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Homework Help: Electric field at the center of an arc

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A charge Q is arranged evenly on a wire bent into an arc or radius R as shown in the
    Figure. What is the electric field at the center of the arc as a function of the opening
    angle theta? Sketch a graph of the electric field as a function of theta for 0<theta<180 degrees.

    2. Relevant equations

    E= KQ/R^2 cos theta

    3. The attempt at a solution

    Is the above equation correct? If not, I'm not really sure what to do, also I'm not sure how the graph should look.
  2. jcsd
  3. Dec 2, 2009 #2
    where is the figure?
  4. Dec 2, 2009 #3
    Where'd you get that equation?

    Also, if you're worried about how the graph should look, start by figuring it out for some easy points. What's the E field going to be at the center if theta is zero (so all the charge is concentrated in a single point)? How bout if Theta is 360 degrees (so a full circle)? It's a bit harder to figure out for 180 degrees, or 90 or 270 degrees, but you should be able to make a very rough estimate. That will give you some insight into the shape of the graph.
  5. Dec 3, 2009 #4
    sorry this is the best I can do: http://i848.photobucket.com/albums/ab41/tag16/problem1.jpg?t=1259862072 [Broken]

    The equation I found in my physics book except the cos theta part, I was just guessing there. Though it would probably be integral cos theta or sin theta, if it's suppose to be in there at all.
    Last edited by a moderator: May 4, 2017
  6. Dec 4, 2009 #5

    dE= kdQ/R^2
    dE= (kdQ/R^2) cos[tex]\theta[/tex]

    dl= Rd[tex]\theta[/tex]


    E=[tex]\int[/tex](k[tex]\lambda[/tex]Rcos[tex]\theta[/tex]/R^2)d[tex]\theta[/tex](from [tex]\pi[/tex]/2 to -[tex]\pi[/tex]/2)
    E=k[tex]\lambda[/tex]/R[tex]\int[/tex] cos[tex]\theta[/tex]d[tex]\theta[/tex]
    E= 2kq/[tex]\pi[/tex]R^2

    Is this right?
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