Electrostatics of ping pong ball question

[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?

Homework Equations

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

 

[nrqed]

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?

Homework Equations

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.

 

[ploppers]

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

 

[nrqed]

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.

 

[ploppers]

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...