# Final velocity of proton after being repelled?

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1. Feb 25, 2016

### Brennen berkley

1. The problem statement, all variables and given/known data
What will be the final velocity of a proton (when it is very far away) if it it released from rest at the center of a uniformly charged hoop and given a slight push in one direction. assume it follows the axis of the hoop.
mass of proton: m = 1.67x10-27 kg
charge of proton: q = 1.6x10-19 C
radius of hoop: R = 1 cm
charge of hoop: Q = 1 nC

2. Relevant equations
λ = Q/2πR
k = 9 x 109
E = kλx2πR/(x2+R2)3/2
F = Eq
3. The attempt at a solution
λ = 9x10-9 / 2π(.01)
= 1.59 x 10-8 C/m

I know that the force and acceleration will vary with the distance from the hoop x, but I don't know how to find the final velocity. I was thinking maybe you have to find the potential energy of the proton before it starts moving and then use that as its final kinetic energy to find velocity? I don't know how to find electrical potential energy though.

2. Feb 25, 2016

### TSny

Welcome to PF!
Energy is definitely the best approach. Can you find the electric potential V at the center of the ring due to the charge on the hoop?

Last edited: Feb 25, 2016
3. Feb 25, 2016

### Brennen berkley

Looking through my book I found the energy equation U = qE, and using that I found the energy to be 1.503 x 10-20, and I got 4242 m/s for the velocity. does that sound about right?

4. Feb 25, 2016

### TSny

Did you mistype this? U is not equal to qE. Instead, F = qE. Check your book again for the correct formula for U.

5. Feb 25, 2016

### Brennen berkley

6. Feb 25, 2016

### Brennen berkley

maybe I'm reading the equation wrong, but there it is.

7. Feb 25, 2016

### TSny

Note that equation (23.5) reads U = qEy which is different from U = qE.

Equation (23.5) is only valid for a uniform electric field. In the hoop problem, E is not uniform. You should have studied a basic equation that relates the potential energy U to the electric potential V.

8. Feb 25, 2016

### Brennen berkley

Thanks for clarifying. In class we have only covered the material through chapter 21, and all the electric potential stuff is in 23, so I haven't read about it very in-depth yet. I'm not sure why it's on our homework. Anyway, I think I understand what I need to do now. Potential Energy U = qV where V is potential (potential energy per Coulomb) and V = kQ/√(x2 + a2) where x is the distance from the hoop (0 in this case).

9. Feb 25, 2016

### Brennen berkley

Using the new equation I got 1.44 x 10-16 W for the energy and 415,277 m/s for the velocity.

10. Feb 25, 2016

### TSny

Yes, I believe that's the correct answer. (I'll let you think about significant figures.) It is strange that you were expected to work this problem based on what you've covered in class. Good work!

11. Feb 25, 2016

### TSny

Chapter 23 has a similar problem: #34.