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Homework Help: Two problems gettin on my nerve

  1. Feb 11, 2007 #1
    4. A thin rod of length ℓ and uniform charge per unit length λ lies along the x axis, as shown in the Figure. (a) Show that the electric field at P, a distance y from the rod along its perpendicular bisector, has no x component and is given by E = 2ke λ sin θ0/y. (b) What If? Using your result to part (a), show that the field of a rod of infinite length is E = 2ke λ /y. (Suggestion: First calculate the field at P due to an element of length dx, which has a charge λ dx. Then change variables from x to θ, using the relationships x = y tan θ and dx = y sec2 θ dθ , and integrate over θ.)

    for this problem, im not gettin how u arrive at the E field. (ie part a)


    the other problem:

    7. A line of charge with uniform density 35.0 nC/m lies along the line y = 15.0 cm, between the points with coordinates x = 0 and x = 40.0 cm. Find the electric field it creates at the origin.

    i get the value but not the irection of the E field.


    can u please help me out
     
  2. jcsd
  3. Feb 11, 2007 #2
    PF requires that you show some work before we help you--sorry!
    Now for your other problem, you can perform a vector summation pretty easily. just sum it up like Ex=int((Kdq/distance^2)sin@) where @ is the angle between the line x=0 and the line from the origin to the charge dq. find a way to sum this from one side of the bar to the other. Then get your y-component and then you should have no problem getting your direction.
     
  4. Feb 11, 2007 #3

    Astronuc

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  5. Feb 11, 2007 #4
    I proved that E has no x component but i cant find the y component . can some1 show me a procedure (specially help withg the integration part)

    as for ur suggestion for the other problem. i dont think im familiar with the kind of math ur using... it loox weird
     
  6. Feb 11, 2007 #5

    HallsofIvy

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    Once again, what have you DONE to set up the integral. Show us what you have tried.
     
  7. Feb 11, 2007 #6
    ooo ok (i will denote lambda by @

    dq = @ dx

    dE = k @ dx / r^2 = k @ dx /(x^2+y^2)

    E = int (k @ dx / (x^2 + y^2)) from -.5L to .5 L

    i cant integrate it and i cant find the y component
    too
     
  8. Feb 11, 2007 #7
    pls also help me out for the other problem. i get the general notion of it already, but its the math where i go wrong. show me the INTEGRATIOn
     
  9. Feb 11, 2007 #8

    cristo

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    Have you tried the substitution given in the hint in the question?
     
  10. Jan 28, 2009 #9
    Hello all! I'm trying to work on this problem as well but I'm a bit lost.
    So far, I know that the Ex component will cancel, so only Ey will remain.
    Then I know that dq = lamda dx.
    I'm defining theta as the angle adjacent to y, so cos(ᶿ)= y / [√(L/2)^2 + (y)^2], where [√(L/2)^2 + y^2] = r.
    and I know that dE = (kdq) / r^2

    Substituting that into the equation for dE gets me:
    dE = k (lamda dx) / (r)^2 * (y / r)

    From here we can take an integral to get E, so:
    E = k(lamda)y ∫ dx / ((L/2)^2 + (y)^2)^1.5
    E = k(lamda)y (L/2) / y √((L/2)^2 + y^2)
    E = k(lamda)L / 2y√((L/2)^2 + y^2), integrating from L to 0?

    I'm actually not sure if integrating from L to 0 is correct... when I tried it I got
    (k(lamda)L) / (2√((L^2/4) + L^2) - k(lamda)
    which does not resemble the final answer, E = 2k*lamda / y

    Any guidance at all would be most appreciated. Thanks!
     
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