Final velocity of a proton through parallel plates

[Serendipitydo]

Homework Statement

A moving proton has 6.4x10^-16 J of kinetic energy. The proton is accelerated by a potential difference of 5000 V between parallel plates.

http://members.shaw.ca/barry-barclay/Self-Tests/test09/elecst17.gif


The proton emerges from the parallel plates with what speed?

A. 8.8x10^5 m/s B. 9.8 x10^5 m/s C. 1.3x10^6 m/s D. 1.8x10^6 m/s

I know that the answer is C, I just don't know how to get there.

Homework Equations

q=1.60x10^-19 C
m=1.67x10^-27 kg
E_k=6.4x10^-16 J
5000V electric potential difference (I don't really fully grasp what this means. I have something that says electric potential difference = work done in moving test charge/magnitude of test charge [V=ΔE/q])


I know that Kinetic Energy=q|E|d
V=ΔE/q
W=Vq
Fe=kq/r²
|E|=V/d
Then I know the basic kinematics formulas

The Attempt at a Solution

I found work by using W=Vq. I get (5000V)(1.60x10^-19C)= 8.00x10^-16J

Then I'm lost as to what to do. But:

Using E_k=0.5mv² I get the answer A. Is that actually the correct answer and the answer key from the thing he copied/pasted is wrong?

I saw this thread when googling https://www.physicsforums.com/showthread.php?t=168634 , which made me think that.

 

[wukunlin]

5000V of potential difference means for every coulomb of your charged particle, they will gain 5000J of energy so...

how much energy is gained/loss for a single particle?
how much energy does the particle end up with?
what speed does that translate to?