Is the m/e Equation Valid When F = 0 in J. J. Thomson's Experiment?

  • Thread starter Thread starter Samanko
  • Start date Start date
  • Tags Tags
    Experiment
Click For Summary
SUMMARY

The discussion centers on the validity of the mass-to-charge ratio (m/e) equation in J. J. Thomson's experiment when the net force (Fnet) is zero. It is established that if Fnet equals zero, the velocity (v) can be expressed as v = E/B, allowing for the measurement of m/e. However, a critical question arises regarding the assumption that a zero net force implies a zero change in the vertical position (y), which would render the denominator of the m/e equation zero, thus invalidating the equation. The participants debate the implications of these assumptions on the equation's validity.

PREREQUISITES
  • Understanding of J. J. Thomson's experiment and its significance in physics.
  • Familiarity with the concepts of electric fields (E) and magnetic fields (B).
  • Knowledge of the mass-to-charge ratio (m/e) and its calculation.
  • Basic principles of force and motion in physics.
NEXT STEPS
  • Explore the derivation of the mass-to-charge ratio (m/e) in Thomson's experiment.
  • Investigate the implications of zero net force in classical mechanics.
  • Learn about the behavior of charged particles in electric and magnetic fields.
  • Review the mathematical treatment of forces and motion in physics.
USEFUL FOR

This discussion is beneficial for physics students, educators, and researchers interested in classical mechanics, electromagnetism, and the historical context of Thomson's experiments.

Samanko
Messages
14
Reaction score
3
Homework Statement
The statement is below.
Relevant Equations
F = qv × B
20191021_145637.jpg
20191021_145723.jpg
The text showed if Fnet = 0 (with F from magnetic fields and electric field), then v = E/B . And we can measure m/e (last equation in the image).

My question is, by assuming F = 0, change in y should also be 0 , so the denominator of right side of the m/e equation must be zero. And the equation doesn't hold. Am I wrong?
 
Physics news on Phys.org
Samanko said:
My question is, by assuming F = 0, change in y should also be 0 , so the denominator of right side of the m/e equation must be zero. And the equation doesn't hold. Am I wrong?

That's ##y## from the first case with ##E## only.
 

Similar threads

Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
12
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K