Coulomb's force - magnitude of the electric force

In summary: Do you have the same FBD as the original problem? If Ball #1 is below Ball #2, it has a downward force on it and the magnitude of that force is Fe=mg(T/r).
  • #1
zachem62
37
3

Homework Statement


Show that the magnitude of the electric force on each ball is given by
Fe = mgtanθ/ (cosα+ sinαtanθ)
where θ = arcsin(x2/L)
α = arcsin((y2-y1)/r)
r = √((x2-x1)^2+(y2-y1)^2)
This is a question from my coulomb's law lab, where 2 styrofoam balls where charged, one was hung on a string and the other was brought closer to it from 20cm away and the change in the hanging ball position was recorded. (x2,y2) is the position of the hanging ball, and (x1,y1) is the position of the other ball that is moving closer to the hanging ball. r is the separation distance between the 2 balls. L is the length of the string that the ball is hung on. If this is not clear enough, please see these links:
question: https://imgur.com/a/IrkZl
diagram: https://imgur.com/a/UzzMu

2. Homework Equations

The Attempt at a Solution


I am pretty confused about how to approach this problem. I drew the free body diagram for the ball and split the forces into their x and y components so derive Fe, and that hasn't got me anywhere. Whats a better way to do this?
 
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  • #2
zachem62 said:

Homework Statement


Show that the magnitude of the electric force on each ball is given by
Fe = mgtanθ/ (cosα+ sinαtanθ)
where θ = arcsin(x2/L)
α = arcsin((y2-y1)/r)
r = √((x2-x1)^2+(y2-y1)^2)
This is a question from my coulomb's law lab, where 2 styrofoam balls where charged, one was hung on a string and the other was brought closer to it from 20cm away and the change in the hanging ball position was recorded. (x2,y2) is the position of the hanging ball, and (x1,y1) is the position of the other ball that is moving closer to the hanging ball. r is the separation distance between the 2 balls. L is the length of the string that the ball is hung on. If this is not clear enough, please see these links:
question: https://imgur.com/a/IrkZl
diagram: https://imgur.com/a/UzzMu

2. Homework Equations

The Attempt at a Solution


I am pretty confused about how to approach this problem. I drew the free body diagram for the ball and split the forces into their x and y components so derive Fe, and that hasn't got me anywhere. Whats a better way to do this?
You didn't include Coulomb's Law as the main Relevant Equation, but maybe just because you figured it was obvious. :smile:

So you get Fe from Coulomb's Law and the separation distance r, right? Once you have that, the rest is just trigonometry to solve the sum of the forces on the hanging ball in the x & y directions are each zero. Can you show us your work so far?
 
  • #3
berkeman said:
You didn't include Coulomb's Law as the main Relevant Equation, but maybe just because you figured it was obvious. :smile:

So you get Fe from Coulomb's Law and the separation distance r, right? Once you have that, the rest is just trigonometry to solve the sum of the forces on the hanging ball in the x & y directions are each zero. Can you show us your work so far?
You're right it all comes down to trig and sum of forces in x and y directions. I figured it out now. Thanks.
 
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  • #4
zachem62 said:
You're right it all comes down to trig and sum of forces in x and y directions. I figured it out now. Thanks.
Great! Looks like a fun problem. :smile:
 
  • #5
berkeman said:
You didn't include Coulomb's Law as the main Relevant Equation, but maybe just because you figured it was obvious. :smile:

So you get Fe from Coulomb's Law and the separation distance r, right? Once you have that, the rest is just trigonometry to solve the sum of the forces on the hanging ball in the x & y directions are each zero. Can you show us your work so far?
Years later, I'm stuck on this very problem. I'm confused about how to derive the above equation for Fe from Coulomb's Law. How can I use Coulomb's Law if I don't know the charges?

I think I understand the trig aspect, and I know I could simply plug values into the given equation for Fe, but that seems like skipping ahead. The lab says specifically to include detailed derivation steps and supporting diagrams.

I also tried breaking the forces into x and y components but I don't think I can find the tension force without knowing Fe, unless T=mg? I've been spinning my wheels for a couple days now, any help is appreciated!
 
  • #6
ngiro0862 said:
How can I use Coulomb's Law if I don't know the charges?
Just leave the charge as a variable Q for now...
ngiro0862 said:
I also tried breaking the forces into x and y components but I don't think I can find the tension force without knowing Fe, unless T=mg?
Do you have the same FBD as the original problem? If Ball #1 is below Ball #2, it will unload some of the string tension according the angles...

1641511598586.png
 
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  • #8
berkeman said:
Just leave the charge as a variable Q for now...

Do you have the same FBD as the original problem? If Ball #1 is below Ball #2, it will unload some of the string tension according the angles...

View attachment 295207
Yeah it's identical! I thought that was the case, but I was trying anything.
 
  • #9
Can you show your work so far so that we can check it?
 
  • #10
IMG_20220106_192810.jpg
IMG_20220106_192823.jpg

Hopefully that's clear enough. The last bit is where I went ahead assuming T=mg.
 
  • #11
ngiro0862 said:
Hopefully that's clear enough.
No sorry, dim cell phone pictures are no help at all.

Please type your work into the forum Edit window to post it. For math equations you can click into the "LaTeX Guide" link below the Edit window to see the basics of posting math on Internet forums (it's a good general skill to have, not just for the PF). :smile:
 
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  • #12
Gotcha sorry, first time.

I know ##m = 0.7g, L = 0.32m, \text{and } g=9.8~\rm{m/s^2}##.

I can obviously solve ##mg=6.9*10^{-3}N##.

Using data points ##X1=0.009m, Y1=-0.024m, X2= 0.050m, Y2=-0.019m## I can solve for $$\begin{align}
\theta & = \sin^{-1}\left( \frac {X2} L \right) \nonumber \\
& = 0.16 rad \nonumber
\end{align}$$
$$\begin{align}
\alpha & = \sin^{-1}\left( \frac {Y2 - Y1} r \right) \nonumber \\
& = 0.085 rad \nonumber
\end{align}$$
$$\begin{align}
r & = \sqrt {(X2 - X1)^2+(Y2-Y1)^2} \nonumber \\
& = 0.059 m \nonumber
\end{align}$$

Because the system is in equilibrium the magnitudes $$T_x =T\sin\theta = Fe_x = Fe\cos\alpha$$ and $$T_y = T\cos\theta = mg - Fe_y = mg - Fe\sin\alpha$$

As for the derivation from Coulomb's Law I am quite lost. Obviously I have the value r, which is the distance between the charges, but I still have two unknowns, Fe and Q, and I don't see a substitution, since T is also unknown.
 
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FAQ: Coulomb's force - magnitude of the electric force

1. What is Coulomb's force?

Coulomb's force, also known as the electric force, is a fundamental force of nature that describes the interaction between electrically charged particles. It follows the inverse square law, meaning that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. How is the magnitude of the electric force calculated?

The magnitude of the electric force can be calculated using Coulomb's law, which states that the force is equal to the product of the two charges divided by the square of the distance between them. Mathematically, it can be expressed as F = k * (q1 * q2)/r², where F is the force, k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.

3. What is the unit of measurement for Coulomb's force?

The unit of measurement for Coulomb's force is Newtons (N). This is the same unit used to measure other types of forces, such as gravitational force. It is named after Sir Isaac Newton, who first described the laws of motion.

4. How does the distance between charged particles affect the magnitude of the electric force?

The distance between charged particles has a direct impact on the magnitude of the electric force. As the distance between them increases, the force decreases according to the inverse square law. This means that the force becomes weaker the further apart the particles are, and it becomes stronger as they move closer together.

5. What are some real-life examples of Coulomb's force?

Coulomb's force is present in many everyday phenomena, such as when a balloon sticks to a wall after rubbing it against hair, or when a person gets an electric shock after touching a doorknob. It is also responsible for the attraction and repulsion of charged particles in atoms, the functioning of electronic devices, and the behavior of lightning and thunderstorms.

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