Where Am I Going Wrong in Solving Repulsive Charges Problem?

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The discussion centers on solving a physics problem involving two charged spheres that repel each other. The initial calculations for tension and force were incorrect due to misunderstanding the relationship between tension and the forces acting on the spheres. A key correction was made regarding how to calculate the charge, leading to a revised value of 747 nC. The final consensus confirms that this revised charge is accurate. The participants successfully clarified the misunderstanding and arrived at the correct solution.
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Homework Statement



Two 3.0g spheres on 1.0m-long threads repel each other after being equally charged, as shown in the figure.
jfk.Figure.20.P64.jpg


Homework Equations



F = k*q1*q2/r2

The Attempt at a Solution



sin 70° = Ty/T
T = (m*g)/sin 70°
T = 0.003*9.81/sin 70°
T = 0.03132 N

Tx = T*cos 70°
Tx = (0.03132)*cos 70°
Tx = 0.01071 N

Since both objects experience this force, the force is doubled at 0.02142 N.

1/2*r = sin 20°
r = 0.6840 m

T2x = k*q2/r2
q = (T2x*r2/k)1/2
q = [(0.02142)(0.6840)2/(8.99*109)]1/2
q = 1056 nC = 1.056* 10-6 C

Where am I going wrong?
 
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Dr. S said:

Homework Statement



Two 3.0g spheres on 1.0m-long threads repel each other after being equally charged, as shown in the figure.
jfk.Figure.20.P64.jpg


Homework Equations



F = k*q1*q2/r2

The Attempt at a Solution



sin 70° = Ty/T
T = (m*g)/sin 70°
T = 0.003*9.81/sin 70°
T = 0.03132 N

Tx = T*cos 70°
Tx = (0.03132)*cos 70°
Tx = 0.01071 N

Since both objects experience this force, the force is doubled at 0.02142 N.

1/2*r = sin 20°
r = 0.6840 m

T2x = k*q2/r2
q = (T2x*r2/k)1/2
q = [(0.02142)(0.6840)2/(8.99*109)]1/2
q = 1056 nC = 1.056* 10-6 C

Where am I going wrong?

I suggest that the line in red is where you went wrong.

during a "tug-of-war", when two teams pull on each end of a sturdy rope, each team pulls with the same force [Newton's third law will show why the forces are equal] but the tension in the rope is just that same value. ie. if each team pulls with 10 000 N, the tension in the rope is 10 000 N [not 20 000 N]
 
I agree
 
sin 70° = Ty/T
T = (m*g)/sin 70°
T = 0.003*9.81/sin 70°
T = 0.03132 N

Tx = T*cos 70°
Tx = (0.03132)*cos 70°
Tx = 0.01071 N

1/2*r = sin 20°
r = 0.6840 m

Tx = k*q2/r2
q = (Tx*r2/k)1/2
q = [(0.01071)(0.6840)2/(8.99*109)]1/2
q = 747 nC = 7.47* 10-7 C

How about now?
 
looks good to me.
 
Got it, thanks guys. :)

(747 nC is correct).
 
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