MasteringPhysics: Find the angle between the point charges.

  • Thread starter danielhep
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  • #1
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Homework Statement


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.

Homework Equations


Fe=kq2/r2
F=ma
Fg=mg

The Attempt at a Solution


Equation 1:[/B]
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
 

Answers and Replies

  • #2
cnh1995
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I didn't check your equations, but
You can assume that θ is a small angle.
..this means you can use,
sinθ=tanθ≈θ (in radian).
 
  • #3
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I didn't check your equations, but

..this means you can use,
sinθ=tanθ≈θ (in radian).
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.
 
  • #4
cnh1995
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Hm, I tried that and it didn't seem to work unfortunately.
Why is it that I can make that assumption anyway?
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.

it didn't seem to work unfortunately.
Did you convert the answer from radians into degrees?
 
  • #6
vela
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Your work is fine. You should get the correct answer if you use cnh1995's suggestion. (I did.)
 
  • #7
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Your work is fine. You should get the correct answer if you use cnh1995's suggestion. (I did.)
Hmmm. 4.1 degrees? Can you look at my link here and see what I entered wrong?
Yes, here's my link.

Thank you all for helping me through this btw!
 
  • #8
vela
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##\sin^2 \theta \,\tan\theta \approx \theta^3##
 
  • #9
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##\sin^2 \theta \,\tan\theta \approx \theta^3##
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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