Basic particle accelerator - escape speed (grade 12)

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SUMMARY

The discussion focuses on the mechanics of a basic particle accelerator designed to fire electrons through a positively charged plate. It addresses modifications needed to accelerate protons and compares their escape velocities to those of electrons under the same electric field conditions. The key conclusion is that protons, due to their larger mass, will achieve a lower exit speed than electrons when subjected to the same electric field strength. Additionally, without modifications, protons are likely to be attracted to the negatively charged plate, preventing them from exiting the device.

PREREQUISITES
  • Understanding of basic electrostatics and electric fields
  • Familiarity with particle physics concepts, specifically charge and mass
  • Knowledge of the equation a = qε/m for particle acceleration
  • Basic principles of particle accelerators and their operation
NEXT STEPS
  • Research modifications for particle accelerators to accommodate different particle types
  • Study the relationship between mass, charge, and acceleration in electric fields
  • Explore the concept of escape velocity in the context of charged particles
  • Learn about the design and function of electric fields in particle accelerators
USEFUL FOR

Students studying physics, particularly those in grade 12 or higher, educators teaching particle physics concepts, and anyone interested in the principles of particle acceleration and electrostatics.

krbs
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Homework Statement


The device has two parallel plates, one positively charged and one negatively. The positive plate has a hole to allow electrons to fire out of it.

b) How can the device be modified to accelerate protons?
c) How does the escape velocity of a proton compare to that of an electron, if the electric field were the same? Explain.

Homework Equations


a = qε/m

The Attempt at a Solution


Part c is what I'm unsure about. Are they assuming that I've modified the device, but the electric field has the same magnitude?

If so, my educated guess is as follows:

The particles have equal but opposite charges and experience the same field, but a proton has a larger mass and will be accelerated less before exiting the device; therefore, it will have a lower exit speed.

On the other hand, assuming the device has not been modified, the proton would still have a lower exit speed (probably won't exit at all) because it's attracted to the negative plate.
 
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You're doing fine. Yes, part c question comes after the modification. (otherwise the protons would indeed go the other way and fall on to the negative plate).
 
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Thank you,
 

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