# Hanging pith balls on a thread...charge, Fg, Fe and tension

• Brittany Anderson
In summary: So you would have to take the square root to get Q.In summary, two pith balls with equal mass and charge are placed at a horizontal distance of 3.0cm from each other, with one hanging from an insulating thread at an angle of 30.0° from the vertical. The gravitational force on the hanging pith ball can be calculated using the formula Fg=mg, where m is the mass of the pith ball and g is the gravitational acceleration. To find the electric force on the hanging pith ball, vector addition can be used with the formula Fe= kq1q2/r^2, where k is the Coulomb constant, q1 and q2 are the charges on the two p
Brittany Anderson

## Homework Statement

Two pith balls each have a mass of 1.0g and equal charges. One pith ball is hung from an insulating thread. The other is fixed at a horizontal distance of 3.0cm from the first, The first hangs at an angle of 30.0° from the vertical. Find:
a) the gravitational force on the hanging pith ball
b) the electric force on the hanging pith ball
c) the charge on each of the balls
d) the tension in the thread

m=0.001kg
dx=0.03 m
θ=30°

## Homework Equations

g=Fg/m
Fe=kq1q2/r^2
E=Fe/q
E=ΔV/Δd
ΔV=ΔEp/q

(all of these should have vectors except for the last equation! I didn't know how to put vectors on.. sorry

## The Attempt at a Solution

SO. We know from the problem,
m=0.001kg
dx=0.03 m
θ=30°

a) if I'm not mistaken, in order to find Fg we must use g=Fg/m or when rearranged mg=Fg
If we use this, we can determine that the Fg= (0.001kg)(9.81m/s^2)
Which would give us 0.00981 as an answer for Fg.. correct?

b) For this one I'm at a bit of a loss.. Based on formulas we don't exactly have one that matches. I have made a triangle however and so I'm wondering if this was the right of executing it... Please view the insert below on the triangle. If I used tan(30)=Fe/Fg or 0.00981tan(30)= Fe, would that work? does the triangle even correspond or make sense in this scenario??

c) for charges, I am assuming that I rearrange the Fe=kq1q2/r^2 formula to find the charges.. as in Fe⋅r^2/k= q1q2 and then just divide by 2 tog et each of the charges because they are equal?

d) For tension, do I use the triangle again with c^2=a^2+b^2 rule?

Please just help me confirm because I might have this all wrong :) Hope all my attempts are clear and make sense! Let me know! Thanks in advance.

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First of all, please use LaTeX to write your equations. It's easy and it makes it a lot more likely that someone will respond to your posts.
Brittany Anderson said:
Which would give us 0.00981 as an answer for Fg.. correct?
This part looks good.

Brittany Anderson said:
b) For this one I'm at a bit of a loss.. Based on formulas we don't exactly have one that matches. I have made a triangle however and so I'm wondering if this was the right of executing it... Please view the insert below on the triangle. If I used tan(30)=Fe/Fg or 0.00981tan(30)= Fe, would that work? does the triangle even correspond or make sense in this scenario??
You have the right approach here. You have done vector addition to get the resultant force on the pith ball. That is, you drew the vectors nose-to-tail. I'm not sure if you knew you were graphically adding the vectors. You have correctly written the equation relating the two vectors to the direction of the resultant force vector and you can solve it to find the electric force on the hanging pith ball.

Brittany Anderson said:
c) for charges, I am assuming that I rearrange the Fe=kq1q2/r^2 formula to find the charges.. as in Fe⋅r^2/k= q1q2 and then just divide by 2 tog et each of the charges because they are equal?
This is almost right, but think of it this way. Both pith balls have the same charge ##Q##. So ##q_1=Q## and ##q_2=Q##. Substitute Q for ##q_1## and ##q_2## in your equation for ##F_e##. Now you have a ##Q^2##. How would you solve for Q?

Brittany Anderson said:
d) For tension, do I use the triangle again with c^2=a^2+b^2 rule?
Yes.

Brittany Anderson said:
then just divide by 2 tog et each of the charges because they are equal?
That would make sense if the equation were Fe⋅r^2/k= q1+q2, but it is Fe⋅r^2/k= q1q2.

## 1. How does hanging pith balls on a thread demonstrate electric charge?

When a pith ball is suspended on a thread, it can be easily moved by an object with an electric charge. This shows that the pith ball itself has acquired an electric charge, either positive or negative, which allows it to be attracted or repelled by other charges.

## 2. What is the relationship between the electric charge and the force of gravity in this experiment?

In this experiment, the force of gravity is not affected by the electric charge of the pith ball. The weight of the pith ball is still determined by its mass and the acceleration due to gravity. However, the electric charge on the pith ball can cause it to move or be affected by other electric forces, such as the force between two charged objects.

## 3. How does the electric force (Fe) compare to the force of gravity (Fg) in this experiment?

The electric force, Fe, is a type of non-contact force that is caused by the interaction of electric charges. In this experiment, Fe is responsible for the movement of the pith ball due to its electric charge. The force of gravity, Fg, is a contact force that is caused by the mass and acceleration due to gravity of the pith ball. In this experiment, Fg is responsible for keeping the pith ball suspended on the thread.

## 4. How does tension in the thread play a role in this experiment?

The tension in the thread is what keeps the pith ball suspended in the air. If the tension is too weak, the pith ball will fall to the ground. The tension is also responsible for keeping the pith ball in equilibrium, meaning that the forces acting on it are balanced. This allows for accurate measurements of the electric force and the force of gravity.

## 5. Can this experiment be used to determine the exact amount of charge on the pith ball?

No, this experiment alone cannot determine the exact amount of charge on the pith ball. It can only show the presence of an electric charge and the direction of the force it experiences. To determine the exact amount of charge, other experiments and calculations are necessary.

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