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3 charges arranged in the form of an equilateral triangle

  1. Jun 26, 2005 #1
    Hello everyone, I'm confused on how you find the distance between the charges. They want you to find the force on q3 due to q1 and q2. They are arranged in an equilateral triangle. It looks like:



    2
    |
    |.............(q3)
    |
    1
    |
    |
    (q1)-------1--------(q2)

    q1 is at the orgin, q2 is 2m away from q1 on the x-axis. q3 is alittle below 2m on the y axis. They use a distance of 2m to find F31, but i don't get how they got 2m. I see q3 is 1m away from q1 if u are looking at the x-axis, and its not qutie 2m away from q1 on the y axis, so how did they figure the distance? :bugeye: Thanks.
     
  2. jcsd
  3. Jun 26, 2005 #2

    Doc Al

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    Staff: Mentor

    Since the charges form an equilateral (which means equal-sided) triangle, the distance between any two is the same.

    When finding the net force on q3, be sure to take direction into account.
     
  4. Jun 26, 2005 #3

    jtbell

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    Further hint: how big are the angles in an equilateral triangle? (All three of them are equal, of course!)
     
  5. Jun 26, 2005 #4
    Since all the three are positive charges , we will take the force due to one charge on second to be directed away from the first.Since this is an equilateral triangle, the distances remain the same between any two charges.Find the force on one due to two others and use vectors to solve further.

    BJ
     
  6. Jun 26, 2005 #5
    Thanks for the responces, that makes sense. Does it make a difference that they arn't all positive charges? q1 and q3 are positive, q2 is a negative charge. I'm trying to find the resultant force of q3. They did the following: Rx = F31cos(60) + F32cos(60) I understand why they are using cos, but i don't get how they found the angle to be 60. No angle was ever given. I suck in geometry maybe this is why i can't see it.
     
  7. Jun 26, 2005 #6

    Doc Al

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    Staff: Mentor

    As jtbell hinted: the three angles in this triangle are equal. And the sum of the angles in any triangle is ???
     
  8. Jun 26, 2005 #7
    Alright i'm getting closer and closer to the answer. Thanks for the reminder that the sum must equal 180. What I did was the following:
    I'm summing up the resultant forces in the x direction on F31 and F32.

    I found the force in the x direction of F32 to be .0169N which is what the book has, but I used: F32*sin30 = Fx. They used, F32*cos60. So I thought I was doing the problem semi right even though I got the answer a different way. So i tried to find F31 in the x direction.


    But now i'm confused, because I thought I would use an angle of 30 but insteed i should use an angle of 60. Is there some geometric rule about exterior angles that i'm forgettting?

    Heres a picture to try and simply what i'm saying:

    ...........^F31
    ........./
    ...|.../
    ...| / 60
    (q3)------->
    30| 30\ 60
    ...| ....\
    ...|......\ F32
    .............v



    Wow that picture just made things suck even more hah, damn these white spaces! So the question is, where are they getting 60, once i figure that out i think i got the problem.
     
  9. Jun 26, 2005 #8

    Doc Al

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    Realize that [itex]\sin 30 = \cos 60[/itex], so either way is correct.

    The typical rule is that [itex]F_x = F \cos \theta[/itex], where [itex]\theta[/itex] is the angle that F makes with the x-axis. In this problem, [itex]\theta = 60[/itex] degrees.
     
  10. Jun 26, 2005 #9
    Thanks!

    Thanks for the help!
     
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