Where Am I Going Wrong in Solving Repulsive Charges Problem?

  • Thread starter Thread starter Dr. S
  • Start date Start date
  • Tags Tags
    Charges
AI Thread Summary
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.
Dr. S
Messages
13
Reaction score
0

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?
 
Physics news on Phys.org
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).
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top