Two balls carrying the same charge

In summary: It may not reach the cylinder at all.In summary, the conversation discusses solving a physics problem involving two point charges and using the concept of total mechanical energy to find the maximum height of one of the charges. The conversation also considers the effects of friction and the possibility of the charge continuously oscillating above the other charge or reaching an equilibrium position.
  • #1
jgridlock
15
0

Homework Statement


The entire problem is in the attachment.

Homework Equations


F = k*(|q||q0|)/(r^2)

The Attempt at a Solution


F = 8.99e9(.450uC*.450uC)/(3cm^2) = 6.068e8

This problem has me completely lost and I am turning to you guys as my last resort. I solved for F and have no idea what to do after this. Any sort of insight would be appreciated and I can provide formulas upon request. Thank you.
 

Attachments

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  • #2
Think in terms of energy, not force, for most of the questions.
 
  • #3
I could use a little more guidance than that. What kind of process should I follow to solve part A?
 
  • #4
Apply conservation of energy. What forms of energy are relevant here and how would you calculate them?
 
  • #5
PE = mgh?
 
  • #6
jgridlock said:
PE = mgh?
OK, that's gravitational PE.

What other kinds of energy are involved here?
 
  • #7
Kinetic energy? KE = 1/2mv^2
 
  • #8
jgridlock said:
Kinetic energy? KE = 1/2mv^2
Yep, that's another.

Keep going.
 
  • #9
I wouldn't know any others. A guess would be charge?
 
  • #10
jgridlock said:
I wouldn't know any others. A guess would be charge?
Due to their electric field, those charges will have electric potential energy. Look it up!
 
  • #11
Electric Potential Energy: PE = k((q1*q1)/r)

A pair of point charges separated by a distance
 
  • #12
You may also want to check the units you are using for r.
 
  • #13
Convert cm to m. 3cm = .03m
And g to kg. 12g = .012kg

Alright, next step?
 
  • #14
jgridlock said:
Electric Potential Energy: PE = k((q1*q1)/r)

A pair of point charges separated by a distance
Good!

Now see if you can come up with an expression for the total mechanical energy.
 
  • #15
TME = 1/2mv^2+mgh+k((q1*q1)/r)
 
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  • #16
jgridlock said:
TME = 1/2mv^2+mgh+k((q1*q1)/r)
Good. But use the same distance r for both gravitational and electrostatic PE.
 
  • #17
So your saying that both h and r should be .03m? And I am solving for velocity?

1/2(.012)(v^2) + (.012)(9.8)(.03) + (8.99e9(.45uC*.45uC)/(.03))?
 
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  • #18
jgridlock said:
So your saying that both h and r should be .03m?
I would use r for both, where r is the distance between the centers of the balls.
And I am solving for velocity?
What's the velocity at the initial position? And at the highest position? (You're solving for that highest position.)

You'll take advantage of the fact that the total mechanical energy remains constant. Evaluate it at the initial position.
 
  • #19
Okay so the r is going to be equal to .06m.

The velocity at the initial position is zero, so I set v = 0?

1/2(.012)(0) + (.012)(9.8)(.06) + (8.99e9(.45uC*.45uC)/(.06)) = .03739725
 
  • #20
jgridlock said:
Okay so the r is going to be equal to .06m.
Yes, at the initial position.
The velocity at the initial position is zero, so I set v = 0?
Right.
1/2(.012)(0) + (.012)(9.8)(.06) + (8.99e9(.45uC*.45uC)/(.06)) = .03739725
I didn't check your arithmetic, but that's the right idea. Now set up a general equation and solve for the value of r at the final position.
 
  • #21
So the number I just calculated is for the TME at r initial?
 
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  • #22
Thanks for all the help, sorry I need so much support.
 
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  • #23
Should the new equation be:

1/2(.012)(0) + (.012)(9.8)(.06) + (8.99e9(.45e9*.45e9)/(r)) = .03739725

Where I solve for r?
 
  • #24
jgridlock said:
So the number I just calculated is for the TME at r initial?
Yes, but since TME is conserved it is the value for any r.
 
  • #25
jgridlock said:
Should the new equation be:

1/2(.012)(0) + (.012)(9.8)(.06) + (8.99e9(.45e9*.45e9)/(r)) = .03739725

Where I solve for r?
Almost. But don't plug in any values for r. (Like your second term.)

(And make sure you use the right charges. .45μC = .45e-6C, not .45e9C.)
 
  • #26
Alright, my final question is about the non mathematical part. Letters b-f (Correct me if I am wrong). When the balls are released the top one should reach a max height and then float above the second one because they repel each other. But then what happens when the friction comes into play?
 
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  • #27
jgridlock said:
When the balls are released the top one should reach a max height and then float above the second one because they repel each other.
I'd say that the top ball will reach a max height then fall back down to the lowest point again. It will keep oscillating, as energy is conserved.
But then what happens when the friction comes into play?
Eventually friction will dissipate the energy and the ball will end up at some equilibrium position.
 
  • #28
Doc Al said:
jgridlock said:
...
But then what happens when the friction comes into play?
...

Eventually friction will dissipate the energy and the ball will end up at some equilibrium position.
This depends upon the height of the cylinder versus the equilibrium position of the top ball.

It may continue to oscillate above the cylinder.
 
  • #29
SammyS said:
This depends upon the height of the cylinder versus the equilibrium position of the top ball.

It may continue to oscillate above the cylinder.
Good point.
 

FAQ: Two balls carrying the same charge

What is the concept of "Two balls carrying the same charge"?

The concept of "Two balls carrying the same charge" refers to two objects that have the same type and amount of electric charge. This means that they will repel each other if they have the same charge, or attract each other if they have opposite charges.

How can two balls have the same charge?

Two balls can have the same charge if they have gained or lost the same number of electrons. This can happen through various processes such as rubbing two objects together, which can transfer electrons from one object to another.

What happens when two balls carrying the same charge are brought close together?

When two balls carrying the same charge are brought close together, they will repel each other due to the repulsive force between like charges. This is because the electric field of one ball will push against the electric field of the other ball.

Can two balls carrying the same charge have different sizes or masses?

Yes, two balls carrying the same charge can have different sizes or masses. The amount of charge an object has is independent of its size or mass. However, the distance between the two objects will affect the strength of the repulsive or attractive force between them.

How does the charge of two balls affect their behavior?

The charge of two balls affects their behavior by determining whether they will attract or repel each other. Like charges will repel, while opposite charges will attract. The strength of the charge also affects the strength of the electric force between the two objects.

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