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Calculating force on the charges placed in the corners of a triangle

  1. Dec 25, 2009 #1
    1. The problem statement, all variables and given/known data

    We place three charges in the corners of the equilateral triangle; the size of the sides is 1 m. Charges are of sizes: e1, e2 = −2e1, and e3 = 3e1, where e1 = 10 ^-6 As. What is the force on the first (second, third) charge and in which direction are they showing?

    http://item.slide.com/r/1/20/i/uNrFtWB4tT_sS7zkuV1ExJS_gf3980kJ/ [Broken]
    http://item.slide.com/r/1/130/i/1G12EjLq5D8Lo4lU8yYKxmOQfVTQfoSl/ [Broken]
    http://item.slide.com/r/1/142/i/dpHAZtcg7j-o-U_uDNFiH60YH-Do8Dwr/ [Broken]

    2. Relevant equations

    F= k[q(1)q(2)/ d²]

    3. The attempt at a solution

    Calculating for e(1):

    d= 1 m

    F(1,2)= k[q(1)q(2)/ d²] → F(1,2)= 1.8 N

    F(1,2-y)= -F(1,2) sin 60º= - 1.56 N
    F(1,2-x)= -F(1,2) cos 60º= - 0.4 N

    F(1,3)= k[q(1)q(3)/ d²] → F(1,3)= 2.7 N

    F(1,3-y)= f(1,3) sin 60º= 2.34 N
    F(1,3-x)= -F(1,3) cos 60º= - 1.35 N

    F(1-x)= -2.25 N
    F(1-y)= F(1,2-y) + F(1,3-y)= 0.78 N

    F(1)= sqrt[F(1-x)² + F(1-y)²]= 2.39 N

    Are my calculations for e(1) correct?
    Thank you for helping!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 25, 2009 #2
    Have't looked at them all but cos 60=0.5 so not sure where F(1,2-x) came from.
     
  4. Dec 25, 2009 #3


    My mistake, it was a typo. here is the correct calculation as I made it:

    F(1,2-x)= -F(1,2) cos 60º= - 0.9 N

    Now are my calculations correct for e(1)?
    Thank you for helping!
     
  5. Dec 25, 2009 #4
    My results are different:

    Here is how I did force of particle 2 on 1: k= 9.0 E9

    total force (2 on 1) = K (1E-6)(2E-6)/1m^2=18 E-3. Since this is attractive force, it is directed upward and rightward.

    Fx = F(2-1) cos 60= 9 E-3N
    Fy = F(2-1) sin 60 = 0.87 * 18E-3= 15.6 E-3

    Do the same for 3 on 1, and then add x components, then Y components and use Pythags formula for total. Direction can be found using tan(angle)=sum y/ sum x
     
  6. Dec 26, 2009 #5
    Thank you for helping, I think I understand now :smile:
     
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