Electromagnetic fields and waves

Discussion in 'Homework Help' started by mentaaal, Oct 31, 2009.

  1. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Hey guys, I was hoping someone could clear up what seems to me to be an apparent contradiction in my course notes, let me explain:

    From the very beginning of our notes on electromagnetic fields and waves, the electric flux intensity vector E is defined to be the force a charge q would experience when in the presence of this field such that E is equal to kq1/r^2 where k is a constant which is dependent on the medium that the field is in and r is the distance between q and q1. It was found that the total flux of E ψE is equal to 4∏kq. A new term ε (permittivity) is defined to be 1/(4∏k) such that the electric flux density vector D is equal to εE. This means that the total flux of D ψD is q.

    From the above, my understanding was that the D field was independent of the medium through which is travelled. My notes then go on to describe "the distribution of flux through non uniform media" which uses as example of two large blocks of different permittivities ε1 and ε2. In this example, the problem that I have with it is that the the E field in medium 1 - E1 is equated to E2 in medium 2 such that E1 = E2 = E. All further derivations in this section stem from this. From what I have mentioned above, I dont see how it is possible for E1 to have the same intensity as E2 if the media have different permittivites? I would have thought that a good starting point would have been to equate the D vectors instead.
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    You must be misunderstanding the problem, or the problem is a special case. Feel free to post the problem if you want further clarification. Study material on electrostatic boundary conditions. There is the famous "pillbox" derivation which shows that the tangential E field is continuous across the boundary, and the normal D field is continuous across the boundary, if there is no surface charge.

    The statement you made is clearly wrong if you consider the case of a perfect conducting sphere. All electric field lines meet the sphere normal to the surface, and it is well known that the electric field in a perfect conductor is zero. The field is normal at the surface because the tangential electric fields must be equal inside and outside, and the field inside is zero. Hence, the electric field outside the sphere must also be zero at the boundary. The normal component of electric field can be nonzero, since E does not have to be continuous.
     
  3. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Hi Steve, yes I agree with everything you say, hence my confusion!

    As you kindly suggested that I post the problem, here are the notes that are causing me hardship:

     
  4. steveb

    Senior Member

    Jul 3, 2008
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    So, this appears to be a special case. The electric field at the boundary of the two media is always tangential, except precisely where the point charge is. Hence, it is correct to say that the electric field E is uniform in this particular case.

    I now see what you are really asking here, and it's a good question. Once you answer this to yourself, you take one of the many steps needed to understand field theory.

    I would say that your confusion stems from your assumption that D is unaffected by the medium, in general. This assumption is too much of a simplification. All you can really say is that the sum total effect of integrating D over the surface is independent of the medium since this is equated to enclosed charge which is constant. As we see in this example, it's possible for E to be uniform and D to be nonuniform. There are other examples in which both E and D are non-uniform.
     
  5. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Ok well at least you understand me so thats a start! Whilst I realise the need to work some things out on my own, I have tried putting some thought into this and cannot reconcile it. I have learnt that E depends on a constant K which is dependent on the medium and D does not depend on this constant by our choosing not to be. If there is more complexity t this I have not seen it so far in the course of my studies. In the notes I have provided above, the continuity of the E vectors are merely stated without any clue as to why this is so. Would you mind giving me a nudge as to WHY the magnitude of the E vectors are the same? What adds to my confusion is that just a page before this in my notes, in a similar example the "pill box" you mentioned earlier it was shown that the normal components of D at an interface between two different media were continuous. I realise that this case is different as there is a point charge on the boundary but I dont see why it should be different.
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    OK, to start, lets realize that it's not correct to say that E depends on K and D does not depend on K. The factor K, or more traditionally epsilon, is the ratio between the two (note this is true for isotropic materials only). Both E and D depend on a total solution of the problem, which varies from case to case.

    In general, determining the total spatial distribution of E and D is not necessarily easy, although some simple problems with symmetry are easy. The total solution depends on the charges, materials and boundary conditions. If we restrict ourselves to electrostatic problems like this, we have Gauss's law for electric fields, and the simplified version of Faraday's Law that says that the line integral of electric field around a closed path is zero. These two equations allow you to solve all electrostatic problems.

    From these equations you can derive the important boundary conditions as follows.

    1. Tangential component of E is continuous across a boundary.

    2. Normal component of D is continuous across a boundary, unless there is a surface charge distribution in which case it is discontinuous by the surface charge density.

    If any of the above does not make sense, you need to come to terms with this through study. This is all contained in any fields book or can be tracked down on the internet.

    Once the above makes sense, you can look at simple examples.

    1. Spherically symmetric geometry with a point charge and a sphere with permitivity  \epsilon_1 , surrounded by a medium with permitivity  \epsilon_2 .

    Here the D vector is normal to the surface of the sphere, and there is no surface charge density on the sphere. Hence D is continuous across the boundary of the sphere. This case shows that D is not affected by the boundary, while E is affected by the change in material across the boundary.

    2. The example you provided has the boundary cutting the sphere is half.

    Here the electric field is tangential to the boundary surface everywhere. Hence E is continuous across the boundary plane. This case shows that E appears unaffected by the material, but actually its magnitude is affected because the surface integral of D is Q, and E must be scaled to make this true. However, E appears uniform over a spherical surface, while D does not.

    If you take more complicated cases with boundaries that are not symmetrical, you'll find that both E and D vary in complicated ways. The general rules that surface integral of D is Q and that D=\epsilon E, at any point, are still valid though.

    Here is a link that shows the derivation of the boundary conditions.

    http://www.cems.uvm.edu/~oughstun/LectureNotes141/Topic_13 (EstaticBoundaryConditions).pdf
     
    Last edited: Oct 31, 2009
  7. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    If you aren't lecturing this somewhere, you should be! Yes after reading and re-reading your post and looking up a chapter on boundary conditions this problem is much easier to grasp. I did not realise that the E components were tangential and in my notes, our "pillbox" example lacked the detail of the surface charge density at the interface.

    Thanks very much for your time in writing that answer which I will be enclosing in my notes on the topic! You cleared the problem right up!
     
  8. steveb

    Senior Member

    Jul 3, 2008
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  9. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    I intend on watching them all thank you.
     
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