# Speed of a Proton in an Electric Field

1. Apr 19, 2007

### Boozehound

A uniform electric field has a magnitude of 2.41E+3 N/C. In a vacuum, a proton begins with a speed of 2.27E+4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 1.85 mm.

alright i looked for some kinematics equations and i used v=d/t. then when i figured out time to be 8.149E-8 i plugged it in to d=1/2(a)(t^2)+(v)(t). i found acceleration to be 2. once i get here im stuck. any help to point me in the right direction would be a great help.

2. Apr 19, 2007

### hage567

v=d/t does not work during acceleration. You need to think about what force is working on the proton and use Newton's second law to find the acceleration. After that, you can use kinematics. So, how do you think you can find the force of the electric field on the proton?

3. Apr 19, 2007

### Boozehound

alright so if i use newtons second law (F=ma) i would think that F is the electric field. so do i then take 2.41E3N/C and multiply it by 1.60E-19 so i would have just newtons? and if i do that then would i use 1.672E-27kg for the mass of the proton?

4. Apr 19, 2007

### hage567

Yes, that's the proper approach to find the acceleration.

5. Apr 19, 2007

### Boozehound

alright thanks!

6. Apr 19, 2007

### Boozehound

well i used newtons 2nd law to find acceleration to be 2.306E11m/s^2. then i too that and plugged it into d=1/2(a)(t^2)+(v)(t) to solve for velocity. but when i got to put the answer i get into my program for school it tells me its wrong. so where am i going wrong?

7. Apr 19, 2007

### hage567

What did you use for t??? The value you found for t before is not valid. There is another kinematic equation that is more suitable.

BTW, you don't actually need to find t, there is an equation you can use to find the final velocity that doesn't involve time.

Last edited: Apr 19, 2007
8. Apr 19, 2007

### Boozehound

ok so i looked up even more kinematics equations and i came up with this

v^2(final)=v^2(initial)+2ax

v^2(final)=2.27E4m/s+2(2.306E11m/s^2)(.00185m)

and i get final velocity to be 2.92E4m/s then i took those answers and plugged them into x=1/2(v(initial)+v(final))t. and i get t to equal 7.129E-8s. and that doesn't sounds right at all. i mean granted its only moving 1.85mm but its a proton and thats a really fast time. this problem has been kicking my ass.

9. Apr 19, 2007

### Boozehound

i saw your last post after i posted that. and i found the formula vf^2=vi^2+2ax and i plug in the numbers and i get 2.92E4m/s which is wrong. so im thinking my acceleration is off.

10. Apr 19, 2007

### hage567

v^2(final)=v^2(initial)+2ax

This is all you need though. You want to find v(final), right? You know everything else in this equation. a you found by Newton's second law, v(initial) and x were given in the question.

I am thinking you didn't square v(initial) in your calculation. Try it again.

11. Apr 19, 2007

### Boozehound

yeap..that was my problemo. ah how the little things can make things seem impossible. thank you very much!

12. Apr 19, 2007

### hage567

You're welcome!