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

AI Thread Summary
The discussion centers on calculating the electric potential two-fifths of the way through a parallel-plate apparatus with a separation of 5 cm and an electric field strength of 5000 N/C. The initial attempt at the solution incorrectly calculated the potential as 150V, based on a misunderstanding of how electric potential changes across the plates. The correct potential at that point is actually 250V, as the total potential difference across the plates is 250V. Clarifications indicate that the potential does not remain constant and decreases from the positive to the negative plate. The final consensus confirms that the initial calculation was incorrect, and the potential at two-fifths of the distance should reflect the total potential difference.
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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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