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