1. Not finding help here? Sign up for a free 30min 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!

Electric Force: Equilateral Triangle

  1. Jul 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Three charges, +Q, +Q, and -Q are placed at the vertices of an equilateral triangle of length "s" on a side. Find the magnitude and direction of the force on one of the +Q charges.


    2. Relevant equations
    ForceElectric=(K*q1*q2/d2)


    3. The attempt at a solution
    I've set the triangle up so that the bottom side is parallel to the x-axis. My +Q charges are in both the bottom left (charge 1) and bottom right (charge 3) corners, while the -Q charge (charge 2) is at the top. Charge 3 is the charge that I have chosen to do the calculation for.

    Force2 on 3 = (2QK/s2)

    Force2 on 3 = Force1 on 3, so Force1 on 3 = (2QK/s2)

    x-component: Forcex:2 on 3 = (2QK/s2)*cos(60°) = (QK/s2)

    y-component: Forcey:2 on 3 = (2QK/s2)*sin(60°)



    ∑Fx = Force1 on 3 - Forcex:2 on 3

    ∑Fx = (2QK/s2) -(QK/s2)


    ∑Fy = Forcey:2 on 3

    ∑Fy = (2QK/s2)*sin(60°)




    Fnet = ∑Fy + ∑Fx (vector equation)

    Fnet = √((∑Fy)2 + (∑Fx)2)

    Simplified this (hopefully) goes to:



    Fnet = (2QK/s2)

    This was my final answer for the magnitude, but I'm not even sure if it is correct, or if my procedure is right. I'm also not sure how to find the direction. Any help would be greatly appreciated! Thank-you!
     
  2. jcsd
  3. Jul 17, 2014 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Hi Sirmeris. Welcome to Physics Forums.

    Check your algebra ....

    Q x Q = Q2
     
  4. Jul 17, 2014 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    Once you have ∑Fx and ∑Fy you can determine the magnitude using Pythagoras, as you show, and the angle, θ, of that vector is given as

    θ = tan-1 (∑Fy / ∑Fx)

    Always wise to sketch these, so you can be sure to give your final answer in the correct quadrant.
     
  5. Jul 18, 2014 #4
    Hi Sirmeris
    Use this diagram to find net force on +Q.
    Net.F=√(F^2+F^2+2F*Fcosθ)
    θ=120.In this problem force on +Q due to -Q and other +Q charge is equal.
     
  6. Jul 18, 2014 #5
    Untitled.png this is the figure.
    What should be the direction of the resultant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electric Force: Equilateral Triangle
Loading...