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Homework Help: Electric field at a point on the perpendicular bisector of a finite line charge

  1. Jul 6, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform line charge of linear charge density [tex]\lambda[/tex] = 5.00 nC/m extends from x = 0 to x = 10m. The magnitude of the electric field at the point y = 12m on the perpendicular bisector of the finite line charge is?


    2. Relevant equations
    E = [tex]\int[/tex]dE = [tex]\int[/tex]k(dq) / r2
    Ex = [tex]\int[/tex]k(dq)cos[tex]\theta[/tex] / r2
    Ey = [tex]\int[/tex]k(dq)sin[tex]\theta[/tex] / r2
    dq = [tex]\lambda[/tex](dx)

    3. The attempt at a solution
    The perpendicular bisector of the finite line charge occurs at x =5m. I have placed a point (P) at the coordinates on the xy plane at P = (5, 12). With P at this point, the radius from (0, 0) to P is:
    r = [tex]\sqrt{(5^2) + (12^2)}[/tex] = [tex]\sqrt{169}[/tex] = 13

    If 'b' is the side of the Pythagorean triple equal to 12m then sin[tex]\theta[/tex] = b / r = 12 / 13

    With point P bisecting the finite line charge along the x-axis from 0 to 10m I'm arguing by symmetry that Ex = 0, thus there is only a y-component to the electric field.

    I'm new to LaTeX and the forums and can't quite seem to get the integration limits to appear properly so the integral is from 0 to 5.

    Ey = [tex]\int[/tex]k(dq)sin[tex]\theta[/tex] / r2
    Ey = [tex]\int[/tex]k([tex]\lambda[/tex]dx) / r2
    Ey = [tex]\int[/tex]k([tex]\lambda[/tex]dx)(12/13) / (132)
    Ey = k[tex]\lambda(12/13) / (132) * [/tex][tex]\int[/tex]dx
    Ey = 12k[tex]\lambda[/tex] / (133) * [5 - 0]
    Ey = (9 x 109)(5 x 10-9)(12 * 5) / (133)
    Ey = 1.23 N/C

    Using the idea of symmetry again, E = 2 * Ey
    E = 2 * (1.23 N/C)
    E = 2.46 N/C


    Is my work/thought process and ultimately answer correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 6, 2010 #2

    vela

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    You're getting there, but remember that both r and θ change as x varies. You can't just use their values for when x=0.
     
  4. Jul 6, 2010 #3
    So I shouldn't compute a value for sin[tex]\theta[/tex] and instead leave it in terms of b and r (sin[tex]\theta[/tex] = b / r = 12 / r)?

    Then I have (integral from x=0 to x=5 again):
    Ey = [tex]\int[/tex] k(dq)sin[tex]\theta[/tex] / r3
    ...
    Ey = 12k[tex]\lambda[/tex] / r3 * [tex]\int[/tex]dx
    ...
    Ey = 60k[tex]\lambda[/tex] / r3

    Then E = 2 * Ey
    E = 120k[tex]\lambda[/tex] / r3

    I feel like I'm missing something here...
     
  5. Jul 6, 2010 #4

    vela

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    Let's back up a bit. Say you're looking at the segment of charge between x and x+dx. What is its distance r to the point (5,12) and what is sin θ?
     
  6. Jul 6, 2010 #5
    I'm not sure what you mean by "segment of charge between x and x+dx". Do you mean a segment of charge, say at x = 3m?

    This new side to the triangle formed by points (3, 0), (5, 0), and (5, 12) would then be a = 5 - x where x is any value along the x-axis from 0 to 5.

    I'm using theta to be the angle between the x-axis and the hypotenuse of this triangle. Wouldn't sin [tex]\theta[/tex] remain to be 12 / r? Should I be using cosine instead?
     
  7. Jul 6, 2010 #6

    vela

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    There's zero charge at x=3 m because a point has zero size. The interval [x, x+dx], on the other hand, has a length dx, which, while infinitesimally small, is not 0. This is what you're doing when you write dq=λ dx.
    Yes, so you need an expression for r, which will depend on x. Since it depends on x, you won't be able to just pull it out of the integral like you did in your previous attempts.
    Sorry, you had already gotten sin θ correctly. I missed it when I looked over your posts too quickly.
     
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