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## Homework Statement

An electron is originally moving at 6.8 x 10

^{7}m/s in positive x direction as it enters a region of confined, uniform, electric field. Magnitude of field is 2.6 x 10

^{6}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

_{0y}t+1/2a

_{y}t

^{2}

L=v

_{0x}t+1/2a

_{x}t

^{2}

## The Attempt at a Solution

F

_{x}=0

F

_{y}=eE

a

_{x}=0

eE=m

_{e}a

_{y}

a

_{y}=eE/m

_{e}

v

_{0y}=0 m/s

y=1/2a

_{y}t

^{2}=1/2eE/m

_{e}t

^{2}

L=v

_{0}t

t=L/v

_{0}

y=1/2eE/m

_{e}t

^{2}=1/2eE/m

_{e}(L/v

_{0})

^{2}=1/2(1.6*10

^{-19}C)(2.6*10

^{6}N/C)/(9.10938188*10

^{-31}kg)(6.2*10

^{-4}m/6.8*10

^{7}m/s)

^{2}=1.9*10

^{-5}m

arctan(1.9*10

^{-5}m/6.2*10

^{-4}m)