MasteringPhysics: Find the angle between the point charges.

[danielhep]

Homework Statement

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

F_e=kq²/r²
F=ma
F_g=mg

The Attempt at a Solution

Equation 1:[/B]
r = 2sinθ
F_e=kq²/(2sinθ)²
Equation 2:
tanθ=F_e/F_g
F_gtanθ=F_e
Combining Them:
F_gtanθ=kq²/(2sinθ)²
F_gtanθ=kq²/(4(sinθ)²)
(sinθ)²tanθ=kq²/(4F_g)
(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

 

[cnh1995]

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).

 

[danielhep]

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.

 

[cnh1995]

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?

 

[danielhep]

Did you convert the answer from radians into degrees?

Yes, here's my link.

 

[vela]

Your work is fine. You should get the correct answer if you use cnh1995's suggestion. (I did.)

 

[danielhep]

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!

 

[vela]

$\sin^2 \theta \,\tan\theta \approx \theta^3$

 

[danielhep]

$\sin^2 \theta \,\tan\theta \approx \theta^3$

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!