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Four Equidistant Particles in electric fields

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    In Fig.22-30, the four particles form a square of edge length a = 6.50 cm and have charges q1 = 7.59 nC, q2 = -10.9 nC, q3 = 11.5 nC, and q4 = -6.06 nC. What is the magnitude of the net electric field produced by the particles at the square's center?

    http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c22/fig22_32.gif


    2. Relevant equations

    E=kq/r^2



    3. The attempt at a solution

    First thing, I converted charges from nC to C as well as distances from cm into m

    q1= 7.59 E-9 C
    q2= -1.09 E-8 C
    q3= 1.15 E-8 C
    q4= -6.06 E-9 C

    a=0.065m


    after that, I formed triangles and used Pythagorean theorem to get distance for each particle to the center.

    Center distance (hypotenuse) = 0.045962


    So what I thought was find the Electric field produced by each particle with r= the hypotenuse and their sum would be the the net electric field

    E1= (8.99 E 9)*(7.59 E -9)/ (0.045962)^2 = 32300.1

    E2=(8.99 E 9)*(-1.09 E-8)/ (0.045962)^2 = -46386.2

    E3= (8.99 E 9)*(1.15 E-8)/ (0.045962)^2 = 48939.5

    E4= (8.99 E 9)*(-6.06 E-9)/ (0.045962)^2 = -25789

    E net= 32300.1 + -46386.2 + 48939.5 + -25789 = 9064.45


    I entered that in but my online homework is marking it as incorrect. Could someone please help. Thanks :smile:
     
  2. jcsd
  3. Feb 4, 2012 #2

    tiny-tim

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    Hi C6ZR1! :smile:
    nooo :redface:

    fields are vectors, so you can't just add their magnitudes (scalar addition) …

    you must add them as vectors :smile:
     
  4. Feb 4, 2012 #3
    so then for each point would I use coordinates and subtract that from the center, C (0.0325,0.0325) and go about getting each particle in vector form?
     
  5. Feb 4, 2012 #4

    tiny-tim

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    i'm not sure what you mean by that

    do it the easy way …

    each vector points along a diagonal, so add the opposite pairs first, and that will give you two perpendicular vectors to add :wink:
     
  6. Feb 4, 2012 #5
    but how do we know its components if we are only given the points? Sorry if I'm over looking it. I literally just learned about vectors/cross and dot products this week in school. :redface:
     
  7. Feb 4, 2012 #6

    tiny-tim

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    technically, a vector isn't components, it's a line (with an arrow at the end of it)

    so draw the lines (with arrows) …

    each line will start at the centre, and point towards (or away from, if the force is repulsive) each of the four corners

    (you could write out the components, but there's no point when the vectors themselves are so easy to draw)
     
  8. Feb 4, 2012 #7
    ok, assuming its a positively charged point charge in the center then the arrows are are pointing towards, the center from Q1, and Q3. and towards Q3 and Q4
     
  9. Feb 4, 2012 #8

    tiny-tim

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    (you meant …)
    yes :smile:

    and now add each pair :wink:
     
  10. Feb 4, 2012 #9
    so adding
    q1 and q3: 1.909E-8 C
    q2+q4: -1.696 E-8

    but I'm still having a little trouble trying to understand why we do this. Is it because those pairs are on a same line?
     
  11. Feb 4, 2012 #10

    tiny-tim

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    sorry, where do those figures come from? :confused:
    yes, it's easy to add vectors that are in the same line
     
  12. Feb 4, 2012 #11
    They were the charges of each particle in C. I dont think I'm understanding this too well. :/ What would I add?
     
  13. Feb 4, 2012 #12

    tiny-tim

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    ohhh

    you should be adding the vectors (the fields)
     
  14. Feb 4, 2012 #13
    Vector fields = Electric field produced by each particle?
     
  15. Feb 5, 2012 #14

    tiny-tim

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    (just got up :zzz: …)

    yes :smile:
     
  16. Feb 5, 2012 #15
    lol, well good morning.

    So if the vector fields are electric fields produced by each particle then was my original attempt correct my finding the electric fields and adding them up?
     
  17. Feb 5, 2012 #16

    tiny-tim

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    you didn't add the fields, you only added (or subtracted) their magnitudes :redface:
     
  18. Feb 5, 2012 #17
    ohhh, but isnt electric field E=(kq)/r^2?
     
  19. Feb 5, 2012 #18

    tiny-tim

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    no, its (kq)/r2 radially away from the source

    it's a vector!!

    potential is a scalar, field is a vector
     
  20. Feb 5, 2012 #19
    so is K the positive test charge in the center of the field?
     
  21. Feb 5, 2012 #20

    tiny-tim

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    you mean the k in kq/r2?

    no, that's just a constant
     
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