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Adding electric field vectorially

  • Thread starter shemer77
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  • #1
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Homework Statement


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]

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Answers and Replies

  • #2
dynamicsolo
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Homework Statement


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?
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.
 
  • #3
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thanks!!
 
  • #4
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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?
 
  • #5
dynamicsolo
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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?
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.
 
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  • #6
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thanks for your help!!!!! Really appreciate it, i got it and understand it now!!
 

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