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Homework Help: Calculating magnitude of electric field at center of square

  1. Sep 9, 2015 #1
    1. The problem statement, all variables and given/known data

    Find the magnitude and direction of net electric field at the center of the square array of charges. Find [itex] E_x and E_y [/itex]

    The square array of charges http://postimg.org/image/4gf94ymmf/

    3. The attempt at a solution

    My attempt at drawing in the force vectors http://postimg.org/image/mae0fm1d9/ . Now the +3q and +q's should contribute a net of zero to the electric field and they can be ignored. So, we have [itex] E = k [ -2q^2 / d^2 + q^2/d^2 + 5q^2/d^2] = 4k q^2/d^2[/itex]

    Taking the x-component of the field we have
    [itex] E_x = (4k q^2/d^2)cos(45)[/itex]

    the y-component:
    [itex] E_x = (4k q^2/d^2)sin(45)[/itex]

    This isn't right or I would not be posting here... so what am I doing wrong? I'm really not understanding how to do these types of problems for net fields in squares. Have I even drawn the vectors correctly?
  2. jcsd
  3. Sep 9, 2015 #2


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    You're missing a term.
  4. Sep 9, 2015 #3
    The missing term is the -5q charge, yes? So it should be
    [itex]E=k[−2q/(\sqrt2 * d)+q/(\sqrt 2 * d) + 2(5q/(\sqrt 2 * d))= k[9q / (\sqrt 2 * d)][/itex]

    Also, changed q^2 to q since this is the electric field not force... silly mistake, and the bottom term should be [itex] \sqrt 2 * d [/itex] instead of d^2?

    This is still incorrect is not? I'm really not seeing what I'm missing here.
  5. Sep 9, 2015 #4


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    Methinks that should do almost do it (dumped the picture, so I'm going by memory); the "two" and "one" are both on axes? In which case "d."
  6. Sep 9, 2015 #5
    Ah fair point forgot about those being strictly on the x-axis, so need to fix that then resolve into E_x and E_y for the respective terms and then take the square root of those squared for the magnitude.
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