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Electric fields and a square.

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the electric field in the middle of the square in magnitude and direction?

    four charges are arranged at the outer corners of the square in order from left to right , then top top to bottom respectively

    +q,-2q,-q,+2q.





    2. Relevant equations



    3. The attempt at a solution

    after looking at the diagram for a while it appears as if the vertical components of the four charges will cancel. Also the horizontal components will partly cancel, so your left with the electric fields from the two chrages on the right side of the square with only their horizontal components and also with only half their original magnitude. Since they partly cancelled with the charges on the left.

    each side of the square has length a.

    equating the length of a diagonal from a charge to the centre call it a length x which
    x=a/√2.

    electric fields add...

    so... E=1/4∏ε0(2q/x2)

    the (2) comes about since I add the horizontal components of both electric fields.

    also since its only the horizontal comp. I think I should take then cos(45) so now my expression becomes

    E=1/4∏ε0(4q/√2a2)

    since a2=2x2.

    does this seem right?

    and is my logic sound?
     
  2. jcsd
  3. Sep 3, 2012 #2

    SammyS

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    What charge is in the upper right corner?

    What charge is in the lower left corner?
     
  4. Sep 3, 2012 #3
    upper right = -2q

    lower left = -q
     
  5. Sep 3, 2012 #4

    SammyS

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    That is what I at first thought, but in that case, your analysis is incorrect.

    For the electric field at the center of the square:

    The vertical component is not zero.

    Considering the electric field at the center of the square only, there is some cancelling of the electric field. The result at the center is the same as if there were only an upper right charge, -q, and a lower right charge, q , with no charges at the other two corners.
     
  6. Sep 8, 2012 #5
    hmm i got the right answer thoe?
     
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