Gauss's Law problem

1. Feb 18, 2008

awvvu

1. The problem statement, all variables and given/known data
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 $E_1$ to $E_4$ in the regions 1 to 4?

Code (Text):
1
-------------------------
2  /  /  /  /  /  /  /  /
-------------------------
3
_________________________
4
3. The attempt at a solution
$E_2 = 0$, as the charges move to the surface.

A Gaussian pillbox that extends a bit out of the conductor has a net flux $E A = \delta A / \epsilon_0$, and so $E = \delta / \epsilon_0$. So, $E_1 = E$ and $E_3 = -E$. 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: Feb 18, 2008
2. Feb 18, 2008

awvvu

Bump?

3. Feb 18, 2008

Gear300

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

Last edited: Feb 18, 2008
4. Feb 18, 2008

awvvu

What does that E represent? The region 4? So it has half the field strength as regions 1 and 3?

5. Feb 18, 2008

Gear300

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.

Last edited: Feb 18, 2008