Gauss' Law & Surface Charge Density

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Homework Help Overview

The discussion revolves around a problem involving Gauss' Law and surface charge density related to a square copper plate placed in a uniform electric field. The original poster attempts to determine the charge density on each face of the plate and the total charge, given the dimensions and characteristics of the electric field.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the properties of conductors in electric fields, including the behavior of charge distribution on the surface and the implications of having zero electric field inside a conductor. Questions arise regarding the setup of equations and the interpretation of charge density.

Discussion Status

Participants are actively engaging in clarifying concepts related to electric fields and charge distribution in conductors. Some guidance has been offered regarding the behavior of charges in response to external fields, and there is an exploration of how to express the total electric field inside the conductor.

Contextual Notes

There is mention of the original poster's struggle to catch up due to missed time, which may affect their understanding of the problem setup and the relevant equations. The discussion also highlights the need for clarity on the application of Gauss' Law in this context.

Shigun
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Homework Statement


A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find:
(a) the charge density of each face of the plate and
(b) the total charge on each face


Homework Equations


[tex] E = \frac{\sigma}{2\epsilon {0}}[/tex]

[tex] \sigma = \frac{q}{a} [/tex]

The Attempt at a Solution


I missed about a week due to unforeseen circumstances, so I'm struggling with this to catch up, but so far:

Area, converted to meters, is 0.25 m^2. Electric field, E, converted to nano units would be 80x10^-9. My understanding was that [tex]\Phi _E = 2EA[/tex] so [tex]\Phi _E = 2 \times \frac{\sigma}{2\epsilon {0}} \times \frac{q}{a}[/tex] which comes out to [tex]\Phi _E = 2 \times \frac{\sigma}{8.854\times(10^-12)}\times\frac{q}{0.50\times0.50}[/tex], however I'm a bit confused on this. How do I find the value of q, and am I correct in my setup such that [tex]\sigma[/tex] is setup within the second fraction, over 8.854*10^-12?

Any help is greatly appreciated
 
Last edited:
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What do you know to be true about a metal (conductor) when it comes to electric fields?
 
Gokul43201 said:
What do you know to be true about a metal (conductor) when it comes to electric fields?

Charge is only ever on the surface, and the interior has a 0 charge. I also know, as mentioned in the equation (and from my reading) that a conductors equation is:

[tex]E=\frac{\sigma}{2\epsilon{0}}[/tex]

While a nonconductor is

[tex]E=\frac{\sigma}{\epsilon{0}}[/tex]

I also know that finding the surface density is supposed to involve using a Gaussian cylinder.

Correct?
 
Shigun said:
Charge is only ever on the surface, and the interior has a 0 charge.
Correct. Also there is no net electric field in the interior of a conductor. So when a conducting plate is placed in an external electric field, what needs to happen to the positive and negative charges in the conductor for this condition to be satisfied? (Hint: If you place freely moving positive and negative charges in an electric field, what happens to them?)
 
Gokul43201 said:
Correct. Also there is no net electric field in the interior of a conductor. So when a conducting plate is placed in an external electric field, what needs to happen to the positive and negative charges in the conductor for this condition to be satisfied? (Hint: If you place freely moving positive and negative charges in an electric field, what happens to them?)

I'm kind of confused by the wording, but my guess of what you are saying is referring to the charges separating out to each side of the plate, one being negative and one being positive. Yes?
 
Shigun said:
I'm kind of confused by the wording, but my guess of what you are saying is referring to the charges separating out to each side of the plate, one being negative and one being positive. Yes?
Yes. You can figure out which side will get what kind of charge based on the direction of the external field. But more importantly, each of these two sheets of charge will now produce their own electric fields. The negative sheet will produce field lines towards it and the positive sheet will produce field lines away from it. The field from each of these sheets will be given by the first equation in your opening post.

Inside the metallic plate, the total field is found by adding all three fields (including the applied external field). Can you now write an expression for the total field inside, in terms of the external field (call it Eo) and the charge density on each sheet (say, [itex]+\sigma,~-\sigma[/itex])? Draw a diagram to help yourself account for all the fields with their correct directions.
 

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