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Adding electric field vectorially

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data
    I just wanted to make sure I am doing this right, my problem is that I think the electric field in the x direction should be the same for each one because the rix is the same for each point. Or am I just thinking too hard?
    http://dl.dropbox.com/u/265697/IMG.jpg [Broken]

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 25, 2011 #2

    dynamicsolo

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    You are correct that rix for the distance from the point at which the field is being measured to each of the cells is just the same. But the x-component of the electric field from each cell, [itex]\Delta E_{ix}[/itex] , is going to differ with each cell for two reasons: the total distance to the point from each cell is getting larger, so the strength of the electric field from each cell is getting successively smaller; and the angle that the direction from each cell makes with the x-axis is opening up further for each successive cell, so [itex]\cos \theta[/itex] is also getting smaller for each cell down the line. So [itex]\Delta E_{ix}[/itex] should get smaller for each cell down the list.
     
  4. Sep 25, 2011 #3
    thanks!!
     
  5. Sep 25, 2011 #4
    Hey one more quick question for the net field box under [itex]\Delta E_{ix}[/itex] and [itex]\Delta E_{iy}[/itex] I just add up all the values right? and then what should I put in for the 2 boxes of the field magnitude and direction?
     
  6. Sep 25, 2011 #5

    dynamicsolo

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    Those are perpendicular components of the field [itex]\Delta E_{i}[/itex] from each cell, so they must be added "in quadrature", that is, according to the Pythagorean Theorem, since the field represents the hypotenuse of a right triangle with the two components being the triangle's legs. So [itex]\Delta E_{i} = \sqrt{(\Delta E_{ix})^{2} + (\Delta E_{iy})^{2} } [/itex].

    EDIT: I had to go back and look at your table again. Yes, you would first add up the columns for [itex]\Delta E_{ix}[/itex] and [itex]\Delta E_{iy}[/itex] to get the totals of each component. Then you would combine the total x-component and y-component of the field in the way I described, but with [itex]\Delta E = \sqrt{(\Delta E_{x})^{2} + (\Delta E_{y})^{2} } [/itex]. (I didn't remember what your table entries looked like...)

    EDIT: For the direction of the total field, you need [itex]\tan \theta[/itex], which is equal to [itex]\frac{\Delta E_{y}}{\Delta E_{x}}[/itex] with the values taken from your sums . You would then calculate the inverse tangent (tan-1) of that value to find [itex]\theta[/itex], which is the angle the field direction makes with the x-axis.
     
    Last edited: Sep 25, 2011
  7. Sep 25, 2011 #6
    thanks for your help!!!!! Really appreciate it, i got it and understand it now!!
     
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