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    Please someone explain what's going on to me, I have no clue where 8.99*10^9 comes from and why they're only dividing by the separation squared rather than 4(pi)(permittivity of free space)(separation squared)

    I also don't understand where the Fr = 2cos30 comes from...

    Thanks in advance
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    basically 1/(4x(pi)x(permittivity of free space)) = 8.99x10^9
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    and then they have left r^2 under the equation
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    the Fr = thing is just vertically resolving the force
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    (Original post by Heisenberg 23)
    the Fr = thing is just vertically resolving the force
    Thanks a lot for explaining that - I still don't understand the resolving part though, where does the 2 come from??

    Thanks in advance.
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    (Original post by Wolfram Alpha)
    Thanks a lot for explaining that - I still don't understand the resolving part though, where does the 2 come from??

    Thanks in advance.
    There is a 2 because there is two forces. On the diagram there is two F's
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    (Original post by Heisenberg 23)
    There is a 2 because there is two forces. On the diagram there is two F's
    But if you're resolving vertically you'll split the triangle into a right angled triangle meaning you'll only have one F, or not?
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    so from the equation above you figured out that the force of one of those 4.18 charges in 14.1. so you can create a right angled triangle with the force acting as the hypotenuse and the adjacent being the vertical force. Because it an equilateral that means all the angles are 60 degrees so you can halve that for the angle between hyp and adj. But there are two potential triangles so you times it by 2. Hope this helped
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    (Original post by Wolfram Alpha)
    But if you're resolving vertically you'll split the triangle into a right angled triangle meaning you'll only have one F, or not?
    This is an equilateral triangle, and both the charges at the other 2 edges of the triangle have the same charge. So you have 2 forces in different directions, and you want the resultant force. The horizontal force from the two charges cancel out, so only the vertical component is what we need to compute. The vertical component from each particle is the projection on to the centre of the triangle i.e. F_v=\vec{F} \cdot \vec{y} and thus for the two particles together this is F_v=2\vec{F} \cdot \vec{y} =2F \cos(\theta)=2F \cos(30)
 
 
 
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