Two balls carrying the same charge

[jgridlock]

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.

 

[Doc Al]

Think in terms of energy, not force, for most of the questions.

 

[jgridlock]

I could use a little more guidance than that. What kind of process should I follow to solve part A?

 

[Doc Al]

Apply conservation of energy. What forms of energy are relevant here and how would you calculate them?

 

[jgridlock]

PE = mgh?

 

[Doc Al]

PE = mgh?

OK, that's gravitational PE.

What other kinds of energy are involved here?

 

[jgridlock]

Kinetic energy? KE = 1/2mv^2

 

[Doc Al]

Kinetic energy? KE = 1/2mv^2

Yep, that's another.

Keep going.

 

[jgridlock]

I wouldn't know any others. A guess would be charge?

 

[Doc Al]

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!

 

[jgridlock]

Electric Potential Energy: PE = k((q1*q1)/r)

A pair of point charges separated by a distance

 

[cbasst]

You may also want to check the units you are using for r.

 

[jgridlock]

Convert cm to m. 3cm = .03m
And g to kg. 12g = .012kg

Alright, next step?

 

[Doc Al]

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.

 

[jgridlock]

TME = 1/2mv^2+mgh+k((q1*q1)/r)

 

[Doc Al]

TME = 1/2mv^2+mgh+k((q1*q1)/r)

Good. But use the same distance r for both gravitational and electrostatic PE.

 

[jgridlock]

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))?

 

[Doc Al]

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.

 

[jgridlock]

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

 

[Doc Al]

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.

 

[jgridlock]

So the number I just calculated is for the TME at r initial?

 

[jgridlock]

Thanks for all the help, sorry I need so much support.

 

[jgridlock]

Should the new equation be:

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

Where I solve for r?

 

[Doc Al]

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.

 

[Doc Al]

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.)

 

[jgridlock]

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?

 

[Doc Al]

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.

 

[SammyS]

...
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.

 

[Doc Al]

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.