# Coulomb's Law And Electric Field between 2 Parallel Plates

## Homework Statement

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.59cm distant from the first, in a time interval of 1.60×10−6s .

A). Find the magnitude of the electric field.
B). Find the speed of the proton when it strikes the negatively charged plate.

## Homework Equations

F=ma
E=F/q
y=y_0+v_0t+1/2at^2

## The Attempt at a Solution

Usually, to solve for the E field between two parallel plates, you use sigma/epsilon_0. However, I am not given sigma nor am I given anything about the dimensions of the plate, so I took the old kinematics route. First, I used the equation y=y_0+v_0t+1/2at^2. I said it starts from y_0=0 and it says it starts from rest, so I am left with y=1/2at^2. I plugged in the distance between the plates (.0159m) and the time (1.6*10^-6s) and solved for the proton's acceleration, which I found to be 1.24 m/s^2 in the negative y direction (towards the negative plate).

Next, I used F=ma - using the acceleration I just found (1.24 m/s^2) and the mass of the proton (1.6*10^-27kg) and solved for F. I got 1.984*10^-27N

Lastly, for part A, I used the equation E=F/q. Substituting the numbers in, I got that E=1.24*10^-8 N/C and it said that it was incorrect.

For part B, i used the acceleration I found above and plugged it into v=v_0+at and got V=1.984*10^-6, which was also wrong.

Any ideas what I did wrong?

## Answers and Replies

nasu
Gold Member
The value you obtained from acceleration seem to be off by many orders of magnitude.

rude man
Homework Helper
Gold Member
The value you obtained from acceleration seem to be off by many orders of magnitude.

'Many' is right! 