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Gauss's Law

  1. Aug 22, 2004 #1
    Hi everyone, I'm very sorry for my first two posts being rather long problems, but could any of you possibly give me a hand with the following...? :uhh:

    the question is as follows;
    "Consider an infinately long cyliner of radius [tex]R[/tex], and uniform positive charge per unit volume [tex]\rho[/tex]. Derive expressions for the electric field (i) outside the cylinder [tex]r>R[/tex], and (ii) inside the cylinder [tex]r<R[/tex]"

    let me show you what I have...
    I'll be calling r outside the cylinder [tex]r_{1}[/tex] and inside [tex]r_{2}[/tex]. also the length of the cylinder will be called l.

    (i) for [tex]r>R[/tex]

    [tex]\phi_{E}=\oint E\cdot dA = E2\pi r_{1}l[/tex]
    [tex]\phi_{E}=\frac{Q}{\varepsilon_{0}}[/tex]
    [tex]E=\frac{Q}{2\pi\varepsilon_{0} r_{1}l}[/tex]

    now [tex]Q=\rho\pi R^2 l[/tex]

    so [tex]E=\frac{\rho R^2}{2\varepsilon_{0} r_{1}}[/tex]


    (ii) for [tex]r<R[/tex]

    [tex]\phi_{E}=\oint E\cdot dA = E2\pi r_{2}l[/tex]
    [tex]Q' = \varepsilon_{0}\pi r_{2}^2 l = Q \frac{r_{2}^2}{R^2}[/tex]

    [tex]\Longrightarrow \frac{Q'}{\varepsilon_{0}} = \frac{Qr_{2}^2}{\varepsilon_{0}R}[/tex]

    and finally, we have...

    [tex]E=\frac{Qr_{2}}{2\pi\varepsilon_{0}R^2}[/tex]


    phew... sorry everything gets a bit tiny towards the end, I'm not sure how to change that :blush:

    would someone be so kind as to confirm that this is correct for me? or, (in the likely event that), if it's wrong, offer some advice as to where I've gone wrong?

    thank you all very much in advance :smile:
     
  2. jcsd
  3. Aug 24, 2004 #2
    for what it is worth

    I gave it a quick once-over in my head, and it looks correct.... but math in my head is not my strong point!
     
  4. Aug 24, 2004 #3
    hey, thank you very much for giving it the once over though! it's very much appreciated :smile:
     
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