Parallel Plates - Electron Liberation

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SUMMARY

The discussion focuses on calculating the time it takes for an electron to travel between two parallel plates separated by 5mm, with a potential difference of 1000V. The initial calculations provided by the user, Mike, yield a time of approximately 5.335 x 10-10 seconds. Another participant suggests using relativistic mass and differentiation to find a more accurate time of 4.5 x 10-11 seconds. The conversation emphasizes the importance of understanding electric fields and forces acting on charged particles.

PREREQUISITES
  • Understanding of electric fields and forces (F = qE)
  • Basic knowledge of kinematics (s = ut + 1/2at2)
  • Familiarity with potential difference and its relation to electric field (E = V/d)
  • Concept of relativistic mass and its implications in physics
NEXT STEPS
  • Study the concept of relativistic mass and its effects on particle motion
  • Learn about electric field calculations and their applications in particle physics
  • Explore advanced kinematics, including differentiation techniques for motion analysis
  • Research the principles of electron dynamics in electric fields
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and kinematics, as well as educators seeking to clarify concepts related to electric fields and particle motion.

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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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