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Why infinite conducting rod - Gauss's Law , uses lambda?

  1. Feb 6, 2016 #1
    hi, i still don't understand why infinite thin-walled cylindrical shell or conducting rod use lambda rather than sigma ?
    lambda = C/m ,,, sigma = C/m^2

    i mean when we look at conducting rod, the charges inside the conductor is zero, so the charges spread on the surface of conducting rod(have same form as thin-walled cylindrical shell), which is when we calculate the electricfield use gauss's Law using formula :

    integral(E . dA) = q enclosed / vaccum permitivity

    the q enclosed is sigma times the surface area , rather than lambda times length ?

    i find this from "Fundamental of Physics" Halliday Resnick
     
  2. jcsd
  3. Feb 6, 2016 #2

    SammyS

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    Hello
    Hello annoyingdude666 . Welcome to PF.

    Using linear density, λ, is often more convenient than using surface charge density, σ . If the problem can be solved with either method, I would use the simpler method.
     
  4. Feb 7, 2016 #3
    hello SammyS
    thank you for answering my question
    btw, i still don't understand.
    what do you mean by "more convenient" or "more simpler" ? can u explain why ?

    because from what i know, the charges of conducting rod spread on its surface (surface = area), so why we use lambda rather than sigma ?
    it doesn't make sense to me hehe
     
  5. Feb 7, 2016 #4

    SammyS

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    This is similar to the case of a charged conducting sphere. You could go through the trouble of finding the surface charge density, However, the electric field external to the sphere is the same as if were to replace the sphere with an equal charge placed at the sphere's center.

    Similarly, if the charge on the surface of the infinitely long cylindrical conductor is replaced by the corresponding linear charge placed along the cylinder's axis, the electric field external to the conductor remains the same.
     
  6. Feb 7, 2016 #5
    Hello annoyingdude666. I am a beginner at this, so please bear with me:

    According to what you originally asked, integral(E . dA) = q enclosed / vaccum permitivity

    So yes, we can say E = q(in) / (permitivity*2πrl) = sigma*A / permitivity. However, at this stage, we are dealing with interactions between these charged wires/plates, and particles; therefore, we only need to know how far the particle is from the wire, rather than how long it is, since it has 0 length. That is why SammyS was saying it is more convenient to say E = lambda*l/permitivity = lambda /(2πr*permitivity*r), where r is perpendicular distance from wire, then to leave it as E = sigma/permitivity
     

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  7. Feb 7, 2016 #6
    ^Since the wire only has length, and not area, it also makes more sense to use lambda rather than sigma
     
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