Coulomb's Law and point charges

AI Thread Summary
Two point charges of mass 6.0g and charge 120nC repel each other, creating an angle θ with the vertical when suspended by 1.0m threads. The forces acting on the charges must balance, leading to equations involving tension and electric force. The discussion highlights the importance of correctly determining the distance between the charges, which is half the length of the string due to the geometry of the setup. A simplification using small angle approximations for sinθ and tanθ is suggested to solve for θ. Participants share their calculations and seek clarification on the parameters used in their equations.
absolutezer0es
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Two m = 6.0g point charges on 1.0-m-long threads repel each other after being charged to q = 120nC , as shown in the figure.

What is the angle θ? You can assume that θ is a small angle.

25_P58.jpg


Hopefully the image will work.

I feel like something is missing. I've gotten 7.4 and 3.4 degrees - both wrong. I know the sum of the forces in all directions must equal zero. My equations then are:

Fx = -Tsinθ+Felectric
Fy = Tcosθ - mg

We know m, g, and q. I know that coulomb's law is:

F = Kq^2/r^2 ... but we don't know r.

Any ideas?
 
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You are given the length of the string and the angle. Do you see a right triangle you could make in order to determine r?
 
Ah, yes ... a little clearer now.

So sinθ = r/1 = r. Alright let me run with that. Let's see if I can get it now.

Thanks for the nudge!
 
Wait! Are you sure ##\sin{\theta}=r##? Remember ##r## is the distance between the charges...
 
When I replied, I realized I made a mistake, like you said. It's not r, but rather r/2!

I was able to boil down the equation to:

(sinθ)^2*(tanθ) = kq^2/(4mg)

I'm pretty sure you can't boil it down anymore than that, so I used my old grapher to find the intersection point between both sides.

Got it!

Thanks a ton Zeta! I appreciate the insight!
 
absolutezer0es said:
When I replied, I realized I made a mistake, like you said. It's not r, but rather r/2!

I was able to boil down the equation to:

(sinθ)^2*(tanθ) = kq^2/(4mg)

I'm pretty sure you can't boil it down anymore than that, so I used my old grapher to find the intersection point between both sides.

Got it!

Thanks a ton Zeta! I appreciate the insight!
You can actually do it withOUT using a graph. Note the hint: you may assume that the angle is small. In that approximation, what can you say about ## \tan \theta ## and ##\sin \theta ## ?
 
nrqed said:
You can actually do it withOUT using a graph. Note the hint: you may assume that the angle is small. In that approximation, what can you say about ## \tan \theta ## and ##\sin \theta ## ?
Can you help? I worked through this and plugged my numbers in but when I try to solve for theta I don't get any kind of real number. Here's what I put in Wolfram Alpha:
http://www.wolframalpha.com/input/?...*(8*10^-9)^2/(4*4*10^-3*9.8)+for+x+in+degrees

Thanks!
 
danielhep said:
Can you help? I worked through this and plugged my numbers in but when I try to solve for theta I don't get any kind of real number. Here's what I put in Wolfram Alpha:
http://www.wolframalpha.com/input/?i=(sin(x))^2*tan(x)=8.99*10^9*(8*10^-9)^2/(4*4*10^-3*9.8)+for+x+in+degrees

Thanks!
Wolfram says:

upload_2017-3-10_4-10-49.png


Can you explain those numbers? I recognize k and g but the rest is a mystery given the data in the problem statement of post #1.

If you are actually working on a different problem with different data please start a new thread of your own. Hijacking old threads is against the rules.
 
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