- #1

- 354

- 2

Firstly, what is the radius of the sphere use for? Secondly, I calculated the electric potential energy of the system but what is this energy equivalent to? Lastly, how can we solve this question if the spheres were conducting spheres?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter semc
- Start date

- #1

- 354

- 2

Firstly, what is the radius of the sphere use for? Secondly, I calculated the electric potential energy of the system but what is this energy equivalent to? Lastly, how can we solve this question if the spheres were conducting spheres?

- #2

- 2

- 0

Sorry for the long, useless response, but I would like an answer to your question as well!

Kelly

- #3

- 354

- 2

- #4

ideasrule

Homework Helper

- 2,266

- 0

Firstly, what is the radius of the sphere use for? Secondly, I calculated the electric potential energy of the system but what is this energy equivalent to? Lastly, how can we solve this question if the spheres were conducting spheres?

The radii are used for calculating the potential energy of the system after the collision. The electric potential energy isn't equal to anything, but the

If the spheres were conducting, the separation of charges would make the problem much harder to solve. I don't think the solution is trivial.

- #5

ideasrule

Homework Helper

- 2,266

- 0

That's what I had in mind at first, so do you think the energy lost by the system is equivalent to the kinetic energy gain by the sphere? So can we treat one sphere as stationary and just calculate the relative velocity of the moving sphere?

The potential energy lost by the system is equal to the kinetic energy gained by TWO spheres. You can't treat one sphere as stationary because in any inertial reference frame, it isn't.

I think if the spheres are conducting they would polarize each other as they move together?

Yes. That would make the problem extremely difficult to solve.

- #6

- 354

- 2

So if we use [tex]\Delta[/tex]U = [tex]\frac{1}{2}[/tex]m_{1}v_{1}^{2} +[tex]\frac{1}{2}[/tex]m_{2}v_{2}^{2} and conservation of momentum m_{1}v_{1}= -m_{2}v_{2} simultaneously is able to solve it? I am not able to get the answer?

Last edited:

- #7

ideasrule

Homework Helper

- 2,266

- 0

- #8

- 354

- 2

The distance r is 1m?

I gotten v

Share: