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How did they divide the equations ?

  1. Mar 1, 2014 #1
    How did they "divide the equations"?

    1. The problem statement, all variables and given/known data

    Two 5.0g point charges on 1.0m-long threads repel each other after being charged to +100nC.
    What is angle theta? You can assume theta is a small angle.

    8SSFQJJ.png

    2. Relevant equations
    [itex]K=9.0*10^9 Nm^2/C^2[/itex]
    [itex]g=9.8m/s^2[/itex]
    [itex]q=\frac{K|q_1||q_2|}{d^2}[/itex]

    3. The attempt at a solution

    I determined my unknowns and translated everything into standard units, and drew out the free-body diagram.

    eQsK9SG.png

    Translated everything into standard units.
    5.0g = 5.0*10^(-3) kg
    100nc = 100*10^(-9) C

    [itex]\frac{1}{2}d=\sin{\theta}[/itex]
    [itex]d=2\sin{\theta}[/itex]
    [itex]T_x=T\sin{\theta}=F_{2on1}=\frac{Kq^2}{d^2}[/itex]
    [itex]T_x=T\sin{\theta}=F_{2on1}=\frac{Kq^2}{(2\sin{\theta})^2}[/itex]
    [itex]T_x=T\sin{\theta}=F_{2on1}=\frac{Kq^2}{4sin^2{\theta} }[/itex]
    [itex]T_y=T\cos{\theta}=mg[/itex]

    The "missing" step:
    The solutions manual says that the next step from here is to "divide the two equations and solve for q". What I don't know is how they "divided the two equations" to get this:

    [itex]\sin^2{\theta}\tan{\theta}=\frac{Kq^2}{4L^2mg}=4.59*10^{-4}[/itex]
     
  2. jcsd
  3. Mar 1, 2014 #2

    Doc Al

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    Staff: Mentor

    Divide this equation:
    By this one:
    Divide the left sides by each other and the right sides by each other.
     
  4. Mar 1, 2014 #3

    tiny-tim

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    Homework Helper

    hi aleksbooker! :wink:
    you divide Tsinθ by Tcosθ, to give you Ttanθ = … ? :smile:

    (hmm … there seems to be an L2 missing :confused:)
     
  5. Mar 1, 2014 #4

    Doc Al

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    Staff: Mentor

    Apparently, L is the length of the string, which is given as 1 m.
     
  6. Mar 1, 2014 #5
    Mind blown. Thanks @Doc Al. Okay, that works - I'm assuming it only makes to do something like this when you can use it to eliminate a variable (like T in this problem).
     
  7. Mar 1, 2014 #6
    Yea, L was the length of the string.
     
  8. Mar 1, 2014 #7

    Doc Al

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    Staff: Mentor

    Think of it as just another tool in your bag of tricks. A useful one.
     
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