(Very Basic) Movement of Charged Particles Through Parallel Plate Apparatus

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SUMMARY

The discussion centers on calculating the electric potential in a parallel-plate apparatus with a total separation of 5 cm and an electric field strength of 5000 N/C. The correct electric potential two-fifths of the way through the apparatus is determined to be 250 V, not 150 V as initially calculated. The confusion arose from the misunderstanding that electric potential decreases as the current flows, which is not applicable in this context. The potential difference across the entire distance is 250 V, confirming that the potential at any point must be less than this total value.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with the formula V = ed for calculating electric potential
  • Knowledge of parallel-plate capacitor configurations
  • Basic principles of electrostatics
NEXT STEPS
  • Study the relationship between electric field strength and potential difference in parallel-plate capacitors
  • Learn about the concept of equipotential surfaces in electrostatics
  • Explore the effects of varying plate separation on electric potential
  • Investigate the implications of electric potential in circuit design and analysis
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Students in physics, electrical engineering majors, and anyone studying electrostatics or parallel-plate capacitor systems will benefit from this discussion.

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Homework Statement


What is the electric potential two-fifths of the way through a parallel-plate apparatus (from the positive plate) if the plates have a total separation of 5cm and field strength of 5000N/C.

Homework Equations


Difference Potential Energy = Electric Field Strength x Distance
V = ed

The Attempt at a Solution


= 5000 x (0.05 - ((2/5)(0.05))
= 5000 x (0.05-0.02)
= 5000 x (0.03)
= 150

Real answer is 250, so basically the whole distance of 0.05.
I thought that electric potential decreases as the current flows from positive to negative?
That is why, in my solution, I took the two-fifth of the whole distance away.
Edit: Please let me know if my solution as infact correct :)
 
Last edited:
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Your solution looks fine to me. As you say, the total potential difference is 250V, so a correct answer for the potential partway between the plates can't be 250V also.
 

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