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Two lines of charge, net electric field

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Short sections of two very long parallel lines of charge are shown, fixed in place, separated by L = 8.0cm. The uniform linear charge densities are [tex]+6.0\mu[/tex]C/m for line 1 and [tex]-2.0\mu[/tex]C/m for line 2. Where along the x axis shown is the net electric field from the two lines zero?

    The known data is:
    [tex]\lambda_{1} = 6 \times 10^{-6} C[/tex]
    [tex]\lambda_{2} = -2 \times 10^{-6} C[/tex]
    [tex]L = 0.08m[/tex]

    [​IMG]
    (sorry, scanner doesn't like big books)

    2. Relevant equations
    line of infinite charge: [tex]\frac{\lambda}{2\pi \epsilon_{0}r}[/tex]
    permittivity constant: [tex]\epsilon_{0} = 8.85*10^{-12}[/tex]

    3. The attempt at a solution
    I didn't get very far with this one. From what I can tell, I need to sum the electric fields, and figure out when it's zero.

    I started out like this:
    0 = E1 + E2
    E1 = -E2

    Obviously, at this point substituting E for the line of infinite charge equation proved fruitless. I don't know if I'm overcomplicating, undercomplicating, or just plain clueless. Any help is appreciated. :)

    Also, the given answer makes no sense to me:
    [tex]x = \frac{\lambda_{1} - \lambda_{2}}{\lambda_{1} + \lambda_{2}}\left( \frac{L}{2} \right)[/tex]
     
  2. jcsd
  3. Oct 4, 2007 #2

    Doc Al

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    Staff: Mentor

    Call the coordinate of the zero-field point x. How would you write the distance to each line charge (in terms of x and L) so that you could use the infinite line charge equation?
     
  4. Oct 4, 2007 #3
    I'm assuming that the 0 point is somewhere in the positive x region (because the first line has a larger charge - please let me know if my thinking is off).

    With that assumption, line 1 would be L + x away from the point, and line 2 would be L/2 + x away?

    Is this on the right track?

    [tex]\frac{\lambda}{2\pi\epsilon_{0}(L + x)} = -\frac{\lambda}{2\pi\epsilon_{0}(L/2 + x)}[/tex]
     
  5. Oct 4, 2007 #4

    Doc Al

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    Staff: Mentor

    Good! Keep going.

    Edit: Oops, looks like your equation is a bit off. See comment in next post.
     
    Last edited: Oct 4, 2007
  6. Oct 4, 2007 #5
    Ok, I algebra'd it out and got x = -3L/4.. which should give me x = -6cm. I guess that means my original assumption of 0 occurring in the positive side was incorrect?

    The given answer is very confusing - why would it be in that form? I never actually came across it while finding x.
     
  7. Oct 4, 2007 #6

    Doc Al

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    I think there's an error in your distances in your equation. Assuming you measure x from the origin, then the distance to line 1 will be x + L/2 and the distance to line 2 will be x - L/2.
    To get that answer, solve the problem symbolically. Don't plug in numbers for L, [itex]\lambda_1[/itex], and [itex]\lambda_2[/itex].
     
  8. Oct 4, 2007 #7
    Phew, got it. I had measured my distances in a weird way, but I fixed it now. :)

    The answer should have been 8cm, correct?

    Thanks so much! I have one more question about the etiquette on here. Is it bad form to post more than one question in a day? There's one other problem I'm banging my head against, but will hopefully figure out on my own.. I'm asking just in case. :)
     
    Last edited: Oct 4, 2007
  9. Oct 4, 2007 #8

    Doc Al

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    Yes. Good work.
    Of course not! Post as many as you want. As long as you're showing your work, why not? (Better to post them in separate threads, of course.)
     
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