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Charges...please help

by BuBbLeS01
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BuBbLeS01
#1
Jan13-08, 07:36 AM
P: 602
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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Doc Al
#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.
BuBbLeS01
#3
Jan13-08, 08:32 AM
P: 602
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?

Doc Al
#4
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
#5
Jan13-08, 08:41 AM
P: 602
Tension
Doc Al
#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
#7
Jan13-08, 08:46 AM
P: 602
Tension and weight are the forces
BuBbLeS01
#8
Jan13-08, 08:48 AM
P: 602
So do I include the electric force? Like...
T - W + Fel = 0
Doc Al
#9
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
#10
Jan13-08, 08:50 AM
P: 602
Isn't that Fel?
Doc Al
#11
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.
BuBbLeS01
#12
Jan13-08, 08:58 AM
P: 602
X.) TSINθ + qE = 0
y.) TCOSθ - mg = 0

T = mg/COSθ
mg/COSθ * SINθ + qE
q = -mg*TANθ/E
Doc Al
#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.
BuBbLeS01
#14
Jan13-08, 09:22 AM
P: 602
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)
Doc Al
#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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