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Homework Help: Equilateral triangle coulombs law problem

  1. Jun 12, 2010 #1
    [1. The problem statement, all variables and given/known data
    1.) Three charges are placed 25 cm apart in an equilateral triangle. Q1 has a charge of -6.0 nC, q2 has a charge of 4.0 nC and q3 has a charge of 4.0 nC. what is the force on q2?

    q1 is the top point, q2 is the bottom left leaving q3 on the bottom right

    2. Relevant equations

    F = K q1q2/r^2

    3. The attempt at a solution

    Fnet2 = F12 + F32

    using the above equation for f12 i get: 3.45 * 10^-6

    components:(3.45 * 10^-6) sin60 = 2.99 * 10^-6 North
    :(3.45 * 10^-6) cos60 = 1.72*10^-6 East

    F32 using the equation above = 2.3 *10^-7 West

    combining the x components = 5.81 *10^-6 West

    so using pythagoras: Root((5.81 * 10^-6)^2 + (2.99*10^-6)^2))

    = 3.05 * 10 ^-6 C

    angle= arctan((2.99*10^-6/ 5.81*10^-6))
    = 79 degrees North of west

    so my final answer is 3.05 mirco coulombs (79 degrees N of W)

    Is this correct? I want to make sure i have a grasp on these problems before my final exam on the 18. Any help is greatly appreciated!
  2. jcsd
  3. Jun 12, 2010 #2
    i used 10^-9 and meters, and i see i made a mistake in the answer it should be 3.05 newtons not micro coulombs.
  4. Jun 12, 2010 #3
    OK firstly I'd say that drawing a quick diagram really helps with visualising a problem like this. Here's one if you don't have one already http://yfrog.com/e4pf3ij


    Using that equation:


    You can indeed work out each of the forces, i.e. of each charge on each other charge, as you've done already.


    [tex]F_{1 on 2} = \frac{k\times (-6)\times {4}}{(25)^{2}} = \frac{-24k}{625}[/tex]

    Then you need to account for the angle between them (since not on a straight line) so should actually be:

    [tex]F_{1 on 2} = \frac{-24k}{625}sin(30)[/itex]

    Then you can use the same method to calculate [itex]F_{3 on 2}[/itex] noting that the angle between them is 0 i.e. don't need to account for an angle between these.

    Hopefully this helps a bit more :smile:
  5. Jun 12, 2010 #4
    Just to add that you should find that [itex]F_{2 on 3}=F_{3 on 2}[/itex] so that you only need to worry about the components [itex]F_{1 on 2}[/itex] and [itex]F_{2 on 1}[/itex].

    Also you can simplify the calculations using that [itex]sin(30)+sin(60)=\frac{1+\sqrt{3}}{2}[/itex] and that [itex]\frac{kq_{1}q_{2}}{r^{2}}[/itex] components of both are equal.

    Hence you should be able to work out the resulting force on charge [itex]q_{2}[/itex]
  6. Jun 12, 2010 #5
    okay im slightly confused, why did you use 30 degrees instead of 60? and is my overall answer correct?
  7. Jun 12, 2010 #6

    Shouldn't it be:

    [tex]F_{1 on 2}=\frac{k\times q_{1}\times q_{2}}{r^{2}}=\frac{(9\times10^{9}N.m^{2}.C^{-2})(-6.0\times 10^{-9}C)(4.0\times 10^{-9}C)}{(25m)^{2}}=-3.45\times 10^{-10}N[/tex]

    Then account for the angle between them, so:

    [tex]F_{1 on 2}=-3.45\times 10^{-10}sin(30)N = -1.73\times 10^{-10}N[/tex]

    Also, since [itex]F_{3,2}=F_{2,3}[/itex] then these forces just cancel each other out right.

    Then the other force to consider:

    [tex]F_{2 on 1}=\frac{k\times q_{2}\times q_{1}}{r^{2}}=\frac{(9\times10^{9}N.m^{2}.C^{-2})(4.0\times 10^{-9}C)(-6.0\times 10^{-9}C)}{(25m)^{2}}=-3.45\times 10^{-10}N[/tex]

    Then account for the angle:

    [tex]F_{2 on 1}=-3.45\times 10^{-10}sin(60)N = -2.99\times 10^{-10}N[/tex]

    Therefore the force on [itex]q_{2}[/itex] is:

    [tex]F=4.6\times 10^{-11}N[/tex]

    Well, that's what I've just got anyway. Your whole method is a bit different. Hopefully that's clearer now :smile: (for both of us, was getting bit confused as well :wink:)
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