Where Am I Going Wrong in Solving Repulsive Charges Problem?

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Homework Help Overview

The discussion revolves around a physics problem involving two charged spheres that repel each other while suspended from threads. Participants are analyzing the forces acting on the spheres and the calculations related to their charges.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of trigonometric functions to resolve forces and question the interpretation of tension in the context of Newton's third law. There is also a focus on the calculation of charge based on the forces involved.

Discussion Status

Some participants have provided guidance on potential errors in the original calculations, particularly regarding the interpretation of tension. There appears to be a progression in the discussion as participants refine their calculations and check each other's reasoning.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can provide or the methods they can use. The original poster is seeking clarification on specific steps in their solution process.

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]
 
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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