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Electric Field of a circular arc at a point

  1. Nov 7, 2014 #1
    1. The problem statement, all variables and given/known data
    zlkok7R.png

    Given that the circular arc wire with radius 'r' has a linear charge density ##\lambda##. What is the Electric field at the origin?

    2. Relevant equations
    ##\vec{E}=\frac{kq}{r^2}## where ##k=9\times10^9## is a constant.

    3. The attempt at a solution

    I took a small segment dy ##\theta## above the x-axis with charge ##dq=\lambda dy##. Therefore ##d\vec{E}=\frac{k\lambda cos\theta dy}{r^2}## as all other charges along the y-axis cancel out each other.

    Now ##cos\theta=\frac{x}{r}##. And ##x^2+y^2=r^2## is the equation of the arc.
    Therefore ##cos\theta=\frac{\sqrt{r^2-y^2}}{r}##. And then proceeding to integrate ##d\vec{E}=\frac{k\lambda \sqrt{r^2-y^2} dy}{r^3}## and arrive at an answer.

    But my text tackles the question the same way until, at a point it takes ##dy=rd\theta## and then substitutes and integrates ##d\vec{E}=\frac{k\lambda cos\theta d\theta}{r}## and arriving at an answer. But my answer differs from the one arrived by my textbook. Am I wrong somewhere?
     
  2. jcsd
  3. Nov 7, 2014 #2

    mfb

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    Staff: Mentor

    ##dq=\lambda dy## would mean every "step" dy gives the same charge, and equivalently the same length of the wire. This is not true.
     
  4. Nov 7, 2014 #3
    mfb : That is the condition given in the question. Moreover my text starts the solution in the same way. (The wire is uniformly charged)
     
  5. Nov 8, 2014 #4

    rude man

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    Use symmetry to conclude what about the y component of E at the origin?
    For the x component, dEx = k cosθ dq/R2 should be apparent.
    What is dq in terms of arc length ds?
    Go that way & you will wind up agreeing with your textbook.

    As mfb implied, dy = R dθ is incorrect. R dθ is an element of arc length ds, not an element of distance along the y axis.
     
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