Simple Gas Ionization Detector Problem

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SUMMARY

The discussion centers on calculating the voltage pulse produced by a gas-filled counter, modeled as a parallel plate capacitor with a capacitance of 9.1 x 10-9 F. The energy deposited between the plates is 2 x 106 eV, with each ionization requiring 20 eV. The participant correctly determines that 100,000 ionizations occur, resulting in a charge change of 1.6 x 10-14 C. The calculated voltage pulse is 1.8 x 10-6 V, indicating a decrease in voltage, which aligns with the physics of charge movement in the capacitor.

PREREQUISITES
  • Understanding of parallel plate capacitors
  • Knowledge of ionization energy in gas detectors
  • Familiarity with basic electrostatics equations, specifically C = Q/V
  • Concept of charge conservation in electric fields
NEXT STEPS
  • Study the principles of ionization in gas detectors
  • Learn about the behavior of capacitors under varying charge conditions
  • Explore advanced topics in electrostatics, including energy storage in capacitors
  • Investigate the applications of gas-filled counters in radiation detection
USEFUL FOR

Students in introductory physics, particularly those studying electromagnetism and gas detection technologies, as well as educators seeking practical examples of capacitor behavior in real-world applications.

jumbogala
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Homework Statement


I'm not sure if this is really introductory physics... tell me if you think I'd have more luck in the advanced physics section.

The problem: You have a gas filled counter, in the form of a parallel plate capacitor. It has capacitance 9.1 x 10-9 F. It takes 20 eV of energy for each ionization in the capacitor to occur.

2 x 106 eV of energy is deposited between the plates by a particle. What is the size of the voltage pulse produced?

Homework Equations


C = Q/V


The Attempt at a Solution


I took a guess, but I have no idea if it's right.

I said that 2E6 / 20 = # of ionizations = 100 000

Each ionization produces an electron and an ion, with charge e. The electrons move to the positive plate and the ions move to the negative plate. Because the charges on the plates will counteract each other, we're lowering Q by (100 000*e) = 1.6E-14.

Then (change in charge) / C = (change in voltage)

(1.6E-14) / 9.1E-9 = 1.8E-6 is the voltage pulse seen (as a decrease).

Is that right? Does it make sense that the voltage decreases - doesn't pulse imply increase?
 
Physics news on Phys.org
"Pulse" can be either up or down.
Very convincing - looks correct.
 

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