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Uniformly charged rod in an arc

  1. Sep 11, 2008 #1
    If a rod with charge density lambda is bent into a 3/4 circle what is the E field at the center?


    Well if the rod is 3/4 of a circle then neither the x or y components can cancel out, thus

    E= k*lambda/ R (i) + k*lambda/R (j)

    right?
     
  2. jcsd
  3. Sep 13, 2008 #2

    Defennder

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    You need to tell us how the circular arc is oriented. And I don't see how you got that. I got a different answer for magnitude.
     
  4. Sep 13, 2008 #3
    from pi/2 to 2pi

    meaning the test particle would go up and to the right.

    as I drew the lines of E I couldn't tell how any component could cancel, and if you spit it up neither the x or y component cancel out as the r's are different. If the rod was a half circle the y's would cancel out but if it is just a 1/4 circle there is an x- and y-components, thus for a 3/4 circle there would be x- y- components.

    and if you add another 1/4 of a circle you would get a ring at which there is no E field at the center. and if you draw two dq lines from opposite sides of the 3/4 circle arc the resultant would be going up and to the right.

    After that I just integrated x and y components from pi/2 to 2pi as my instructor indicated.
     
    Last edited: Sep 13, 2008
  5. Sep 14, 2008 #4

    alphysicist

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    You can redefine your axes so that one set of components will cancel. For example, if you define the resultant field to be along the y-axis, so that the arc is from (3/4)pi to (1/4)pi, all of the x components will cancel.
     
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