Confined, Uniform, Electric Field

[eestep]

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.

Homework Equations

y=v_0yt+1/2a_yt²
L=v_0xt+1/2a_xt²

The Attempt at a Solution

F_x=0
F_y=eE
a_x=0
eE=m_ea_y
a_y=eE/m_e
v_0y=0 m/s
y=1/2a_yt²=1/2eE/m_et²
L=v₀t
t=L/v₀
y=1/2eE/m_et²=1/2eE/m_e(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)

 

[Redbelly98]

The angle with which something is traveling refers to its velocity, not its displacement.

 

[eestep]

I appreciate the assistance!