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Magnitude and DIRECTION of electric field of a square.

  1. May 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the magnitude of the electric field at the center of a square 42.5cm on a side if one corner is occupied by a −38.2μC charge and the other three are occupied by −26.9μC charges. (This part done)

    Choose the correct direction of the electric field at the center of the square.
    A.outward of −38.2μC charge
    B.perpendicular to the diagonal passing through −38.2μC charge
    C.toward to −38.2μC charge

    Not sure how to approach the directions part.
    2. Relevant equations
    E = kQ/r^2
    Magnitude E = ((Ex)^2 + (Ey)^2)^1/2


    3. The attempt at a solution
    r = 0.425m
    E1 = kQ1/r^2 = ((9*10^9)(38.2*10^-6))/0.3^2 = 3.8*10^6 N/C
    E2 = E3 = E4 = ((9*10^9)(26.9*10^-6))/0.3^2 = 2.7*10^6 N/C

    I then solved each of their x and y components, and added them together. Net Ex is -787000 and net Ey is 787000.

    Resultant = ((-787000)^2 + (787000)^2)^1/2 = 1.11*10^-6 N/C

    I am not sure how I would figure out the direction part.
    *I am thinking towards the - 38.2 charge
     
    Last edited: May 14, 2016
  2. jcsd
  3. May 14, 2016 #2

    haruspex

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    The direction of the field is the direction a positive charge would tend to move if placed in the field. Does that fit with your answer? (I'm not hinting either way.)
     
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