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Electric field question involving 4 charges on a square

  1. Jul 12, 2012 #1
    Four charges are placed at the vertices of a square, centered at the origin, as shown in the diagram. If each side of the square has a length of 0.224 m, what is the strength and direction of the net electric field at the origin? Express your answer in terms of the charge magnitude q.


    physicsquestion.png


    2. Relevant equations
    e = kq/r^2
    a^2+b^2=c^2
    addition of vectors
    also the unit circle
    x=rcos
    y=rsin


    3. attempt at a solution
    seperated them since they were all on a 45 degree angles into pi/4, 3pi/4, 5pi/4, 7pi/4
    I then found each of the lentghs from the origins to be the sqrt of 2(.112^2 ) from the pythagorean theorem which gave me a radius of .158391919, I then plugged that into the equation for e=kq/r^2 which gave me the magnitude 3.58338648*10^11 I then plugged that into the rcos and r sin formulas for each answer and got my final answer to be (0c, 0c) which of course is in a direction of 0. This was wrong, I know I must be something really simple that I am missing. could some one please help me?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 12, 2012 #2

    so why dont you need to worry about the two positive q's on the opposite sides of the diagonal?

    and, what do you think the overall direction of the e- field is at the center just by looking at the setup without any math? Which charge would the e-field be pointed at?
     
  4. Jul 12, 2012 #3
    I think It would be pulled to the Positive quadrant. quadrant 1. And I dont worry about the two positive qs because one is in the x component, and one is in the y component, and there is a negative in the opposite side on each of those.
     
  5. Jul 12, 2012 #4
    Oh shoot never mind, its supposed to terminate at the negative pole.
     
  6. Jul 12, 2012 #5
    The Efield vector points a the negative charge.

    But it has a magnitude as well. Fortunately its the same equation doubled. Both Efield vectors that do not cancel each other out point in exactly the same direction and have the same magnitude, so you double.
     
  7. Jul 12, 2012 #6
    the angle is still a 45 degree angle right, only to the negative side. right?
     
  8. Jul 12, 2012 #7
    Sure its 45 degrees with respect to the -x and -y axis shown because the vector points to the -q charge.

    But r is of course half of the diagonal for the field contributions from the +q charge and -q charge. Since you have the sides of the square you can find half of the hypotenuse or half of the diagonal so that you can give a magnitude to the field with numbers that has q as the unknown quantity or the given quantity (q). ie they dont give you q in coulombs so its just a part of your answer.

    Ho there just a millisecond...

    I see this is a problem that you have to input answers and you only get so many tries? Is this correct? If it is, then maybe they want 180 degrees + 45 more degrees as the direction of the field? Im not sure what they want but it should be obvious without any math that the field direction from the center is towards the -q charge
     
    Last edited: Jul 12, 2012
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