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Homework Help: Magnitude and direction of electric force (file attached)

  1. Jul 28, 2007 #1
    1. The problem statement, all variables and given/known data
    I was kind of confused on how the radius for the charge on q3 by q1 is 2a^2, i attached the file for my notes regarding this. The question is at the bottom of the pdf scan.


    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

    Last edited: Jul 28, 2007
  2. jcsd
  3. Jul 28, 2007 #2

    Doc Al

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    The needed distance is the hypotenuse of the right triangle whose sides are both "a". Hint: Pythagorean theorem.
     
  4. Jul 28, 2007 #3
    c^2=a^2+b^2

    so

    c^2=a^2+a^2
    c^2=2a^2

    what about the ^2 for the c, is it just left there? because it seems like it since r=2a^2
     
  5. Jul 28, 2007 #4

    Doc Al

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    Careful: it's r^2 = 2a^2, not r = 2a^2.
     
  6. Jul 28, 2007 #5
    On this other page of notes which has to do with the introduction of acceleration in electric fields, there are three parts of the problem that im not quite following, i figured it would be easier if i just wrote it down by hand on the actual note sheet than typing it all out.:smile:

    For the cosign question, i dont undestand what happened to it because the sin is still there in the next step but not the cos.
     

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    Last edited: Jul 28, 2007
  7. Jul 28, 2007 #6
    If what im asking isnt really quite clear, i can write it out again more specifically and scan it.
     
  8. Jul 29, 2007 #7

    Doc Al

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    (Be sure to include the full problem statement next time.)

    (1) Why negative? The field points down (thus is negative), the charge is positive: so the force and acceleration are both downward, thus negative.

    (2) They are solving for the value of time that makes y = 0. Set that factor equal to zero and solve for t.

    (3) Regarding what happened to the cosine: note that [itex]\theta[/itex] changed to [itex]2\theta[/itex]. (Review your trig identities for [itex]\sin\theta \cos\theta[/itex].)
     
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