Electrostatics uniform electric field of magnitude

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An electron traveling at a speed of 2.00×10^7 m/s in a uniform electric field of 1.16×10^4 N/C experiences an acceleration of -2.81×10^15 m/s². The distance it travels before stopping is calculated using the equation x = -(V(i))^2 / 2(a), resulting in approximately 0.0712 m. To determine the time before it returns to its starting point, the formula t = 2V(i) / a gives a time of about 1.42×10^-8 seconds. A common error noted in the discussion was the incorrect use of the electric field value, which affected the calculations. Accurate values are crucial for solving physics problems correctly.
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An electron with speed 2.00×10^7m/s is traveling parallel to a uniform electric field of magnitude 1.16×10^4N/C .

#1). How far will the electron travel before it stops?
f=ma=Eq=-eE
a=-eE/m
I got a=-2.81x10^15
Then I used v(f)^2=v(i)^2+2a(x-x(i))
Solved for x...
x=-(V(i))^2/2(a)
x=.0712m

#2). How much time will elapse before it returns to its starting point?
v=v(i) +at
If we want the time BEFORE it returns
t=2V(i)/a
t=1.42x10^-8 s

I got both wrong on Mastering Physics ... Please help I don't know what I am doing wrong
 
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meme177 said:
An electron with speed 2.00×10^7m/s is traveling parallel to a uniform electric field of magnitude 1.16×10^4N/C .

#1). How far will the electron travel before it stops?
f=ma=Eq=-eE
a=-eE/m
I got a=-2.81x10^15

You appear to have used a field of 1.6E4 instead of 1.16E4.
 
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I got threw off with all the sixes! Thanx a bunch!
 
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