Proton Suspension Between Horizontal Parallel Plates: Charge & Voltage

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Homework Help Overview

The discussion revolves around a proton suspended in a vacuum between two oppositely charged horizontal parallel plates, with a separation of 2.60 mm. Participants are exploring the relationship between electric fields, forces acting on the proton, and the implications of charge and voltage across the plates.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the proton and question the direction of those forces. There is mention of using the equation E=vd/d, though some express uncertainty about its application. Questions are raised regarding the sign of the charge on the lower plate and the voltage across the plates.

Discussion Status

The discussion is active, with participants offering hints about identifying forces and checking calculations. There is a mix of attempts to clarify concepts and verify calculations, but no consensus has been reached regarding the final values or interpretations.

Contextual Notes

Participants are required to show their work toward a solution before receiving further assistance, which may influence the depth of discussion and the types of guidance offered.

lizaliiu
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1. A proton is suspended in a vacuum between two oppositely charged horizontal parallel plates. The separation between the plates is 2.60mm



2. I assume it uses E=vd/d, but I don't really know



3. what is the sign of the charge on the lower plate?
what is the voltage across the plate?
 
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2. I assume it uses E=vd/d, but I don't really know
What does that mean?

Some hints how to start:
Can you identify the forces acting on the proton?
If the proton stays at rest between the plates, can you identify the direction of those forces?
 
lizaliiu said:
1. A proton is suspended in a vacuum between two oppositely charged horizontal parallel plates. The separation between the plates is 2.60mm



2. I assume it uses E=vd/d, but I don't really know



3. what is the sign of the charge on the lower plate?
what is the voltage across the plate?

Welcome to the PF.

You are required to show your work toward a solution before we can offer much tutorial help.
 
since gravity pulls the proton downward so the lower plate should be the positive charge to repel the proton.
F=mg=qE
E=mg/q=v/d, m=1.673x10^-27kg, g=9.8m/s2, q=1.602x10^-19 and d=2.60x10^-3m
so we just need to solve q to get the voltage across the plates?
 
You know q, I think you mean V? Right.
 
yes, I solved V= 2.66x10^-10...not positive sure it's correct tho
 
Gravity is weak, and you can check your calculations in WolframAlpha, for example. If the units match, it should be right.
 

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