Charges...please help!!


by BuBbLeS01
Tags: chargesplease
BuBbLeS01
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#1
Jan13-08, 07:36 AM
P: 605
1. The problem statement, all variables and given/known data



Two 3.52 g point charges on 7.49-m-long threads repel each other after being equally charged. What is the charge q? (θ=31.)


2. Relevant equations
F = KQQ/R^2


3. The attempt at a solution
How do I find Q without having a distance r???
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#2
Jan13-08, 07:54 AM
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Quote Quote by BuBbLeS01 View Post
How do I find Q without having a distance r???
You are given the thread length and the angle--use that to figure out the distance.
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#3
Jan13-08, 08:32 AM
P: 605
Oh yea...so the length from the center to one of the masses is 3.858 m.
So I can do...
F = K*Q*Q/r^2
F = K*Q^2/ r^2
An wouldn't F = 0 since they cancel each other out?

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Jan13-08, 08:40 AM
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Charges...please help!!


Since the masses are in equilibrium, the net force on each is zero. What forces act on each mass?
BuBbLeS01
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#5
Jan13-08, 08:41 AM
P: 605
Tension
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#6
Jan13-08, 08:43 AM
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Quote Quote by BuBbLeS01 View Post
Tension
That's one force. List them all. (And draw yourself a free body diagram showing how the force act on each mass.)
BuBbLeS01
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#7
Jan13-08, 08:46 AM
P: 605
Tension and weight are the forces
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#8
Jan13-08, 08:48 AM
P: 605
So do I include the electric force? Like...
T - W + Fel = 0
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Jan13-08, 08:48 AM
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Quote Quote by BuBbLeS01 View Post
Tension and weight are the forces
Don't forget the electrostatic force!
BuBbLeS01
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#10
Jan13-08, 08:50 AM
P: 605
Isn't that Fel?
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Jan13-08, 08:50 AM
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Quote Quote by BuBbLeS01 View Post
So do I include the electric force? Like...
T - W + Fel = 0
Yes, but realize that forces are vectors--direction counts. Set up two equations: One for the horizontal components, one for the vertical components. Combine these to solve for the charge.
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#12
Jan13-08, 08:58 AM
P: 605
X.) TSINθ + qE = 0
y.) TCOSθ - mg = 0

T = mg/COSθ
mg/COSθ * SINθ + qE
q = -mg*TANθ/E
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#13
Jan13-08, 09:16 AM
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Quote Quote by BuBbLeS01 View Post
X.) TSINθ + qE = 0
y.) TCOSθ - mg = 0
I'd write that first equation as:
Tsinθ - qE = 0 (since the force components are in opposite directions)

Realize that E is also a function of q, so rewrite that in terms of k, q, and r (which you figured out).

Otherwise, you are on the right track.
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#14
Jan13-08, 09:22 AM
P: 605
Okay so I end up with...
-q = -(mg*tanθ) / (kq/r^2)
q = (mg*tanθ) / (kq/r^2)
and the q's cancel?
= (mg*tanθ) / (k/r^2)
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#15
Jan13-08, 12:11 PM
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Quote Quote by BuBbLeS01 View Post
Okay so I end up with...
-q = -(mg*tanθ) / (kq/r^2)
q = (mg*tanθ) / (kq/r^2)
and the q's cancel?
The q's don't cancel. And when you divide by a fraction, simplify the result. (Invert and multiply.)


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