Parallel Plates - Electron Liberation

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Homework Help Overview

The problem involves an electron being liberated from the lower of two parallel plates separated by a distance of 5mm, with the upper plate at a potential difference of 1000V relative to the lower plate. The objective is to determine the time it takes for the electron to reach the upper plate.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Newton's second law and equations of motion to find the time taken for the electron to travel the distance. Another participant suggests using relativistic mass and differentiation to solve for velocity and time. The original poster questions the differentiation approach and seeks a comparison of results.

Discussion Status

The discussion includes various approaches to the problem, with some participants exploring different methods of calculation. There is no explicit consensus on the best method or the correctness of the original poster's solution, but multiple interpretations are being examined.

Contextual Notes

Participants are discussing the implications of using classical versus relativistic mechanics in the context of the problem, which may affect the calculations and assumptions made.

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Homework Statement


Hello, I wonder if anyone could check my answer to this problem.

Thanks in advance

An electron is liberated from the lower of two parallel plates separated by a distance of 5mm. The upper plate has a p.d. of 1000v relative to the lower plate. How long does it it take for the electron to reach to upper plate.

Homework Equations


f=ma, f=qe, e=v/d

The Attempt at a Solution


I said F=ma=QE therefore QE=ma and since E=V/d=1000/0.005=2x105

a=3.513x1016m/s2!

Then using s=ut+1/2at2, t=5.335x10-10s.

Thanks again for any help

Mike
 
Last edited:
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i would take

ma = relativistic mass * dv/dt = qE

solve for v and find time taken
 
Hello, thanks for your response.

Is that differentiation? Not too sure about how to do that.

Any chance you could get an answer from your method and compare it to mine.

Thanks again,

Mike
 
4.5*10^-11 s

try it, its really simple
 

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