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Homework Help: Displacement current and conduction current

  1. Jun 18, 2010 #1
    1. The problem statement, all variables and given/known data

    conductivity=1/100 siemens/meter
    at which frequency displacement current is equal to conduction current?

    2. Relevant equations
    The conduction current is LaTeX Code: I =C * \\frac {dV}{dt}

    The displacement current D can be calculated from

    LaTeX Code: \\frac{\\partial D}{\\partial t} = - \\frac{\\epsilon}{w} * \\frac{dV}{dt}

    3. The attempt at a solution

    can not attempt using these formulae.
  2. jcsd
  3. Jun 18, 2010 #2
    Let's look at a conducting element of length [tex]\ell[/tex] and cross section A:


    For simple AC current across a conducting element: [tex]V=V_0 \cos {\omega t}[/tex]

    The electric field across the conductor is proportional to the voltage: [tex]E\ell = V[/tex]

    From here on out, you can find the displacement current and compare it with the previous result.

    My result was (Looking only at the absolute values of the currents, disregarding the phase difference between the two):

    [tex]\omega _c \approx 3.76\cdot 10^8 [sec ^ {-1}][/tex]

    Which agrees well with common sense that says that for most frequencies that we deal with we don't concern ourselves with the displacement current at all.
    Last edited: Jun 18, 2010
  4. Jun 18, 2010 #3
    I also find this problem a little vague in the wording, and hence difficult to answer, but I'll take my best guess. I would approach this problem from the point of view of complex conductivity for sinusoidal fields. For sinusoidal excitation of linear media with nonzero conductivity (i.e. the real world and low field strengths), one can use complex conductivity as follows:

    [tex] \sigma + j \omega \epsilon[/tex]

    where [tex] j = \sqrt{-1}[/tex], [tex] \sigma [/tex] is the conductivity of the medium, [tex] \epsilon [/tex] is the permitivity of the medium and [tex]\omega[/tex] is the field frequency of oscillation of the electric field.

    Solving this problem amounts to setting the real and imaginary parts of the complex conductivity equal to each other. It seems to me that this would result in displacement currents (as typical for dielectrics) equal to conduction current (as typical for conductors). The resulting phase angle (field to current) would be 45 degrees in this case. Normally displacement current is 90 degree's out of phase (i.e. time derivative involved), while pure conduction is in phase, ideally.

    EDIT: I just noticed RoyalCat's update with a numerical answer. I also get 376 Mrad/s for frequency (or 60 MHz). I guess this is the correct answer.
    Last edited: Jun 18, 2010
  5. Jun 18, 2010 #4
    you have said that you have got an answer of 60MHz .
    can you tell me the formulae you have used,to find out that answer?
  6. Jun 18, 2010 #5
    Nice solution, stevenb. I'm not quite as experienced with the P, D and H fields so reading up on complex permittivity was enlightening.

    This is a great example of how in-depth analysis of a certain problem, structure and all, can be circumvented by using the specific concepts specific to that kind of analysis. :)
    (With much less mathematical hassle to boot!)

    As for the formulas we used, stevenb's derivation relies on the concept of complex permittivity, which helps relate the phase difference brought on by displacement current to the conduction current. (Correct me if I'm oversimplifying or flat out wrong here :))

    I analyzed the conduction current and displacement current separately. My conduction current analysis is further up in my first post.

    As for the displacement current, I started with the relation [tex]V=E\ell[/tex] which simply utilizes the definition of the electrical potential for a constant E field: [tex]V_B - V_A=\int_A^B \vec E \cdot \vec d\ell[/tex]

    I then related the current to the electrical flux: [tex]I_{displacement} = \epsilon _ r \epsilon _0 \frac{d\Phi_E}{dt}[/tex]

    Where the electrical flux in this case, assuming uniformity of the E field (Which breaks down substantially at the frequencies we're talking about) is simply [tex]E\cdot A[/tex]

    And lastly, to convert from angular frequency to regular frequency: [tex]\omega = 2\pi f[/tex]
  7. Jun 18, 2010 #6
    Very simply, some people prefer to work with cyclic frequency (i.e. cylcles per second). Angular frequency and cyclic frequecy are related through the circle relation.

    [tex] \omega = 2\pi f[/tex]
  8. Jun 18, 2010 #7
    thanks a lot to both of you,stevenb and RoyalCat.
    I got the answer from complex relation,

    and equating imaginary with real part.

    but can you tell me how does this relation come?
    any reference, if possible.
    thanks again to both of you.
  9. Jun 18, 2010 #8
    I'll look for a good reference on-line. I did a quick search but didn't find anything that I liked. I cobbled together the attached PDF from memory. Hopefully, I didn't make a mistake.

    There is one book I think is good, but it is expensive and probably too specialized for what your are doing. However, if you can find it in your library, it is worth reading the introductory sections. (note that if you find the first edition in the library, that's good too because the fundamentals don't change)

    "Handbook of Biological Effects of Electromagnetic Fields", 2nd edition, Charles Polk and Elliot Postow, CRC Press, 1996

    Attached Files:

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