Solving Gauss's Law Problem: Electric Fields in 4 Regions

  • Thread starter Thread starter awvvu
  • Start date Start date
  • Tags Tags
    Gauss's law Law
Click For Summary

Homework Help Overview

The problem involves determining the electric fields in four distinct regions surrounding an infinitely wide conductor and an infinitely wide plane of charge. The setup includes a conductor positioned parallel to a plane with a specified surface charge density.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of using Gaussian surfaces to analyze electric fields, particularly questioning the behavior of electric fields in different regions and how to apply Gauss's Law effectively. There is confusion regarding the calculation of electric fields in region 4 and the application of superposition principles.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the electric fields in the specified regions. Some guidance has been offered regarding the use of Gaussian surfaces, but there remains uncertainty about the reliability of the proposed answers and the assumptions involved.

Contextual Notes

Participants are considering the implications of the conductor's properties, such as whether it has surface charge density on both sides or if it can be treated as a single plane. There is also mention of the need for clarity on the electric field strengths in different regions.

awvvu
Messages
188
Reaction score
1

Homework Statement


An infinitely wide conductor (shaded thing on top) is parallel to and at a distance d from an infinitely wide plane of charge (line on bottom) with surface charge density delta. What are the electric fields [itex]E_1[/itex] to [itex]E_4[/itex] in the regions 1 to 4?

Code:
1
-------------------------
2  /  /  /  /  /  /  /  /
-------------------------
3
_________________________
4

The Attempt at a Solution


[itex]E_2 = 0[/itex], as the charges move to the surface.

A Gaussian pillbox that extends a bit out of the conductor has a net flux [itex]E A = \delta A / \epsilon_0[/itex], and so [itex]E = \delta / \epsilon_0[/itex]. So, [itex]E_1 = E[/itex] and [itex]E_3 = -E[/itex]. I'm getting a little confused about finding E_4. Is it 2E, since there's twice as much flux as there was through a pillbox just above the conductor?

Actually, I'm getting confused about the whole thing. When I have a Gaussian surface that encompasses two different electric fields I want to find, which one am I "getting" when I do the calculations?

Also, is there some way to check this by using superposition or something?
 
Last edited:
Physics news on Phys.org
Bump?
 
E1 and E3 at the surfaces? If that's the case you're right, they should be equal but opposite. What do you mean by a Gaussian pillbox? If the coductor is an infinetely wide plane, then what you need to do is construct a gaussian surface surrounding the plane. To do that for a plane that approaches infinite distances, just surround it by two parallel planes of area A that also approach infinite distances:
___________1 (plane)

-------------
////////////// (Conductor)
-------------
___________2 (plane)

In that sense, set E*2A = q/e... dq = (delta)*dA; therefore, E*2A = [(delta)*A]/e, and E = (delta)/2e, in which delta is the surface charge density
 
Last edited:
Gear300 said:
In that sense, set E*2A = q/e... dq = (delta)*dA; therefore, E*2A = [(delta)*A]/e, and E = (delta)/2e, in which delta is the surface charge density

What does that E represent? The region 4? So it has half the field strength as regions 1 and 3?
 
Well...dont know how reliable my answer is, but that's what I'm thinking. I could be wrong if the conductor has two sides with surface charge density...but if its thin enough to be a single plane, it should be alright.
 
Last edited:

Similar threads

Replies
11
Views
4K
Replies
4
Views
5K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
23
Views
5K
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
8K
Replies
6
Views
2K