# Confined, Uniform, Electric Field

## Homework Statement

An electron is originally moving at 6.8 x 107 m/s in positive x direction as it enters a region of confined, uniform, electric field. Magnitude of field is 2.6 x 106 N/C in negative y direction and its length in x direction is 6.2 x 10-4 meters. If one ignores weight of electron as a force, with what angle will electron be traveling when it leaves field when measured counter-clockwise from positive x axis in degrees? Answer is 3.43.

y=v0yt+1/2ayt2
L=v0xt+1/2axt2

## The Attempt at a Solution

Fx=0
Fy=eE
ax=0
eE=meay
ay=eE/me
v0y=0 m/s
y=1/2ayt2=1/2eE/met2
L=v0t
t=L/v0
y=1/2eE/met2=1/2eE/me(L/v0)2=1/2(1.6*10-19C)(2.6*106N/C)/(9.10938188*10-31kg)(6.2*10-4m/6.8*107m/s)2=1.9*10-5m
arctan(1.9*10-5m/6.2*10-4m)

## Answers and Replies

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Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
The angle with which something is traveling refers to its velocity, not its displacement.

I appreciate the assistance!