# Electrostatics of ping pong ball question

• ploppers
In summary, the ball is subjected to a 420V potential difference and the tension force is 2.92*10^-4 N.
ploppers

## Homework Statement

A ping pong ball of mass 3*10^-4 kg is hanging from a light thread 1 m between 2 vertical parallel plates 10 cm apart. When the potential difference across the plate is 420V, the ball comes to equilibrium 1 cm from one side of the original position. What is the tension of the string, charge and electric force on the ball?

E = V/r
F = Eq

## The Attempt at a Solution

found the force in the horizontal is 2.92*10^-4 through spliting up the force of gravity. Tension force is the same as it would be going straight down so I found the angle with 1 cm and 10cm and cos the force to get the horizontal force. That would be the force the system acts on it.

F = Eq
2.92*10^-4 = Vq/r
2.92*10^-4 * r / v = q

this is where I get my problem...is the potential energy conssistant throughout the system like the force? If it is then I can get to an answer, but if it isn't, what do I do? If you can;'t understand what I wrote i apologize, I'm very confused and frusterated

ploppers said:

## Homework Statement

A ping pong ball of mass 3*10^-4 kg is hanging from a light thread 1 m between 2 vertical parallel plates 10 cm apart. When the potential difference across the plate is 420V, the ball comes to equilibrium 1 cm from one side of the original position. What is the tension of the string, charge and electric force on the ball?

E = V/r
F = Eq

## The Attempt at a Solution

found the force in the horizontal is 2.92*10^-4 through spliting up the force of gravity. Tension force is the same as it would be going straight down

No, the tension is not the same as when the ball is hanging straight down (in which case T=mg). You have to draw a FBD and solve. There is not enough details in the rest of your post to see if you did it right and I did not check your final answer.
so I found the angle with 1 cm and 10cm and cos the force to get the horizontal force. That would be the force the system acts on it.

F = Eq
2.92*10^-4 = Vq/r
2.92*10^-4 * r / v = q

this is where I get my problem...is the potential energy conssistant throughout the system like the force? If it is then I can get to an answer, but if it isn't, what do I do? If you can;'t understand what I wrote i apologize, I'm very confused and frusterated

I am not sure where you got the formula F = Vq/r but that's completely wrong. But the formula "electric Force = qE" is correct.

There is a formula V=kq/r but that's the electric potential produced by a point charge. For two infinite plates, the E field is constant and is given by the potential difference between the plates divided by the plate separation, $E = \Delta V/ d$. Use that to find E and then use that the electric force is equal to q E.

Would you be able to show me how to properly solve this?

ploppers said:
Would you be able to show me how to properly solve this?

First, can you find the value of the E field using the formula I gave you?

The second step is : can you sho wme the steps you followed to find the electric foce? (this force must be equal to the x component of the tension, right? And to find this x component, you must first find the magnitude of the tension using that the net force along y is zero. Before doing this, you need to find the angle made by the string).

So the steps for the second part are

a) find the angle

b) working in the y direction, th enet force along y is zero which should allow you to find the magnitude of th etension.

c) Then go in the x direction to find the electric force

d) finally, use electric force = q E and the E found in the first step to solve for q.

Thnx, I followed and ot the right answer :D Actually I used the wrong variables for my equations. My Vq/r is actually combining the F = (funky E)q with (funky E) = V/d.

However, I still wonder why I would use the 10 cm for d. is the V always the same between 2 plates? That would make no sense...

## 1. What is electrostatics?

Electrostatics is the branch of physics that deals with the study of electric charges at rest and their interactions.

## 2. How does electrostatics apply to ping pong balls?

Ping pong balls can become electrically charged through friction with other surfaces, and electrostatics explains the behavior of these charged particles.

## 3. Why do ping pong balls stick to certain surfaces?

Ping pong balls can stick to certain surfaces due to electrostatic forces, which attract the opposite charges between the ball and the surface.

## 4. Can electrostatics affect the trajectory of a ping pong ball?

Yes, electrostatic forces can cause a ping pong ball to deviate from its expected trajectory if it is charged and interacts with other charged objects or surfaces.

## 5. How can I measure the electrostatic charge of a ping pong ball?

The electrostatic charge of a ping pong ball can be measured using an electroscope or by observing the deflection of a charged ping pong ball in the presence of other charged objects.

• Introductory Physics Homework Help
Replies
12
Views
692
• Introductory Physics Homework Help
Replies
20
Views
2K
• Introductory Physics Homework Help
Replies
23
Views
4K
• Introductory Physics Homework Help
Replies
24
Views
294
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
754
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
7K
• Introductory Physics Homework Help
Replies
1
Views
4K
• Introductory Physics Homework Help
Replies
60
Views
863