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Finding the net field strength due to a line of charge

  1. Feb 9, 2012 #1
    I want to apologize ahead of time for my apparent inability to get LaTeX to display properly.

    1. The problem statement, all variables and given/known data
    prob2-33.png

    λ = 3.8 x 10^(-6) C/m

    R = 32m

    r = magnitude of the hypotenuse (the book's convention, a poor choice IMHO)
    r = [itex] \sqrt{R^2 + x^2} [/itex]

    k = 8.988 x 10^9

    P:(0,-32)

    cosθ = x/r

    a→∞

    2. Relevant equations
    [itex] dE = λk \int_0^a \frac{dx}{r^2}[/itex]

    3. The attempt at a solution

    We're supposed to figure out the overall field strength, which is finite.

    [itex]dE = λk \int_0^a \frac{dx}{r^2} [/itex]

    [itex]dE_x = λk \int_0^a \frac{dx}{r^2} cosθ [/itex]

    [itex]dE_x = λk \int_0^a \frac{dx}{r^2} \frac{x}{r}[/itex]

    [itex]
    dE_x = λk \int_0^a \frac{xdx}{r^3}
    [/itex]

    [itex]
    u = R^2 + x^2 → \frac{du}{2} = xdx
    [/itex]

    [itex]
    dE_x = λk \int_0^a (u)^{-3/2} du
    [/itex]

    [itex]
    dE_x = λk [ -2 u^{-1/2} ]_0^a
    [/itex]

    [itex]
    dE_x = -2λk [ (R^2 + x^2)^{-1/2} ]_0^a
    [/itex]

    [itex]
    dE_x = -2λk [ \frac{1}{\sqrt{R^2 + x^2}} ]_0^a
    [/itex]

    [itex]
    dE_x = -2λk [ \frac{1}{\sqrt{R^2 + \inf^2}} - \frac{1}{\sqrt{R^2}} ]
    [/itex]

    [itex]
    dE_x = -2λk [ 0 - \frac{1}{\sqrt{R^2}} ]
    [/itex]

    [itex]
    dE_x = -2λk [ \frac{1}{R} ]
    [/itex]


    [itex]
    dE_x ≈ -2.13465 x 10^3
    [/itex]
     
    Last edited: Feb 9, 2012
  2. jcsd
  3. Feb 9, 2012 #2
    Looks good, though the sign is important to how you're defining your coordinate system. Make sure your coordinate system is the one the program is looking for.
     
  4. Feb 9, 2012 #3
    I literally pasted all the information I was given. >.< Do you have any guidance as to how to figure out what coordinate system I should use?
     
  5. Feb 10, 2012 #4
    I mean, typically we would define y is being up and x as being right. But you could do any permutation of those. I guess I don't understand what your questions about the problem are.
     
  6. Feb 10, 2012 #5
    So I talked to my Physics professor, and he took a look at it. Evidently I forgot to carry over an x.

    Why the solution is positive still confounds me.
     
    Last edited: Feb 10, 2012
  7. Feb 10, 2012 #6
    Ah, whoops, that is easy to do. Yeah, I didn't really check all your steps, just that the integration was correct.
     
  8. Feb 10, 2012 #7
    The most frustrating part is that I apparently was correct the first time, but the answer they wanted was positive rather than negative. But looking at the problem, the vector is clearly negative in the X and Y directions!
     
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