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Four particles form a square with different charges

  1. Sep 18, 2009 #1
    http://www.webassign.net/hrw/hrw7_21-22.gif

    Four particles form a square. The charges are q1= q4=Q and q2=q3=q. a) what is Q/q if the net electrostatic force on particles 1 and 4 is zero? B) is there any value of q that makes the net electrostatic force on each of the four particles zero? Explain.

    Spending way too much time on this problem!

    I know that Fnet = F1 + F2 +...

    also F = K (absq1)(absq2)/ r^2

    absolutely clueless
     
  2. jcsd
  3. Sep 18, 2009 #2
    Are you familiar with vector notation?
     
  4. Sep 18, 2009 #3
    kind of not really
     
  5. Sep 18, 2009 #4
    i think i got it but i dont understand why you are adding here to get the force of q4 on q1 here using cos and sin.....F 41 = − Q⋅Q4πε 0 ⋅ 2a2 cos 45°iˆ + Q⋅Q/4π ε0 ⋅ 2a2 sin 45°
     
  6. Sep 18, 2009 #5
    i think a sketch would help you to understand:

    http://www.go-krang.de/physicsforums.com[/URL] [Broken] - thread - 338289.png[/PLAIN] [Broken]​

    I assumed that
    [tex]q < 0[/tex]​
    so the forces
    [tex]\vec F_{12}, \vec F_{13}[/tex]​
    are attractive! Of course
    [tex]\vec F_{14}[/tex]​
    is repulsive!


    So if you are not familiar with vectors, we have to calculate the magnitude of
    [tex]F_{12+13} = \sqrt{F_{12}^2 + F_{13}^2} = \sqrt{2} \cdot F_{12} ~~~~~~~~~~~~~~ \rm{with} ~~~~~~ F_{12} = F_{13}[/tex]​


    Cause the forces F_{12+13} and F_{14} are antiparallel the requirement for your question is
    [tex]F_{12+13} = F_{14}[/tex]​



    Because of the symmetrie of this problem, that's the only thing you have to show!



    Hope i could help you!?
     
    Last edited by a moderator: May 4, 2017
  7. Sep 18, 2009 #6
    thanks
     
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