How Fast Does an Electron Move Between Capacitor Plates?

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An electron released from rest at the negative plate of a parallel plate capacitor experiences an electric field calculated using E=σ/ϵ, resulting in E=1.9096E4 N/C. The force on the electron is determined to be F=3.05536E-15 N, leading to an acceleration of a=3.35E15 m/s^2. To find the final velocity before reaching the positive plate, kinematic equations can be utilized, specifically v^2 = v_o^2 + 2ax. The discussion highlights the need to clarify initial conditions and suggests using potential energy and kinetic energy relationships as alternative methods. Ultimately, the problem can be solved using various approaches, emphasizing the electron's initial state as critical.
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An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ = 1.69E-7 C/m2, and the plates are separated by a distance of 1.72E-2 m. How fast is the electron moving just before it reaches the positive plate?

i started by using the equation E=σ/ϵ. so i got E=1.9096E4N/C. then i multiplied that by one coulomb to get force. so F=3.05536E-15N. i then took Newtons seconds law (F=ma) and found acceleration. for the mass of the electron i used 9.109E-31kg. so i was left with a=3.35E15m/s^2. i need to find out velocity, and so i looked for kinematic equations and i can't find one with all the variables that i have. it seems like everyone i try to use is missing 2 variables. am i overlooking something? please point me in the right direction. thanks!
 
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You can use this equation again:

v^2 = v_o^2 + 2ax

You have all of the values you need. Which ones are you unsure of?
 
I would try to equate difference in PE's with KE and forget kinematics entirely.

(edit: pretty much end up with Hages eqn either way)
 
As long as you've computed the acceleration, if you really want to use kinematics, use d=(1/2)*a*t^2 to compute the time and v=a*t to compute the final velocity. You have a superabundance of choices of how to complete the problem.
 
in response to hage567 i don't think that i have the initial velocity. thanks to all three of you.
 
Boozehound said:
in response to hage567 i don't think that i have the initial velocity. thanks to all three of you.

"An electron is released from rest". Think again.
 

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