Gauss's Law problem

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  • #1
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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:

Answers and Replies

  • #2
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Bump?
 
  • #3
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E1 and E3 at the surfaces? If thats 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
 
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  • #4
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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?
 
  • #5
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Well...dont know how reliable my answer is, but thats 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.
 
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