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MasteringPhysics: Find the angle between the point charges.

  1. Mar 11, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-3-10_22-15-30.png
    Two m = 4.0 g point charges on 1.0-m-long threads repel each other after being charged to q = 80 nC , as shown in the figure. What is the angle θ? You can assume that θ is a small angle.

    2. Relevant equations
    Fe=kq2/r2
    F=ma
    Fg=mg

    3. The attempt at a solution
    Equation 1:

    r = 2sinθ
    Fe=kq2/(2sinθ)2
    Equation 2:
    tanθ=Fe/Fg
    Fgtanθ=Fe
    Combining Them:
    Fgtanθ=kq2/(2sinθ)2
    Fgtanθ=kq2/(4(sinθ)2)
    (sinθ)2tanθ=kq2/(4Fg)
    (sinθ)2tanθ=kq2/(4mg)

    Okay, so now I have an equation with all the thetas on one side. So I tried to plug it into Wolfram Alpha to solve for theta, but it is giving me some really weird results. I'm not quite sure where I went wrong. Any help is greatly appreciated! Here's the Wolfram Alpha link. Thanks!

    http://www.wolframalpha.com/input/?...*(8*10^-9)^2/(4*4*10^-3*9.8)+for+x+in+degrees
     
  2. jcsd
  3. Mar 11, 2017 #2

    cnh1995

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    I didn't check your equations, but
    ..this means you can use,
    sinθ=tanθ≈θ (in radian).
     
  4. Mar 11, 2017 #3
    Hm, I tried that and it didn't seem to work unfortunately.
    Why is it that I can make that assumption anyway?

    EDIT: Just used all my attempts and got that the correct answer is 4.1 degrees. Still need to figure this out since it's for studying.
     
  5. Mar 11, 2017 #4

    cnh1995

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    It comes from the Maclaurin series expansion for sinθ and cosθ, where θ is in radian.
    sinθ=θ-(θ3/3!)+(θ5/5!)-..

    cosθ=1+(θ2/2!)+θ4/4!+...

    You can see for small angle θ, sinθ≈θ and cosθ≈1 as higher order terms can be safely neglected.
    Hence, tanθ≈θ.

    Did you convert the answer from radians into degrees?
     
  6. Mar 11, 2017 #5
    Yes, here's my link.
     
  7. Mar 11, 2017 #6

    vela

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    Your work is fine. You should get the correct answer if you use cnh1995's suggestion. (I did.)
     
  8. Mar 11, 2017 #7
    Hmmm. 4.1 degrees? Can you look at my link here and see what I entered wrong?
    Thank you all for helping me through this btw!
     
  9. Mar 11, 2017 #8

    vela

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    ##\sin^2 \theta \,\tan\theta \approx \theta^3##
     
  10. Mar 11, 2017 #9
    Ah yes, good catch.
    I wrapped the whole thing in a cube root and the answer is closer to correct but still not there. I'm probably missing something simple but I am just not seeing it.
    Link

    EDIT: I got it!! I had 8 nC instead of 80. Thank you all!
     
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