What Is the Velocity of an Electron 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/2εo, resulting in a value of 9604.52 N/m. The force on the electron is determined to be 1.537E-15 N, leading to the equation v^2 = 2(F/m)s for calculating its velocity. The initial attempt to find the velocity resulted in an incorrect value due to misapplication of the equations. It was noted that the electric field formula used was intended for a single plate, not accounting for the presence of two plates. The correct approach involves using the total electric field between the plates to accurately determine the electron's final velocity.
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Homework Statement


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 = 1.70E-7 C/m2, and the plates are separated by a distance of 1.59E-2 m. How fast is the electron moving just before it reaches the positive plate?



Homework Equations


E = σ/2εo
F=q X E
v2 - u2 = 2 a s = 2 (F/m) s


The Attempt at a Solution


E= 1.70E-7/2(8.85E-12)= 9604.52 N/m

F= 1.6E-19 * 9604.52= 1.537E-15 N

v^2= 2(1.537E-15/9.1E-31)*1.59E-2
= 5.370E13...this is wrong...what am i doing wrong??
 
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Did you remember to take the square root to get v instead of v²?
 
yes, when i take the sq root, i get 7.328 E6 m/s...still not the answer!
 
missyc8 said:

Homework Equations


E = σ/2εo
That formula is for a single plate. However, there are two plates here.
 
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