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Net electric force in particles in triangular formation

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Three positive point charges are arranged in a triangular pattern in a plane, as shown below. The Coulomb constant is 8.98755 × 109 N · m2/C2.
    Find the magnitude of the net electric force on the 2 nC charge. Answer in units of N.

    2. Relevant equations
    Fe = (kq1q2) / (d2)


    3. The attempt at a solution
    I was looking at this thread with the exact same problem and I followed the directions. However, I don't know what I am doing wrong.
    https://www.physicsforums.com/showthread.php?p=1892198"

    My diagram: (it's like the picture given in the thread I looked at but just with different numbers)

    3nC
    (+)
    |
    2m
    ||---2m---|
    -------------(+) 2nC
    |
    2m
    |
    (+)
    6nC

    What I tried to do:
    ((8.9875e-9)(2e-9)(3e-9))/((sqroot(8))2) = 6.740625e-9
    ((8.9875e-9)(2e-9)(6e-9))/((sqroot(8))2) = 1.348125e-8
    cos(45)6.740625e-9 + sin(45)6.740625e-9 = 9.53268329e-9
    -cos(45)1.348125e-8 + sin(45)1.348125e-8 = 0

    After that I am confused.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 28, 2009 #2

    LowlyPion

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    Homework Helper

    I'd write it out more simply to keep track of what's going on.

    F1 = k*2*nC*(3/(2*4) i - 3/(2*4) j)
    F2 = k*2*nC*(6/2*4) i + 6/(2*4) j)

    Fnet = (k*nC/4)*(9 i + 3 j)

    |Fnet| = (k*nC/4)*(81 + 9)1/2
     
  4. Mar 28, 2009 #3
    I still don't understand what i and j are and where all the other numbers came from.
     
  5. Mar 28, 2009 #4

    LowlyPion

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    Homework Helper

    F is a vector right?

    i,j are the x,y components of the vector.

    The d or r radius is to the charges that are √2*2 away.
    When you square that it's 2*4 right?
     
  6. Mar 28, 2009 #5
    ok, then what about the "2*nC" part? nC is nano-Coulombs though...
    Am I supposed to find the force between 3nC and 2nC and then the force between 2nC and 6nC? After that, do I use the Pythagorean theorem to get the resultant force?
     
  7. Mar 28, 2009 #6
    OK, nevermind. I did what I just posted and I got it right!

    F1 = ((8.9875e-9)(2e-9)(3e-9))/((sqroot(8))2) = 6.740625e-9
    F2 = ((8.9875e-9)(2e-9)(6e-9))/((sqroot(8))2) = 1.348125e-8
    sqroot ((F1)^2 + (F2)^2) = approx. 1.51e-8

    YAY!
     
  8. Mar 28, 2009 #7
    NOw, another question.

    How do I find the direction of this force?

    I did
    inversetan(f1/f2) = 26.6 but it's wrong
     
  9. Mar 28, 2009 #8

    LowlyPion

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    That would be the method.

    I didn't run the math. Apparently it works for you. So good.
     
  10. Mar 28, 2009 #9

    LowlyPion

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    Aren't the components in the ratio (3 j)/(9 i) = 1/3 ?

    Arctan(1/3) is angle with the positive x axis?
     
  11. Mar 28, 2009 #10
    oh i see. thanks so much for your help! i've learned more from you than from my teacher...
     
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