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Electric field due to arc of charge

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data
    A circular arc of charge has a radius R and contains a total charge Q. If the angle of the arc is 90 degrees find:
    a) the charge density of the arc
    b) the electric field at point P in terms of the charge density L and the radius of the arc R

    L should really be lambda and P is the center of the arc.

    2. Relevant equations
    E = (1/4pi Epsilon naught)
    s is the arc length

    3. The attempt at a solution
    a) (I got this part right)
    Q = Ls, L = Q/s
    s = 1/4 circle = (1/2)pi R
    L = 2Q/pi R

    b) (I didn't get this part right)
    I said that because the arc is symmetric, the electric field in the x direction has the same magnitude as the electric field in the y direction, so I only solved for one of them.
    I decided to integrate dE = (1/4pi Epsilon) (dQ/R squared) * sin theta (for the x direction) from 0 to pi/2

    I said that dQ = L * dS and that (this may very well be the part where I went wrong) ds = R*d theta for small theta so that I could integrate in terms of d theta.

    When I integrated I got Ex = L/(4 pi Epsilon R) *-cos theta evaluated from 0 to pi/2
    That last part ended up being 1, so for my answer I got Ex = L/(4 pi Epsilon R), which was wrong.

    Hope you followed that. These are test corrections, so I might have made dumb math mistakes during my panic.

    I believe the correct answer is something like what I got but multiplied by square root of 2.
     
  2. jcsd
  3. Jan 16, 2008 #2
  4. Jan 16, 2008 #3
    I follow your (very well executed) reply to that question but it still leads me to where I got in the first place.
    I come to E = [tex]\int^{\pi/2}_{0}\frac{\lambda}{4\pi\epsilon R}\cdot sin\theta\cdot d\theta[/tex]
     
    Last edited: Jan 16, 2008
  5. Jan 16, 2008 #4
    I figured out exactly what happened. They were asking for the magnitude of the electric field. All you had to do was take your answer,

    [tex]\textbf{E} = \frac{\lambda}{4\pi \epsilon_o R}(\textbf{i} + \textbf{j}),[/tex]

    and find its magnitude! Don't you hate when you do things like that? You can't imagine how many times I've gotten back a test and was like, "Why the hell'd I do that?"
     
  6. Jan 16, 2008 #5
    This may sound incredibly dumb, but how do I exactly go about doing that? We weren't given any values. The answer should be "in terms of the charge density lambda and the radius of the arc R."

    On the test the teacher corrected my answer of [tex]E = \frac{\lambda}{4\pi\epsilon R}[/tex] by adding [tex]\sqrt{2} \chi[/tex] or perhaps [tex]\sqrt{2} x[/tex]
     
  7. Jan 16, 2008 #6
    [tex]\sqrt{\textbf{E}\cdot\textbf{E}}[/tex] where [tex]\textbf{u}\cdot\textbf{v}[/tex] denotes the dot product.
     
  8. Jan 16, 2008 #7
    I doubt she would have added anything besides [tex]\sqrt{2}[/tex] because that would be introducing an extra variable. Plus, she wanted it in terms of [tex]\lambda[/tex] and [tex]R[/tex].
     
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