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Gauss's law and vectors

  1. Sep 12, 2004 #1
    I'm encountering a problem with a practice test and previous homework problem in my electrostatics class.

    Charges exist at three corners of a square as shown. Edge length = L. What is the Electric Field Strength (E) at the corner with no charge.


    -3q
    O------------O +2q
    |
    |
    |
    |
    O-------------
    +2q


    so i break it into vectors using E= (k q)/(r^2) (Form of gauss) But i dont know what the final answer should look like...
     
  2. jcsd
  3. Sep 12, 2004 #2

    Tide

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    Remember, the electric field is a vector - it has magnitude and direction!
     
  4. Sep 12, 2004 #3

    robphy

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    Gold Member

    Using symmetry of this problem,
    you should be able to determine the direction,
    and [almost by inspection] write down the algebraic answer.

    Hint: if you can determine the direction (using symmetry), it is advantageous to use a coordinate system that has an axis along this direction [instead of the standard x- and y- axes]. The only components you'll need to consider are those along this axis.
     
  5. Sep 12, 2004 #4
    Right a vector. So i would come out with an answer in unit vector form...like....



    (x-component)i+(y-component)j. Or is there a simpiler way to write it?
     
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