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More electric fields

  1. Jun 3, 2008 #1
    [SOLVED] more electric fields

    1. The problem statement, all variables and given/known data
    four particles form a square of edge length a = 5.20 cm and have charges q1 = 10.8 nC, q2 = -8.87 nC, q3 = 11.6 nC, and q4 = -10.4 nC. What is the magnitude of the net electric field produced by the particles at the square's center? q1 is in quadrant 2, q2 is in quadrant 1, q3 is in quadrant 4, and q4 is in quadrant 3

    (SORRY THERE IS NO PICTURE, BUT HOPEFULLY THE DESCRIPTION WILL HELP YOU TO ENVISION THE FIGURE)

    2. Relevant equations
    E=k(q/d^2) also multiplied by the sin or cos of a relevant angle


    3. The attempt at a solution

    the value for d is obtained by the pythagoran theorem where d= sq.rt. 5.2^2+5.2^2= 3.68cm

    I have

    E2= 9e9(-8.87e-9/.0368^2) = -58948
    E4= 9e9(-10.4e-9/.0368^2) = -69116

    I sum those parts as the vector diagrm shows that the field for these two particles is pointing the same direction and get E-= -128064

    E3= 9e9(10.8e-9/.0368^2) = 71775
    E4= 9e9(11.6e-9/.0368^2) = 77091

    Again summing the parts to get E+= 148866

    I try to find the components by

    E+(cos45)= 105264
    E+(sin45)= 105264

    E-(cos45)= -90555
    E-(sin45)= -90555

    then I use vector addition and obtain 14709i + 14709j
    to find the magnitude of the field I used

    sq.rt 14709^2 + 14709^2 = 20803 N/C (wrong according to the online program)

    Where did I go wrong? The only example in our text uses symmetry to cancel out all but one charge, so when I look at it to compare why I'm getting the answer wrong, I can't complete my problem because I still have four charges to contend with, and don't know where I'm messing up. any help is great! Thanks!
     
  2. jcsd
  3. Jun 3, 2008 #2

    nrqed

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    Wait...you want the E field at the center of the square, right? So the distance you must use is
    [tex] \sqrt{2.6^2 + 2.6^2} cm [/tex] not what you wrote!

    Also be sure to put everything in starndard units (covert back the distance to meter and the charges to Coulombs)
     
  4. Jun 3, 2008 #3
    ummm, ok, so I messed up what I wrote in my post.....I took the sq.rt. 2.6^2 + 2.6^2 = 3.68cm which is equal to .0368m, so everything IS in standard units in the equations that I posted
     
  5. Jun 3, 2008 #4
    "I sum those parts as the vector diagrm shows that the field for these two particles is pointing the same direction and get E-= -128064"

    Are you sure the field for (any) two of the particles is pointing in the same direction?

    q1 q2 Look at your diagram and draw the vectors.
    .
    q4 q3
     
  6. Jun 3, 2008 #5
    But even when I just did each field individually and added them up component-wise, I still ended up with 20801 N/C.

    In other words, I used

    E1x=E1(cos45) = 50752 (same for E1y)
    E2x=E2(cos45) = -41683 (same for E2y)
    E3x=E3(cos45) = 54512 (same for E3y)
    E4x=E4(cos45) = -48873 (same for E4y)

    when the components are added up, I come up with 14708i + 14708j
    So, I'm still stuck....did I do it wrong still?
     
  7. Jun 3, 2008 #6
    oooooh! I figured it out! Thanks for the little tip there fantispug....though cryptic, I went back and actually looked at which direction the vectors were pointing, and realized that certain components required certain signs to be correct....hmmm, not so tricky now, are you physics problem.... :))
     
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