1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook