Calculating Electron Speed and Energy in a Parallel-Plate Apparatus

AI Thread Summary
The discussion focuses on calculating the speed of an electron released from a parallel-plate apparatus under a 450 V potential difference. The energy gained by the electron is calculated using the formula Eel = V x q, resulting in 7.2 x 10^17 J. The speed of the electron when it hits the positive plate is derived from the kinetic energy equation, yielding a velocity of approximately 1.26 x 10^7 m/s. There is confusion regarding the separation of the problem into parts a) and b), with a request for clarification on the first question about the electron's speed one third of the way between the plates. The importance of correctly handling exponent signs in calculations is also emphasized.
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Homework Statement


An electron is released from rest from the negative plate of a parallel-plate apparatus, what is the electrons speed one third of the way between the plates. It also asks for the speed at which electron will hit the positive plate if a 450 V potential difference is applied.


The Attempt at a Solution



a)
Eel = V x q
Eel = ( 450 V ) x (1.6 x 10^-19C)
Eel = 7.2 x 10^17 J

Eel = 1/2mv^2
V = √((2 x (7.2 x 10^17 J))/9.1 x 10^31Kg)
V = 1.26 x10^7 m/s

b) i don't know how to do this?
 
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I don't understand your answer's division into a) and b). The a) section looks like a response to the second question in the problem statement. What happened to the first question? Or is that now part b)?

Also, watch out for the signs associated with exponents; there's a rather large difference between 10^17 J and 10^-17 J. Same goes for the electron's mass. Your final velocity for the 450V excursion looks okay though.
 
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