Electric Field and Uniformly Charged Planes (conceptual)

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Homework Help Overview

The discussion revolves around the electric field generated by uniformly charged planes, particularly in the context of capacitors with unequal charge distributions on the plates.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how to calculate the electric field between capacitor plates that have unequal charges and questions whether the field remains uniform in such a case. Other participants suggest considering the net electric field from arbitrary charge distributions and inquire about the correctness of the calculations presented.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the electric field due to charged plates. Some guidance has been offered regarding the approach to finding the net electric field, but no consensus has been reached on the correctness of the calculations provided.

Contextual Notes

Participants are working within the constraints of conceptual understanding and are questioning the assumptions related to charge distributions on capacitor plates.

dimpledur
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Homework Statement


I am pursuing conceptual help regarding electric field due to uniformly charged planes.


Homework Equations


[tex]E=\frac{\sigma}{2\epsilon}[/tex]

The Attempt at a Solution



I understand that a capictor has plates that are +/-Q. However, how would you calculate the elctric field between the plates if the plates are not equal in charge? Would the electric field between the plates still be uniform?

[tex]E=\frac{\sigma _1}{2\epsilon} + \frac{\sigma _2}{2\epsilon}[/tex]
 
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There's a simple way to find that out!
Consider an arbitrary charge distribution on the two plates, of the capacitor. Find out the expression for the net electric field, at any point on the interior of a plate. What should it be equal to?
 
The force on a test charge q, due to plate Q1, would be:

F= (kqQ1)/r^2

F=[kq(sigma1)A]/r^2

therefore, E=[kq(sigma)A]/(r^2*q) =[k(sigma1)A]/r^2

If there was an addition plate with charge Q2, we would have E=[k(sigma1)A]/r^2

Therefore, the electric field due to the two plates would be:
E=[k(sigma1)A]/r^2 + [k(sigma1)A]/r^2
 
Does that look correct?
 

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