electric flux density and electric field

Discussion in 'Homework Help' started by PG1995, Aug 15, 2012.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi

    Would you please help me with the query included in the attachment?

    For high resolution image, please use this link: http://img825.imageshack.us/img825/4709/electricfluxdensity3.jpg

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    Please help me. Thank you.

    Regards
    PG
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    You pretty much answered the question yourself. You were careful to point out that these are two different and distinct documents and you noted that they used different symbols. Probably your confusion stems from the fact that they refer to two different things as "electric flux", but "flux" is a general term and can apply to many things.

    In general, be careful of definitions. People will often refer to different things using the same symbol or the same words. But the definition tells the true story. Although these two forms of flux are very closely related, they are not exactly the same thing.

    In my opinion, document 2 is closer to the traditional definitions and presentation. But, these are just matters of convention, not substance.
     
  3. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, Steve.

    I'm sorry. Though, as you say, I have already answered it myself indirectly, it's still confusing me. I have added another question in the attachment. Please have a look on the attachment and kindly help me. Thank you.

    For higher resolution image please use this link: http://img825.imageshack.us/img825/1961/electricfluxdensity3x.jpg
    If you can't see the image, then please use this username: imgshack4every1, and password: imgshack4every1. Thanks.

    Regards
    PG
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    The place where you have it wrong is where you say D=E.

    This is only true in a system of units where ε=1, and even then only in vacuum. Generally, E and D are different things. Neither document says that D=E, so you must have arrived at this by mixing the two documents incorrectly. It all comes back to being careful about definitions. Don't worry about what something is called, or what symbol is used to represent it. Instead worry about the nature of the object in question, as revealed by the fundamental definition. This is the difference between deep knowledge and superficial knowledge.
     
  5. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi

    Now I see where I had it wrong. But part of the blame for this confusion goes to Document #1 here which used 'electric flux' terminology for something which is quite different from the 'usual' electric flux as described in many other texts. Both fluxes use different units, N.m^2/C and C. You also pinpointed this here:
    Could you please help me with the query included in the attachment? Thanks a lot for always being there to help.

    Best wishes
    PG
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    As far as I can see, you are correct that it is a unit vector. This would then be a unit vector in the radial direction of a spherical coordinate system. This vector points outward from the origin, no matter where you are in 3D space. This makes sense since you are basically looking at Coulombs law which always produces an electric field radially outward from a point source. The new vector D (flux density) is then defined such that the integrated value over a closed surface is equal to the enclosed charge (which is Gauss' Law). In vacuum, the constant epsilon sub zero relates the two vectors \vec D=\epsilon_0 \vec E .

    For example,

    at the point x=1, y=0, z=0, ar is pointing in the positive x-direction.

    at the point x=0, y=1, z=0, ar is pointing in the positive y-direction.

    at the point x=0, y=0, z=1, ar is pointing in the positive z-direction.

    at the point x=1, y=1, z=0, ar is pointing in the xy-plane away from the origin with a slope of 1.

    The use of rectangular, cylindrical and spherical coordinates is common in electromagnetics because it allows solutions for problems with rectangular, cylindrical and spherical symmetry to be solved more easily. Hence, you need to get used to these types of varying unit vectors that show up in cylindrical and spherical coordinates.

    The notation "a sub r" ( a_r) seems uncommon to me and usually we see "r" with a hat over it (\hat r), as in this following document.

    http://www.polymerprocessing.com/notes/root91a.pdf
     
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  7. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi Steve

    I'm sorry for asking about it now. Some days ago I believe I read in some textbook the same thing that ε=1 because the way it has been defined. But somehow I can't locate that page in any of the textbooks where it said so. This link also says the same thing:



    [​IMG]



    Could you please tell me how and when the permittivity of free space would be unity? Thanks a lot.

    Regards
    PG
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    This happens with Gaussian units, also called cgs units.

    http://en.wikipedia.org/wiki/Gaussian_units

    Actually, according to this reference, perhaps it is better to say that the εo does not even exist in Gaussian units.

    Please note that your reference is talking about a different constant of proportionality being equal to one. In SI units permittivity is not one, but that flux to charge scale factor is one. However, in Gaussian units, permittivity is one, but the flux to charge scale factor is 4∏
     
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  9. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you.

    I'm sorry but I couldn't get what you really meant by the line in bold? Could you explain it a bit? I hope it's not too obvious and self-explanatory!

    Kind regards
    PG

    PS: The Wikipedia does talk about it too.


    [​IMG]
     
    Last edited: Aug 31, 2012
  10. steveb

    Senior Member

    Jul 3, 2008
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    In your reference you seemed to be saying that the expression ψ=Q was the same as ε=1, but this is a different scale factor. The fact that ψ=Q in SI units is a nice thing, but is different than the nice thing that ε=1 in Gaussian units. In Gaussian units ψ does not equal Q, but there is a 4pi scale factor between them.

    I just wanted to make sure you were not confusing the two different things.

    By the way, stick with SI units while you are learning. This units thing is very very confusing in electromagnetics and much worse than in mechanics. There are many benefits to Gaussian units, but they are mostly important from a theoretical side for physicists. For an engineer and all practical problems, SI units are exclusively used. Even in physics, people are switching over. The latest edition of the very famous book by Jackson switched over to SI units. In my view this was a mistake to change a great classic text, but it shows just how prevalent the SI system is.
     
    Last edited: Aug 31, 2012
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  11. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Once again, my thanks for the help.

    In Sear's and Zemansky' book it is:

    \Phi _{E}=\frac{Q_{encl}}{\epsilon _{0}}

    I don't see how in that reference I quoted previously they have \psi =Q. I mean what happens the permittivity. I hope I'm not sounding that dumb as usual.

    By the way, where you referring to this book? As per Amazon's listing, there is no recent edition, the last one was released in 1998. Just curious! :)

    Best wishes
    PG
     
    Last edited: Sep 17, 2012
  12. steveb

    Senior Member

    Jul 3, 2008
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    In SI units, the units of D are such that when you calculate flux by integrating over a closed surface you get the total enclosed charge. This rule is called Gauss' Law and in SI units it looks as follows.

    \Psi=\oint \vec D \cdot d\vec s = Q

    The permittivity shows up in the equation \vec D=\epsilon_0 \vec E

    In other words \Phi_E is not the same as \Psi, and in vacuum \Psi=\epsilon_0 \Phi_E.



    Hahaha, I think you just made a very good joke, whether intentional or not. Yes, that's the one. I'm old, so that 1998 edition is recent to me. I used the 2'nd edition in school, published in 1975.

    This book is a classic graduate level text - hated by some and loved by some. The problems are very challenging, but the writing and other content is very illuminating.
     
    Last edited: Aug 31, 2012
  13. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Actually I did look for the 4th edition. I thought perhaps it's there but Amazon doesn't have it. Then, after some googling it was confirmed that the 3rd edition was the last one. Then, I got curious to know that how come you consider the 1998's edition a recent one! :) And I don't think being about 50 years old is old enough to classify yourself as old. So, cheer up! You need to wait for some more years!:D Perhaps, there is some issue with your biological clocks!;)

    Best wishes
    PG
     
    Last edited: Sep 1, 2012
  14. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    Could you please help me with the queries? Thank you.

    Regards
    PG
     
  15. steveb

    Senior Member

    Jul 3, 2008
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    Ok to be honest there are a great number of conceptual mistakes here. I can try to track down a source that gives a comprehensive viewpoint.

    First D is flux density and it is a vector like quantity that represents the outward flow of the field due to charge. Don't worry that you feel that the field E is the better quantity to base flux on. Eventually you will appreciate why D is more convenient.

    Note that D is NOT surface charge density. Charge density, usually denoted as the Greek letter rho, surface charge density, sigma, and linear charge density are scalar functions that have no direction.

    There are other errors too, but let come back to this after I figure out a good viewpoint and reference to present.
     
    Last edited: Sep 1, 2012
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  16. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you. This related thread would help you to help me. Thanks.

    Regards
    PG
     
  17. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you very much, Steve, for getting along with me so patiently to help me.

    As you know I'm new to this electromagnetic stuff so I think it won't hurt me if 'wrong understanding' of some concepts helps me a bit to overcome conceptual obstacle and as I learn more about the electromagnetism these 'wrong understandings' will get corrected themselves. Please have a look on the attachment and kindly help me with the queries. Thanks a lot.

    Regards
    PG
     
  18. steveb

    Senior Member

    Jul 3, 2008
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    OK, first lets correct a critical error in your description. You describe dA and ds as being area elements that are pointing in the same direction as E and D. This is not correct. They are pointing in a direction normal to the surface at that point. Then the integral is done by doing a dot product of the two vectors E and dA (or D and ds).

    For Q1,

    The difference between D and E can be looked at from different perspectives. I think at this level of study, it's useful to think of D as capturing the effects of matter. I say this because E is more fundamental and in vacuum D is not really needed since it is related to E through a constant scale factor called the permittivity of free space. In free space both vectors always point in the same direction.

    Once you consider matter interactions, D becomes a very useful object, and the permittivity becomes a second rank tensor, with D and E not necessarily in the same direction, in general. In media, D=εo E + P, where P is the polarization of the medium. In other words, by using D, we can add in the D contribution from matter and add it to the D contribution from vacuum. You will study polarization in detail at some point, and then this will make more sense. We usually bury the P into obscurity be defining a relative permittivity for the medium εr, which results in the following.

    \vec D=\epsilon_o \vec E+ \vec P=\epsilon_o \epsilon_r \vec E

    It is too early for you to worry about this, but mathematically D and E can be shown to be different types of objects under the more general general mathematics of differential geometry. Here E is a one-form, and D is a two-form. Vector calculus in 3 dimensions tries to simplify these objects and modifies them into standard vectors, but fundamentally they are very different types of geometrical objects. You don't need to worry about this, but keep the knowledge in the back of your head, and it will be useful in more advanced study.

    For Q2,

    It is important to understand the fundamental nature of and differences between various geometrical objects. Just because entities have the same units, doesn't mean they are the same thing. We spoke about Gaussian units before, and it turns out that D,E,H and B all have the same fundamental units (in that system), even though we call the units different names to keep the context clear.

    Certainly, D contains information about the enclosed charge, as indicated by Gauss' Law, but D is definitely not surface charge density. Actually, surface charge density is an artificial construct that we use to describe very thin layers of charge. Rather than trying to integrate volume charge density over a very thin shell, we pretend that the shell is infinitely thin and just do a surface integral. So, the closed loop integral of D dot ds is related to the volume charge density, and this can be put into a differential form (point relation) as follows.

    \nabla \cdot \vec D=\rho

    where ρ is the volume charge density. Clearly, D and rho are not the same thing at all, since there is a derivative type of operator acting on D. The fact that we might squeeze charge into a shell (for convenience) and call it surface charge density, does not change this concept at all.
     
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