What is the Charge on Two Suspended Pith Balls with Coulomb's Law and Tension?

In summary, the two small pith balls, each of mass 12 g, are suspended by 1.2 m fine (so that we can neglect their mass in this problem) strings and are not moving. If the angle that each string makes with the vertical is θ=42.6°, and the charges on the two balls are equal, what is that charge (in μC)?The two small pith balls, each of mass 12 g, are suspended by 1.2 m fine (so that we can neglect their mass in this problem) strings and are not moving. If the angle that each string makes with the vertical is θ=42.6°, and the charges on the two balls are equal, the
  • #1
dancer2012
7
0

Homework Statement



Two small pith balls, each of mass m = 12 g, are suspended by 1.2 m fine (so that we can neglect their mass in this problem) strings and are not moving. If the angle that each string makes with the vertical is θ = 42.6°, and the charges on the two balls are equal, what is that charge (in μC)?

Homework Equations



F(T)+F(g)+F(E)=0
F(Ty)=F(g)=mg
F(Tx)=k(q^2/r^2)

The Attempt at a Solution



I really don't know how to manipulate these formulas to find an answer. I know you find F(Ty) first, but then I don't know how to get F(Tx). I think I am missing a formula somewhere. Please Help Thank you
 
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  • #2
You're pretty close, but the problem has an additional piece of information that you haven't used yet. Hint: it's the angle!
 
  • #3
Probably best to draw free-body diagrams first perhaps?
Hi Dancer, welcome to PF.
 
  • #4
The force diagram is attached


F(Ty)=cos(θ)*F(T) - - - F(T)=F(Ty)/cos(θ)
F(Tx)=sin(θ)*F(T) - - - F(T)=F(Tx)/sin(θ)

So, F(Ty)/cos(θ)=F(Tx)/sin(θ) - - - F(Tx)=[F(Ty)*sin(θ)]/cos(θ) - - - F(Tx)=F(Ty)*tan(θ)

Did I do all of that correctly?
 
  • #5
Sorry here is the force diagram actually
 

Attachments

  • Force Diagram.jpg
    Force Diagram.jpg
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  • #6
The next step after drawing the diagram is to add up the forces in each direction. Start with just the left-hand ball. eg. [tex]F_T\cos\theta=F_g[/tex]... do the horizontal one, then write out what each of the forces actually are. eg. [tex]F_e=\frac{kq^2}{r^2}[/tex]... and r is related to the angle... see?
 
  • #7
I understand that
 
  • #8
Yeah, that's right. tanØ=Tx/Ty

So you're set, work out the algebra from here. Don't forget, as Simon mentioned, to get the EM force in terms of "x distance."
 
  • #9
F(Tx)=mg*tan(θ)

q=√((F(Tx)*r^2)/k)

would r be mg/tan(θ)??
 
  • #10
Complete:
if the pith balls were at positions (±x,-y) ...
then there is a triangle l:y:x with angle θ between l and y (l=length of the string)
From this picture:

the distance between the balls is ____________.

and tanθ=_x/y_, sinθ=_____, and cosθ=_____.

which of the three is likely to help you?
 
Last edited:
  • #11
dancer2012 said:
F(Tx)=mg*tan(θ)

q=√((F(Tx)*r^2)/k)

would r be mg/tan(θ)??

Nah, that would be the horizontal force, which is the coulomb force you're in part trying to solve for. You've a string length you haven't used yet. ;)
 
  • #12
Ok so sin(θ)*string length=r

SO..
F(Ty)=F(g)=mg
F(Tx)=k(q^2/r^2)
F(Tx)=F(Ty)*tan(θ)
q=√((F(Tx)*r^2)/k)

F(Tx)=mg*tan(θ) - - - - - - - - - F(Tx)=.012kg*9.81m/s^2=.11772N→A
q=√((F(Tx)*(sin(θ)*L)^2)/k)----q=√(A*(sin(42.6)*1.2)^2/9E9)=2.938E-6C or 2.9μC

Is 2.9 μC the correct answer or did I make a mistake?
 
  • #13
Careful, you want the total distance between the pith balls. You just found one half, one leg.
 

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law of electromagnetism that describes the force between two charged particles. It states that the force is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

2. How does Coulomb's Law relate to tension?

Coulomb's Law does not directly relate to tension. However, it can be used to calculate the electric force between charged particles, which can then be used to determine the tension in a string or cable that is supporting the particles.

3. What is tension?

Tension is the pulling force exerted by a string, cable, or other object that is being stretched. It is the force that is required to keep an object in equilibrium when it is attached to one or more other objects.

4. How is tension affected by Coulomb's Law?

Tension is indirectly affected by Coulomb's Law through the electric force between charged particles. As the electric force between particles changes, the tension in the string or cable supporting them will also change.

5. Can Coulomb's Law and tension be applied to real-life situations?

Yes, Coulomb's Law and tension are applicable to many real-life situations involving charged particles. For example, they can be used to understand the behavior of electric charges in circuits, the force between charged particles in the human body, and the stability of structures such as suspension bridges.

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