P}.
    \end{eqnarray}
    [/tex]

    If we define [itex]\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}[/itex] then these become the more familiar,
    [tex]
    \begin{eqnarray}
    \nabla \cdot \mathbf{D} & = & \rho. \\
    \nabla \times \mu_0^{-1} \mathbf{B} & = & \mathbf{J} + \frac{\partial }{\partial t} \mathbf{D}.
    \end{eqnarray}
    [/tex]

    I hope that helps a little!

    jason
     
    Last edited: Mar 3, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook