Where Am I Going Wrong in Solving Repulsive Charges Problem?

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Dr. S
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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]
 
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?
 
Got it, thanks guys. :)

(747 nC is correct).