# Proton movement in an electric field

1. Apr 11, 2015

### Voodoo doodler

A proton initially moves left to right along the x-axis at a speed of 2 x 10^3 m/s. It moves into an electric field, which points in the negative x direction, and travels a distance of 0.15m before coming to rest. What acceleration magnitude does the proton experience?

equations

E = kq/r^2
F=ma
F=qE

mass of a proton = 1.67x10^-27 Kg
charge of a proton = 1.602 x 10^-19 C

Attempt 1

qE=ma

E= (9.00x10^9 N*m^2/C^2)(1.602x10^-19 C/.15m^2 )
= 6.4x 10^-8 N/C

((1.602x10^-19 C)(6.4x10^-8 N/C))/(1.67x10^-27Kg) = a

a = 6.139m/s^2

actual answer = 1.33x10^7

I realize this means that i have to do something with the initial velocity, but i can't seem to find anything that ties the velocity of a proton to its acceleration in an electric field except for maybe a= (vf-vi) / (tf-ti) but I'm not given the time. What am i missing?

2. Apr 11, 2015

### Simon Bridge

Welcome to PF;
I don't follow your reasoning for the calculation of the electric field strength.
In the equation kq/r^2 the "r" would be the distance that the proton is from the field source, you used the distance the proton took to stop. The "q" should be from charges other than the proton - no other charges are mentioned in the problem statement.

How would you normally find the acceleration of any old object object given initial speed and distance to come to rest?

3. Apr 11, 2015

### Staff: Mentor

There are no point-sources involved in creating the field that stops the proton. The problem statement should probably mention that the electric field is homogeneous.

This is purely classical mechanics.

4. Apr 11, 2015

### Voodoo doodler

oh! it's a kinematics question? so vf^2=vi^2*2ax

so v final = 0 because it comes to a rest
v initial = (2 x 10^3)^2
x = .15m
solving for a we get

a = 1.33 x 10^7 which is the correct answer, thank you.

Why do we not into account the electric field in this type of question? is there another method that can be used in respect to coulombs law and the laws of electric fields? just out of curiosity.

5. Apr 11, 2015

### Simon Bridge

The electric field is not taken into account because you have enough information to solve the problem without it. Besides - you are not told anything about the electric field so how could you possibly use it to solve the problem?

Consider the following problem:
A block initially slides left to right along a frictionless surface in the +x direction at a speed of 2 x 10^3 m/s. It moves onto an upwards sloping ramp, and travels a distance of 0.15m before coming to rest. What acceleration magnitude does the block experience?

Do you need the force of gravity or the slope of the ramp to solve this problem?
I could also have written the problem in terms of "a wind starts blowing in the -x direction slowing the block..." see?

Note: the calculation you did was properly for the average acceleration over the distance.

6. Apr 11, 2015

### Voodoo doodler

Yes i see now,

thank you.

7. Apr 11, 2015

### Simon Bridge

BTW: you can use that information to work out how strong a constant electric field would have to be to get that stopping distance ... qE=ma (where a is the acceleration you just calculated). That's how that equation fits into the problem.

One of the things that can happen is that the author provides enough information to do a problem several different ways but messes up the numbers so the different results are inconsistent. Depending on the character of the marker, spotting the inconsistency can be worth extra marks.

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