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Magnitude and Direction of Electric Field at the origin? Please help

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A continuous line of charge lies along the x axis, extending from x=+xo to positive infinity. The line carries charge with a uniform linear charge density lambda. What are the magnitude and direction of the electric field at the origin?


    2. Relevant equations
    E = lambda/2pieor for the magnitude of electric field produced by a uniformly charged infinite line.

    3. The attempt at a solution
    I have drawn a cylindrical Gaussian Surface around it. By the wording of the question, I am not sure where the line of charge is placed (does it start at the origin?). I have placed the line at the origin, with the cylinder starting at the origin. I have concluded (not sure if it is correct) that the field on the end of the cylinder is equal to zero, because the field can only pass through the center of the circle.

    Am I close at all to the answer?
     
  2. jcsd
  3. Sep 28, 2008 #2

    tiny-tim

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    Hi SilverGirl! :smile:

    (have a pi: π and an epsilon: ε and a lambda: λ and an infinity: ∞ :smile:)

    No, the line starts at x0 and goes to +∞.

    hmm … no idea what you mean by a cylindrical Gaussian Surface. :confused:

    Hint: slice the line into bits of length dx, and integrate from x0 to +∞. :smile:
     
  4. Sep 28, 2008 #3
    Is xo at any particular place on the x axis though? Maybe I didn't have to use the cylinder.
     
  5. Sep 28, 2008 #4

    tiny-tim

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    erm … it's at x0 ! :confused:

    isn't that particular enough for you? :wink:
     
  6. Sep 28, 2008 #5
    Not really..lol..because couldn't x0 be at the origin? Also, couldn't it be very very far away from the origin? Do you know of a way to do this without integrating?
     
  7. Sep 28, 2008 #6

    tiny-tim

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    No, because the question says "+x0", and 0 isn't + :smile:
    Nope! :biggrin:
     
  8. Sep 28, 2008 #7
    lol..good point.

    I am guessing the field is not 0 at the origin, even though the line of charge is not there.
     
  9. Sep 29, 2008 #8

    Defennder

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    Yes, you're right. It's not zero at the origin. It would be zero if the line extended from [tex]-\infty[/tex] to [tex]\infty[/tex]. Now tiny-tim's hint was to integrate it. You need to start with [tex]dE = \frac{dq}{4\pi \varepsilon_0 r(x)^2}[/tex], where r(x) is function of x which tells you the distance from any point on the line of charge to the origin.

    You still need to express dq in some other way so that the integration can be done.

    By symmetry, you should be able to determine what the E-field direction at O should be.
     
  10. Oct 24, 2008 #9
    I've been using this problem for reference but I am so lost...I'm doing the same exact thing.

    I am having trouble integrating. Could someone please walk me through it step by step with this equation?
     
  11. Oct 24, 2008 #10

    tiny-tim

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    Hi Doc Z! :smile:

    Show us how far you've got with the equation (and which equation? :confused:), and where you're stuck, and then we'll know how to help. :smile:
     
  12. Oct 24, 2008 #11
    I'm using E=2ke * lambda/r

    I pretty much don't even know how to start integrating this.
     
  13. Oct 24, 2008 #12

    tiny-tim

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    Hint: slice the line into bits of length dx, and integrate from x0 to +∞.

    What do you get? :smile:
     
  14. Oct 24, 2008 #13
    I found the answer to be ke * lambda/x0 in the -i direction but I don't understand how to get it. I don't remember integrals too well.
     
  15. Oct 24, 2008 #14

    LowlyPion

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    You know that E = k*Q/r2 for a point charge.

    So the field at the origin from any point over the range of charges is given by

    E(r) = k*Q/r2

    But Q by the problem statement is

    ΔQ = λ * Δr

    Rewriting the equation:

    E(r) = k * λ * Δr/r2

    Consequently then to sum up all the charge elements from r = xo to r = ∞ you take the definite integral.

    [tex] E_{(r)} = \int_{x_0}^{\infty} \frac{k * \lambda * dr}{r^2} [/tex]
     
    Last edited: Oct 24, 2008
  16. Oct 24, 2008 #15
    Thanks for the help!
     
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