Proton release inside a parallel capacitor Find velocity

[conov3]

Homework Statement

The electric field strength is 47100 N/C inside a parallel-plate capacitor with 8.60 mm spacing. A proton is released from rest at the positive plate. What is the speed of the proton (in m/s) when it reaches the negative plate?

Homework Equations

Vf=√(-2U/m) I think?
ΔU=-.5eEd

The Attempt at a Solution

I tried using that equation to find ΔU=-3.768*10^-15
then plugged it in seem to have the wrong final answer though

 

[tiny-tim]

welcome to pf!

hi conov3! welcome to pf!

ΔU=-.5eEd

why .5 ?

 

[conov3]

Thank you! I figured since I am in a college level physics, I might as well try to get some help if I need it! ha

Using my Physics book.. it says ΔU=Uf-Ui= (Uo+ (-e)Ed)-(Uo+(-e)E(d/2)
Then under that it says =-1/2eEd

I am confused about it also and am really unsure.

 

[tiny-tim]

Using my Physics book.. it says ΔU=Uf-Ui= (Uo+ (-e)Ed)-(Uo+(-e)E(d/2)

i don't understand that … it doesn't seem to fit the question