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Finding the net electrostatic force on particle 1, triangle!

  • Thread starter mr_coffee
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  • #1
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Hello everyone, Did I do the 2nd part of this problem correct? Part B. I boxed in the answer, i think it will just be easier by showing you my drawing so here is the picture -> http://img221.imageshack.us/img221/9340/phsyicss9lb.jpg [Broken]
thanks!
 
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Answers and Replies

  • #2
Doc Al
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I suggest that you recheck your arithmetic for part a. The method you used for part b looks OK (it's a bit hard to follow), but it's not the easiest way to get the answer. (You didn't take full advantage of the symmetry of the geometry.)
 
  • #3
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thanks for the reply but I don't see how part A is wrong...
F = [k(q1)(q2)]/r^2;
F = [9.9E9*(45.0E-6)^2]/(2.70)^2 = 2.75 N
 
  • #4
Doc Al
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The Coulomb constant (k) is about 9.0 E9, not 9.9 E9.
 
  • #5
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thanks alot that woulda sucked!
 
  • #6
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for the second part, i don't even think it's necessary to use trigonometry.

consider point 3 at the origin in R2. put particles 1 and 2 at the appropriate positions in quadrant's 3 & 4. draw your force vectors for each of the forces. add them visually -- they interfere constructively directly in the +y direction. it looks to me like you'd just have to multiply your answer from a by 2, due to the geometry.

i think it's right and a lot easier than breaking it down into components, but it looks fine barring the oofpez business.
 
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  • #7
Doc Al
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teclo said:
consider point 3 at the origin in R2. put particles 1 and 2 at the appropriate positions in quadrant's 3 & 4. draw your force vectors for each of the forces. add them visually -- they interfere constructively directly in the +y direction. it looks to me like you'd just have to multiply your answer from a by 2, due to the geometry.
You would multiply the y-component by 2 to get the answer. But you'd still have to use some trig to find the y-component. (The answer to part a is the full force between two charges, not the y-component.)

Note to mr_coffee: This is the approach I would use, since it takes advantage of the symmetry of the problem.
 
  • #8
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(The answer to part a is the full force between two charges, not the y-component.)
So part A isn't correct? I don't see why I would need to break up part A into components if its a straight line. F = [(9E9)(45.0E-6)^2]/(2.70m)^2 = 2.5N
The way I did my part B isn't it also correct though, even though I didn't do it the best of ways? I got a final answer of 4.33N
 
  • #9
Doc Al
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Your solutions are perfectly OK. My only point was that there's an easier way to get part b.
 
  • #10
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Oh alright, thanks for the help and i'll keep that in mind the next time! :biggrin:
 

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