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Electric Field from a Finite Length, Uniformly Charged Wire (EM, not Gen Phys II)

  1. Sep 28, 2012 #1
    So there's this situation going on:
    http://imageshack.us/a/img826/7398/physicsforums.png [Broken]

    Going from the definition of an electric field:
    (1) [itex]\vec{E} ( \vec{x} ) = \frac{1}{4\pi\epsilon_{0}} ∫ \frac{\vec{x} - \vec{x'}}{| \vec{x} - \vec{x'} | ^3} ρ( \vec{x'}) d^3x'[/itex]


    (2) The [itex]ρ(\vec{x'})d^3x'[/itex] reduces to [itex]λdz'[/itex]. And [itex]\vec{x} - \vec{x'} = x \hat{i} - z' \hat{k} = \sqrt{x^2 + z'^{2}}[/itex].


    (3) Now, plugging this information into the electric field equation yields:
    [itex]\vec{E} ( \vec{x} ) = \frac{1}{4\pi\epsilon_{0}} \int^{l}_{-l} \frac{\sqrt{x^2 + z'^{2}}}{(\sqrt{x^2 + z'^{2}})^3} λdz'[/itex]


    (4) However, the book (Electromagnetism by Pollack and Stump) shows:
    [itex]E_{x}(x,0,0) = \frac{1}{4\pi\epsilon_{0}} \int^{l}_{-l} \frac{x}{(\sqrt{x^2 + z'^{2}})^3} λdz' [/itex]

    How do we get from (3) to (4)? Why is z' only removed from the numerator?

    Thanks in advance.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 28, 2012 #2

    vanhees71

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    Your formula (3) is obviously wrong, because on the left-hand side is a vector and on the right-hand side a scalar quantity.

    From (1) and (2) you immediately write down (4). So your book is correct. There is no z' in the numerator of the x component!
     
  4. Sep 28, 2012 #3
    so the charged wire is along z axis so by symmetry only x component will survive so
    z will be absent in numerator and beware it is a vector.
     
  5. Sep 29, 2012 #4
    I don't understand how I could just go from (1) and (2) to (4)...

    Or why the numerator doesn't have a vector magnitude, whereas the demoninator does...


    I was, however, able to get the correct answer, just not using the full notation from equation (1).

    So I start with: [itex]\vec{E} ( \vec{x} ) = \frac{1}{4\pi\epsilon_{0}} ∫ \frac{\vec{x} - \vec{x'}}{| \vec{x} - \vec{x'} | ^3} ρ( \vec{x'}) d^3x'[/itex]

    And since [itex]d\vec{E}[/itex] generated by [itex]dq[/itex] on the line will be symmetric (i.e. only the x-component will survive), we can say [itex]dE_{x} = d\vec{E}cos(\theta)[/itex].

    Therefore: [itex]dE = \frac{1}{4\pi\epsilon_{0}} \frac{1}{r^2} dq[/itex], where [itex]r = | \vec{x} - \vec{x'}| = \sqrt{x^2 + z'^2}[/itex]

    Therefore: [itex]dE_{x} = \frac{1}{4\pi\epsilon_{0}} \frac{1}{x^2 + z'^2} \frac{x}{\sqrt{x^2 + z'^2}} λdz' [/itex] from [itex]cos(\theta) = \frac{x}{\sqrt{x^2 + z'^2}} [/itex] and [itex] dq = λdz'[/itex]

    Which gives the result in the book: [itex]E_{x}(x,0,0) = \frac{1}{4\pi\epsilon_{0}} \int^{l}_{-l} \frac{x}{(\sqrt{x^2 + z'^{2}})^3} λdz' [/itex].

    So was I supposed to recognize that I needed to use the cos(θ) relationship from the beginning? I'm also still confused on how the vector difference only acts on the denominator, if I were to just use equation (1).

    Thank in advance.
     
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