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Homework Help: Find the electric field inside a circle

  1. Dec 13, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I set up myself to find the electric field inside a charged circle, but not in the center of the circle. Instead, in a point on a diameter, at a distance R/2 from the center, where R is the radius.
    2. The attempt at a solution
    I've tried to set up an integral summing the contribution of all the circle but without success. Is it a good idea? How would you proceed?
    I know that the electric field must be 0, but as I'm not really convinced, I wanted to verify it.
    Any help is appreciated.
    dE=dQ/r².
    [tex]Q=2 \pi R \lambda[/tex].
    [tex]dl=xd\theta[/tex] where dl is a differential arc on the circle, [tex]\theta[/tex] is the angle measured from the horizontal diameter to the dl element and x is the distance between the point where I want to find the electric field and the dl element.
     
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  3. Dec 13, 2009 #2

    AEM

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    I would compute the field dE at your point P due to a bit of charge [itex] dq = \lambda rd\theta [/itex] and integrate around the circle. I would also look to see if there is any symmetry condition that makes solving the problem easier. (Note: this is my first take on how to approach it. There might be a better way.)
     
  4. Dec 13, 2009 #3

    fluidistic

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    Ok perfect! That was the kind of reply I was hoping for. Thanks, I'll try to work something out.
     
  5. Dec 13, 2009 #4

    fluidistic

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    I get [tex]d\vec E=\frac{Qd\theta}{2\pi Rr}\hat r[/tex] where R is the radius and r is the distance between P and dl.
    I'm sure there's a [tex]cos \theta[/tex] that will appear from here, and integrating from 0 to [tex]2 \pi[/tex] I would eventually reach 0... but I don't know how to make the [tex]cos \theta[/tex] to appear. This is not new to me. I always struggle with this. :cry:
     
  6. Dec 13, 2009 #5

    AEM

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    Okay, so you got me hooked enough that I took another look at it. I drew a circle and drew a horizontal line I'll call the x-axis and a vertical y-axis. For each bit of charge [itex] \lambda r d\theta [/itex] there will be one symmetrically placed on the opposite side of the x axis. That means that the y components of the vector dE will cancel. That takes care of your [itex] cos(\theta) [/itex]. You need that to get the x component of dE. What concerns me is that th expression for r is going to be messy, but I'm going to have to sign off soon. I'm curious as to what you're going to come up with...
     
  7. Dec 13, 2009 #6

    fluidistic

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    You drew the same sketch than I. I also realized that the symmetry will automatically tell us that the E field in the y direction is null, but after all I wanted to prove it mathematically.
    I'm also stuck as to write r... I'd have an "x" variable, but it doesn't seem easy to continue.
    For example [tex]r=\sqrt{x^2+y^2}[/tex] or I could choose [tex]r=\frac{x}{\cos \theta}[/tex], though I don't know how to go any further. What limit to take in the integral for example.
    [tex]\theta[/tex], 0 to [tex]2\pi[/tex]. But x...
    I've no idea.
     
  8. Dec 13, 2009 #7

    rl.bhat

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    If you start measuring x from the point R/2, using the properties of circle, you can write
    y^2 = (R/2-x)*(3R/2+x)
    You can take the integration form x = 0 to x = R/2 for smaller part and x=0 to x = 3R/2 for the larger part.
     
  9. Dec 13, 2009 #8

    fluidistic

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    I thank you for your help and time. I don't understand how you reached this. Is it be simple Pythagorean theorem? I only see y²=r²-x² or y=r sin (theta).
    Time for me to go to sleep, it's past 2 am and if I don't sleep now I won't sleep the night before the exam, ahah.
     
  10. Dec 13, 2009 #9

    rl.bhat

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    The relation is true for any intersecting diameter and a cord which is perpendicular to the diameter. Cord bisects the diameter.
    Let AB be the diameter and PQ is the cord. Let O be the point of intersection.
    Triangle OAP and OBQ are similar. So the sides are proportional.
    Hence OP/OB = OA/OQ Or
    OA*OB = OP*OQ. But OP = OQ. So
    OA*OB = OP^2.
     
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