Does a Proton Reach the Other Side of a Charged Capacitor?

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SUMMARY

The discussion focuses on the behavior of a proton shot through a charged parallel plate capacitor with a potential difference of 250 V. The electric field inside the capacitor is calculated to be approximately 166,667 V/m, and the charge on each plate is determined to be 1.5 x 10^-6 C. The critical question is whether the proton, with an initial speed of 1 x 10^6 m/s, can reach the other side of the capacitor. The analysis concludes that the proton does not reach the other side due to the influence of the electric field, which decelerates the proton before it can exit the capacitor.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Knowledge of parallel plate capacitor construction and operation
  • Familiarity with basic kinematics and motion under uniform acceleration
  • Proficiency in calculating electric force and energy of charged particles
NEXT STEPS
  • Study the principles of electric fields in capacitors
  • Learn about the motion of charged particles in electric fields
  • Explore the concept of energy conservation in electric fields
  • Investigate the effects of varying potential differences on particle trajectories
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Physics students, electrical engineers, and anyone interested in the dynamics of charged particles in electric fields.

leospyder
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Ive figured out the first parts (a) and (b) but I am stumped on (c).

A parallel plate capacitor is constructed of two 1.0 cm diameter disks which are spaced 1.5 mm apart. The capacitor is charged to a potential difference of 250 V. (a) Calculate the magnitude of the electric field inside the capacitor. (b) Calculate the charge on each plate. (c) Now assume that a proton is shot through a small hole in the negative plate with a speed of 1 X 10^(6) m/s. Does it or does it not reach the other side? If it does not, where is the turning point? Explain your reasoning and approach fully.

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