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Vector Confusion In Applying Coulomb's Law

  1. Jul 26, 2016 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/48cLE6q

    2. Relevant equations
    Coulomb's law

    3. The attempt at a solution
    I can follow most of this problem, but I am unsure where the constants in front of the trig functions are coming from. Why is it 2cos(135), 1cos(45), 2cos(-45), etc?
     
  2. jcsd
  3. Jul 27, 2016 #2
    pl. write down the electric field due to charges placed at those points on an unit positive charge placed at the coordinate under consideration, you will need the the distance square in the denominator and the field has been resolved in unit vector i and j directions...the factor 2 and i are coming due to those distances....
    e.g. take the charge at a... it is distant sqrt(2)/2 ; take square then it will be 1/2 in the denominator so a factor of 2 in the numerator-resolve the field intensity in i and j direction along x and y respectively.

    similarly check other ones.
     
  4. Jul 27, 2016 #3
    Why are there two distances 2 and 1? Isn't the center point equally distant from all the other points?

    edit: I wrote down the electric field. I see now where the constants in front of the cos are coming from. But is there a sign error for the constants in front of sin?
     
    Last edited: Jul 27, 2016
  5. Jul 27, 2016 #4
    check the full expression!
     
  6. Jul 27, 2016 #5
    I see it now. What I didn't realize is that all of the y vectors are positive because the third vector points away from the positive charge. Thank you.
     
    Last edited: Jul 27, 2016
  7. Jul 27, 2016 #6

    haruspex

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    There are not two distances. Factors 1 and 2 are the charges. The factors for the displacements are 2 (1/r2), the unit vectors ##\hat i## and ##\hat j## and the trig components, sin and cos..
     
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