Electric Field and Acceleration for a Single Charged Plate and Capacitor

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Homework Help Overview

The discussion revolves around understanding the electric field and acceleration related to a charged plate and a capacitor. Participants are exploring the implications of given equations and constants in the context of electric fields and forces acting on charged particles.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive the electric field from given equations and constants, questioning the appropriate geometry for the area involved. There is discussion about the relationship between the electric field and the trajectory of a charged particle, with some suggesting that the acceleration of the particle must align with the electric field.

Discussion Status

Some participants have provided insights regarding the nature of the charged particle, suggesting it is likely a proton based on its mass and charge. There is acknowledgment of the potential scenarios involving a charged plate or a capacitor, but no consensus has been reached on the exact problem statement.

Contextual Notes

Participants note that the concept of capacitors has not been covered in their lectures, which may influence their understanding and approach to the problem. There is a mention of the need for clarification on the assumptions regarding the electric field and its configuration.

Zbud
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Homework Statement


I don't have any statement, i have to find the problem statement from the answers
a) E = Q / [ Eo* (0.020 m)^2 ]
b) a = E (1.60*10-19 C)/(1.67*10^-27 kg) = 2.0 *10^12 m/s^2
R=0,020 m
q=1.60*10-19 C
m=1.67*10^-27 kg

Homework Equations


F=ma => a=qE/m
Gauss Law Integral of (E,ds)=Q/Eo

The Attempt at a Solution


So given E constant everywhere i can assume that E*(integral of ds)=Q/Eo, so if this was a sphere it would be
4piR^2*E=Q/Eo so its out, my equation must be something like E*R^2=Q/Eo , so my area should be a a square ?,
also E and the trajectory of my q must be parallel right ?
Any helpful hints are really appreciated ;d
 
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Zbud said:

Homework Statement


I don't have any statement, i have to find the problem statement from the answers
a) E = Q / [ Eo* (0.020 m)^2 ]
b) a = E (1.60*10-19 C)/(1.67*10^-27 kg) = 2.0 *10^12 m/s^2
R=0,020 m
q=1.60*10-19 C
m=1.67*10^-27 kg

Homework Equations


F=ma => a=qE/m
Gauss Law Integral of (E,ds)=Q/Eo

The Attempt at a Solution


So given E constant everywhere i can assume that E*(integral of ds)=Q/Eo, so if this was a sphere it would be
4piR^2*E=Q/Eo so its out, my equation must be something like E*R^2=Q/Eo , so my area should be a a square ?,
also E and the trajectory of my q must be parallel right ?
Any helpful hints are really appreciated ;d

The Attempt at a Solution


So given E constant everywhere i can assume that E*(integral of ds)=Q/Eo, so if this was a sphere it would be
4piR^2*E=Q/Eo so its out, my equation must be something like E*R^2=Q/Eo , so my area should be a a square ?,
also E and the trajectory of my q must be parallel right ?
Any helpful hints are really appreciated ;d[/QUOTE]
You guess well, it might be the constant electric field of a a charged square-shaped plate, or rather the electric field between the plates of a planar capacitor with charge Q, where the plates of the capacitor are squares of sides 0.02 m.
The trajectory of the particle with charge q need not be parallel to E, but its acceleration has to be. What do you think the particle is?
 
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Given its mass and charge it must be certainly a proton, i have been thinking about the capacitor but truly as we have not talked about it once during the lectures so I wasnt keen on that idea,I'm grateful for your insight into this unusual problem
 
Zbud said:
Given its mass and charge it must be certainly a proton, i have been thinking about the capacitor but truly as we have not talked about it once during the lectures so I wasnt keen on that idea,I'm grateful for your insight into this unusual problem
That is correct, it must be a proton. Well done!
In case of single charged plate, the electric field is Q/(2Aε0), but it is Q/(Aε0) in case of a capacitor.
 

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