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Electrostatic Force of a Triangle

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose that the magnitude of the net electrostatic force exerted on the point charge q2 in the figure is 0.65 N . (Figure 1) http://imgur.com/4lZliPq

    Find the distance D.


    q1 = 2.1 micro C
    q2 = 6.3 micro C
    q3 = -.89 micro C


    2. Relevant equations

    F = k |q1| |q2| / r^2
    k = 8.99 E9

    3. The attempt at a solution

    Normally for this problem I'd try to find the different forces that affect each other but for this problem there is 2 unknowns and I'm not sure how to use the net force of q2 to find the distance.

    Fx 1 on 2 = K|q1||q2|(cos 60) / d^2 =

    Fy 1 on 2 = K|q1||q2|sin60 / d^2=

    Not really sure where to go with this problem :( any help would be appreciated
     
  2. jcsd
  3. Feb 19, 2013 #2

    Simon Bridge

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    Force is a vector - you know how to find the relation for the magnitude of each force individually, and you know what they have to add up to, and you know how to do a vector sum.

    You are doing the trig a little early - just add them head-to-tail.
    (Remember about triangles and parallelograms?)
     
    Last edited: Feb 19, 2013
  4. Feb 20, 2013 #3
    Hmmmm....Not quite following you, so in a way we should work backwards?

    F2= .65 N

    .65 = F 1 on 2 + F 3 on 2

    Is that what you're saying?
     
  5. Feb 20, 2013 #4

    Simon Bridge

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    I'm saying $$|\vec{F}_2|=|\vec{F}_{21}+\vec{F}_{23}|=0.65 \text{N}$$... remember, forces are vectors.
    The net force on 2, is the vector sum of the forces due to 3 and 1 - and the magnitude of the resultant force is 0.65N.

    Try this - draw a free-body diagram for q2.

    Sketch arrows for the directions of the forces (one of them will be a lot bigger than the other, and be careful about the directions since the other two charges have opposite signs.) Concentrate on getting the angles right - use a protractor.

    Draw the parallelogram of vectors to get the resultant.
    This will give you two triangles and some angles - you also know the size of the resultant vector.
    Use your knowledge of triangles - you know, all that geometry you did in math class?
    sum of angles, the sine rule, the cosine rule, that stuff.
     
  6. Feb 20, 2013 #5
    So this is what I have so far...

    .65N=F12 + F32
    .65N = (Kq2 / R^2)(q1cos60 + q1sin60 + q3cos60 + q3sin60)

    Am I on the right track?
     
  7. Feb 20, 2013 #6
    Anyone have any ideas?
     
  8. Feb 21, 2013 #7

    Simon Bridge

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    This does not make sense. Where does all this trig come from? It looks to me that you don't know how to add vectors geometrically. OK then - define an x and y direction and resolve the vectors against them.
     
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