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Homework Help: Trouble calculating the electric field at a point

  1. Jan 5, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    The situation is the one of a hairpin such as in this website : http://www.physics.brocku.ca/Courses/1P22_Crandles/problems/hairpin.jpg.
    I must prove that the electric field in "a" is worth 0. There's a tip. It says to calculate the electric field due to a differential part on the semi circle and to compare it with a differential part on the infinite length wire. With both the same angle [tex]\theta[/tex], where theta is the angle between the point over the semicircle considered, point "a" and the vertical radius (bottom one) of the semicircle.



    2. Relevant equations

    Radius of the semicircle =b.
    Linear charge density of the wire : [tex]\lambda[/tex].
    3. The attempt at a solution
    [tex]d\vec E=\frac{dQ \hat r}{r^2}[/tex]. For the differential portion over the semicircle, I get that [tex]dE=\frac{\lambda \cos (\theta ) d\theta}{b}[/tex]. Obviously I should find the same result for the dE of the differential portion of the straight wire, but I find [tex]\frac{\lambda \left [ \tan (\theta +d\theta) - \tan \theta \right ] }{b \cos ^2 (\theta + d \theta)}[/tex]. I don't think they are the same. At least I've tried to equate them, but with no success.
    Did I do everything wrong?
     
  2. jcsd
  3. Jan 5, 2010 #2

    rl.bhat

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    The current carrying conductor does not produce electric field, because it does not have any net charge. In every differential part of the conductor electrons are drifting in a particular direction but the net charge is zero at any instant.
    In the attached link they have asked the magnetic field at a and b.
    Have you changed it to electric field?
     
  4. Jan 6, 2010 #3

    fluidistic

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    Dear rl.bhat, I should have mentioned that my problem is different from the picture at http://www.physics.brocku.ca/Courses/1P22_Crandles/problems/hairpin.jpg. In my problem there's no current, but a uniformly charged wire whose linear charge density is [tex]\lambda[/tex]. I'm sorry about the confusion, I realize it's entirely my fault. The website I provided is good to understand the shape of the charged wire and the position where I want to calculate the electric field.
     
  5. Jan 7, 2010 #4

    rl.bhat

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    If you consider the electric field due to two symmetric differential elements on the semicircle, you get dE= 2*k*λ*cosθ*dθ/b. If you take the integration between the limits from 0 to π/2, you get E = 2*k*λ/b.
    Similarly if you consider two symmetric differential elements of length dx at a distance x from the starting point of the straight conductors, the electric field at a is given by
    dE = 2*k*λ*x*dx/(x^2 + b^2)^3/2. If you take the integration between the limits from 0 to infinity, you get the same result as above. These two fields are in the opposite direction. So they cancel each other making net field at a zero.
     
  6. Jan 7, 2010 #5

    fluidistic

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    Thanks for helping!
    Ok, I almost got this for dE, except for the k (that I simply forgot) and the factor 2. I follow you until here.

    Wow... I don't know why your dE differs SO much from mine. I'll try to reach your dE. If I have any problem, I will ask for further help. Thank you very much for your time, effort and will to help.
     
  7. Jan 8, 2010 #6

    fluidistic

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    All these differentials, I'm having a very hard time with them and I'm spending a too big amount of time playing with them.

    For the semi circular path, here is my work:
    [tex]d\vec E=\frac{kdQ\vec r}{r^3}[/tex] where [tex]\vec r=r\cos \theta \hat i +r\sin \theta \hat j[/tex].
    [tex]dQ=\lambda bd\theta[/tex].
    Thus [tex]d\vec E=\frac{k\lambda bd\theta (b\cos \theta \hat i + b\sin \theta \hat j)}{b^3}=\frac{k\lambda d\theta (\cos \theta \hat i+ \sin \theta \hat j)}{b}[/tex]. That was for the 0 to pi/2 part. Now I have to multiply by 2 the vertical component and throw off the horizontal one. Then I integrate from 0 to [tex]\frac{\pi}{2}[/tex] to get the desired result:
    [tex]\vec E=\frac{2k\lambda}{b}\int _0 ^{\frac{\pi}{2}} \sin \theta d\theta =\frac{2k\lambda}{b} \hat j[/tex]. I don't understand why I don't reach a negative sign...

    I've tried several ways to solve for the 2 semi-infinite wires, but I can't reach your result "dE = 2*k*λ*x*dx/(x^2 + b^2)^3/2". Can someone guide me a bit further please?
     
  8. Jan 8, 2010 #7

    rl.bhat

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    Your expression for E due to semicircle is correct. Its direction is towards the right side from the semicircle. No need to put negative sign.
    Now for field due to straight wire, consider a small element dx at a distance x from the
    left end of the bottom wire. Field at a due to this element is
    dE = k*λ*dx/r^2 = k*λ*dx/(x^2 + b^2). Similarly find the field due to the identical element on the top wire. dE*sinθ components of these fields get canceled each other.
    So net field due to straight wire is
    E = Intg(2*dE*cosθ) from limit x = 0 to x = infinity. Here cosθ = x /(x^2 + b^2)^1/2.
    Its direction is towards left of a.
     
  9. Jan 8, 2010 #8

    vela

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    Probably because you're thinking of [tex]\vec{r}[/tex] the wrong way. It goes from the charge to the point, not the other way around.
     
  10. Jan 9, 2010 #9

    fluidistic

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    Thanks to both!
     
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