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Unable to solve an approximation

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data

    I'm trying to understand an approximation Griffiths does (in his solutions' manual - exercise 9.18-b) and I'm not quite getting it.
    Let
    $$k = \omega \sqrt{\dfrac{\epsilon \mu}{2}} [\sqrt{ 1 + (\dfrac{\sigma}{\epsilon \omega}})^2-1]^{1/2}$$

    He says that, because ##\sigma >> \omega \epsilon##, we have:$$k = \sqrt{\dfrac {\mu \sigma \omega}{2}}$$

    3. The attempt at a solution

    Essentially, my attempt was to say that ##(\dfrac{\sigma}{\epsilon \omega})^2 \approx \sigma ^2##. After that, and since sigma >> 1, I'd say that ##\sqrt {1 + \sigma ^2} \approx \sigma##, and for the same reasoning ##\sigma -1 \approx \sigma##, which would make everything inside the straight brakets be ##\sqrt{\sigma}##. After all this, ##k## would be ##\omega \sqrt {\dfrac{\epsilon \mu \sigma}{2}}##. Now, sadly, that's no way near the solution I'm supposed to arrive.
     
    Last edited: Jan 10, 2015
  2. jcsd
  3. Jan 10, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    That would correspond to ##\epsilon \omega \approx 1##. The approximation gives ##(\dfrac{\sigma}{\epsilon \omega})^2 \approx 1## which allows to simplify the expression.
    You can use the same approximation for the first ω in the equation and then ignore some prefactor.
     
  4. Jan 10, 2015 #3
    I don't quite understand what you're saying. You're saying that ##(\dfrac{\sigma}{\epsilon \omega})^2 \approx 1## solves my problem? If so, I don't understand how, as it would eliminate my ##\omega ##.
     
  5. Jan 10, 2015 #4

    rock.freak667

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    Homework Helper

    If σ >> ωε then (σ/ωε) >>1.

    And so (σ/ωε)2 +1 ≈ (σ/ωε)2

    You'd then use your same reasoning however carrying the σ/ωε expression instead of just the σ.
     
  6. Jan 10, 2015 #5

    mfb

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    Staff: Mentor

    Oh sorry, I read ##\approx## instead of >>. See the post above this how to do the approximation then.
     
  7. Jan 11, 2015 #6
    Thank you! This works wonderfully. I wonder if there's a text somewhere that explains how to do this type of approximations or if one just picks them up as one goes.

    That was my bad mfb, sorry. Initially, I had ##\approx##, but then edited and by mistake made it ##>>##, and then saw the error and edited again.
     
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