Proton Deflection when traversing a a Parallel-plate Capacitor

[Colton1995]

Homework Statement

A parallel-plate capacitor has 4.0cm × 4.0cm electrodes with surface charge densities ±1.0×10^−6C/m2. A proton traveling parallel to the electrodes at 2.0×10^6m/s enters the center of the gap between them.
Part A
By what distance has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the field is uniform inside the capacitor and zero outside the capacitor.

Homework Equations

Ek=(1/2)mv^2

The Attempt at a Solution

I solved for the kinetic energy using the above equation, Ek=(1/2)(1.67x10^-27kg)(2.0x10^6m/s = 1.67x10^-21J.
I'm pretty sure that I'm suppose to split it up into x and y variables to solve for delta y, but I am completely stuck. I would normally look at the answer and try to work it out from there but I am doing extra practice questions for review for my final and my prof. hasn't posted the answers yet. I'm not looking for an answer, but rather a push in the right direction. Thanks!

 

[paisiello2]

I don't think using energy is the way to solve this problem. I would suggest just a simple kinematic approach using Coulomb's Law.

 

[Elsi b]

By using acceleration formula with electric field.
a= (P x E) / m