# MasteringPhysics: Find the angle between the point charges.

1. Mar 11, 2017

### danielhep

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

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. Mar 11, 2017

### cnh1995

I didn't check your equations, but
..this means you can use,

3. Mar 11, 2017

### danielhep

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. Mar 11, 2017

### cnh1995

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.

5. Mar 11, 2017

### danielhep

6. Mar 11, 2017

### vela

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

7. Mar 11, 2017

### danielhep

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!

8. Mar 11, 2017

### vela

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

9. Mar 11, 2017

### danielhep

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.