Electric Fields Question, What is the speed of the electron?

AI Thread Summary
An electron released in a weak electric field of -2.8 x 10^-10 N/C must account for both electric and gravitational forces to determine its speed after traveling 1.9 µm. The relevant equations include the electric force F=qE and the net force Fnet=Fg=ma. The attempt to calculate acceleration using F=qE resulted in incorrect values, indicating a misunderstanding of the forces at play. The work-energy principle, Work Done = F*d = 1/2(mv^2), was also incorrectly applied. A comprehensive approach considering both forces is necessary to solve for the electron's speed accurately.
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Homework Statement


An electron is released from rest in a weak electric field given by -2.8 x 10^-10 N/C [PLAIN]https://www.flipitphysics.com/Content/smartPhysics/Media/Images/Tipler/Symbols/jhatbold.gif. After the electron has traveled a vertical distance of 1.9 µm, what is its speed? (Do not neglect the gravitational force on the electron.) Answer should be in mm/s.

Any help towards setting up this problem or what formula to use will be greatly appreciated.

Homework Equations


E=F/q
F=qE+qV*B

F= (1/4πE0)(q1q2/r^2)
Fnet=Fg=ma=mg

Work Done = F*d = 1/2(mv^2)

The Attempt at a Solution


I don't really know where to start, this is the last problem on my assignment and I only have a few attempts left, we did not go over this in class yet and our class uses "Flipit Physics" videos instead of a book.

I tried Fe=qE
=(1.602x10^-19)(-2.8x10^-10) = -4.4856x10^-29
then a = Fe/m
= -4..572477064 x 10^-31 which was wrong

I also tried Work Done = F*d = 1/2(mv^2) and solved for v but that was also wrong.
 
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E=1/2mv^2 is equal to the total work done on the particle. You have two forces: the Lorentz force F=q(E+v*B) and gravitational force F=mg, both in the same direction (I assume j-hat is the unit vector in the y direction).
 
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