# Homework Help: Electron traveling parallel to a Uniform Electric Field

1. Sep 4, 2011

### format1998

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

An electron with speed 2.0E7 m/s is traveling parallel to a uniform electric field of magnitude 1.18E4 N/C. How far will the electron travel before it stops? How much time will elapse before it returns to its starting point

q = 1.60E-19 C
mass of electron = 9.11E-31 kg
velocity (initial) = 2.10E7 m/s
velocity(final) = 0 (zero)

2. Relevant equations
Electric Field
E=F/q --> F=qE

Acceleration / Force equation
a=F/m

Constant Acceleration
v(final)^2=v(initial)^2 +2a*(delta x) --> delta x = (v(final)^2 - v(initial)^2) / 2a

3. The attempt at a solution

F=qE = (1.60E-19 C)(1.8E4 N/C) = 2.88E-15N

a=F/m = (2.88E-15N) / (9.11E-31 kg) = 3.16E15 m/s^2

delta x = (v(final)^2 - v(initial)^2) / 2a
delta x = (0 - 2.10E7 m/s) / 2*(3.16E15 m/s^2)
delta x = -6.97785E-2 m <--- ACCORDING TO MASTERING PHYSICS THIS ANSWER IS WRONG!!!

with the 2nd part of the question... I was going to use the equation
delta x = 1/2[v(initial) + v(final)]*t
but since my delta x is wrong, i haven't quite started this part.....

Any and all help will be appreciated! Thank you in advance!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 4, 2011

### cepheid

Staff Emeritus
Welcome to PF,

It looks like you might have slipped up on your arithmetic. The given electric field is 1.18e4 N/C, but you have used 1.8e4 N/C.

By the way, the first part of this question is much easier to solve using conservation of energy. For a uniform electric field, the potential difference between two points is just the electric field strength multiplied by the distance between the two points (V = E*Δx). And the potential energy change for the electron is just the electron charge multiplied by the potential difference. Finally, since the electric force is a conservative force, all of the initial kinetic energy of the electron is converted into electric potential energy (as the electric force does negative work on it to slow it). So you just equate the kinetic and potential energies and solve for Δx.

You'll end up with:

-e*V = (1/2)mv2

where e = 1.60e-19 C is the elementary charge, the smallest unit of charge found in nature (e is the charge of a proton, and -e is the charge of an electron).

-2eEΔx = mv2

Δx = -mv2/(2eE)

Notice that this is just the same formula as the one you obtained using kinematics, since a = F/m = eE/m. However, with this method you arrive at it more quickly. Also, written in this way, it allows you to just plug in all of the given quantities without having to take the intermediate step of computing a.

Last edited: Sep 4, 2011
3. Sep 4, 2011

### format1998

Hmmm... Hahahaha! One digit causing so much trouble. Thank you so much. Your help is much appreciated...