# Homework Help: Final velocity of a proton through parallel plates

1. Feb 3, 2012

### Serendipitydo

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

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.

2. Relevant equations
q=1.60x10^-19 C
m=1.67x10^-27 kg
Ek=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

3. 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 Ek=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?