# Parallel-plate Capacitor

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

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.

**2. Relevant equations**

Ek=(1/2)mv^2

**3. The attempt at a solution**

I solved for the kinetic energy using the above equation, Ek=(1/2)(1.67×10^-27kg)(2.0×10^6m/s = 1.67×10^-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!

http://ift.tt/POp6AR

## Leave a comment