Finding the charge on pith balls

AI Thread Summary
The discussion focuses on determining the charge on pith balls using the electrostatic force formula and gravitational force. The participants clarify that the forces acting on the balls include the electrostatic force, tension, and gravitational force, which act in different directions. A key point is the relationship derived from balancing these forces, leading to the conclusion that the electrostatic force equals the gravitational force multiplied by the tangent of half the angle, expressed as mg tan(θ/2) = F_e. The conversation emphasizes the importance of correctly identifying and balancing the forces to arrive at the desired result. Overall, the participants successfully connect the electrostatic and gravitational forces to find the charge on the pith balls.
hmparticle9
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Homework Statement
Two identical equally charged pith balls of mass ##m## and negligible radius are hanging from the same point on strings of length ##L##. The strings have an angle ##\theta## between them. What is the charge on each and the electric force on each ball?
Relevant Equations
$$F = \frac{q^2}{4\pi \epsilon_0 4 L^2 \sin^2 \frac{\theta}{2}}$$
The force is easily found to be:

$$F = \frac{q^2}{4\pi \epsilon_0 4 L^2 \sin^2 \frac{\theta}{2}}$$

I am trying to think how to get the charge ##q## from this. Since the balls are stationary the force felt on each of the balls must equal the force of gravity on the balls:

The force of gravity on the balls is ##mg \sin \frac{\theta}{2} ##.

However, if I set ##F## equal to this I do not get the answer in the book. The answer in the book is

$$\sqrt{(mg) \tan \frac{\theta}{2}4\pi \epsilon_0 4 L^2 \sin^2 \frac{\theta}{2}}$$

This,to me, suggests that the force on the ball is given by ##(mg) \tan \frac{\theta}{2}##. Could someone explain to me why this is the case?
 
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hmparticle9 said:
The force of gravity on the balls is ##mg \sin \frac{\theta}{2} ##.
Can you tell us how you arrived at this assertion?
 
hmparticle9 said:
However, if I set F equal to this
Why would you set the electrostatic force equal to the gravitational force? Are they in the same direction?
 
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Hi @hmparticle9. To add what's already been said, there are three forces acting on each particle. Can you name them and can you draw a free-body diagram for one of the particles?
 
Screenshot 2025-06-24 14.22.34.webp


##F_e## is the repulsive force from the other charge, ##T## is the tension from the string and ##mg## is the force due to gravity. The above comments are obviously correct: ##F_e## and the other forces point in different directions. I am not sure where to go from here.
 
Last edited:
hmparticle9 said:
View attachment 362505

##F_e## is the repulsive force from the other charge, ##T## is the tension from the string and ##mg## is the force due to gravity. The above comments are obviously correct: ##F_e## and the other forces point in different directions. I am not sure where to go from here.
Nicely drawn.

With the drawing in mind and given a value for ##\theta##, can you find a formula for ##T##?

With the formula for ##T## in hand, can you then find a formula for ##f_e##?
 
So I get
$$T \sin \frac{\theta}{2} = F_e \text{ and } T \cos \frac{\theta}{2} = mg$$
Dividing and rearranging I get
$$mg \tan \frac{\theta}{2} = F_e$$
which is exactly what I want?
 
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hmparticle9 said:
$$mg \tan \frac{\theta}{2} = F_e$$
which is exactly what I want?
Presumably the question mark at the end is not needed?
 
Yes, my bad. I was thinking that @jbriggs444 had a different solution method and I am all ears.
 
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hmparticle9 said:
Yes, my bad. I was thinking that @jbriggs444 had a different solution method and I am all ears.
Nothing significantly different. It is essentially the same algebra the way I had in mind. And delivers the same result.
 
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