# How can I find the velocity of a proton?

1. Sep 20, 2009

### peaceinmideas

1. The problem statement, all variables and given/known data

A charged particle (either an electron or a proton) is moving rightward between two parallel charged plates separated by distance d = 2.20 mm. The charge on the left has a potential of -70V and the charge on the right has a potential of -50V. The particle is slowing from an initial speed of 91.0 km/s at the left plate.
(a) Is the particle an electron or a proton?
(b) What is its speed just as it reaches the plate at the right?

2. Relevant equations

?

3. The attempt at a solution

I understand that the particle is a proton, but I have not the slightest clue as to what equation to use. Any ideas?

2. Sep 20, 2009

### willem2

conservation of energy. (both electrical an kinetic).

you can also calculate the size of th electric field, and then the force on the proton, and make it into a standard projectile motion problem.

3. Sep 20, 2009

### Redbelly98

Staff Emeritus
As you travel to the right, the potential _____(increases or decreases?)

Therefore, the plate on the right is _____(positive or negative?) relative to the plate on the left.

Also, since the particle is slowing down, the plate on the right is ____(attracting or repelling?) the charge -- so the charge must be ____(positive or negative?).

Hope that helps.

4. Sep 20, 2009

### peaceinmideas

Still have no idea, i was thinking about using Vfinal^2= Vinitial^2 + 2a(Xfinal - Xinitial
) but i am not sure how to find a.

5. Sep 21, 2009

### willem2

You know how to compute the electric field in a capacitor? You also know the charge of a proton
so you can find out what the force on it is, and you know the mass of a proton, so you can find out a with F=ma

6. Sep 21, 2009

### Redbelly98

Staff Emeritus
Does the figure show how far apart those charges are? If you know the distance between them, you can use the two potential values to find the electric field.

The electric field can then be used to find the force on the charge.