Calculating the charge densities

[diredragon]

Homework Statement

Lets say you have a infinitely long surface with one side of length $L$ and a surface charge density $ρ_s$ and you need to transform that into a linear charge density $Q'$ so that you can represent the surface along some axis $y$ so the the surface is placed normal to the axis and goes from $y=0$ to $y=L$ how would you make that transition? Take the infinitely long dimension to be $h$.

Homework Equations

3. The Attempt at a Solution [/B]
I tried this line of thinking:
$$Q'=\frac{dQ}{dh}=\frac{ρ_sdS}{dh}=\frac{ρ_sdhdl}{dh}=ρ_sdl$$
Could this be correct and how would it be in a more complicated case? Is there a pattern here?

 

[haruspex]

I don't understand the arrangement. It is an infinite strip width L, and the y-axis is normal to the strip? So say it is L in the x-axis and infinite in z. But now you have y from 0 to L? And in what sense are you wanting to "represent" the surface?

 

[diredragon]

I don't understand the arrangement. It is an infinite strip width L, and the y-axis is normal to the strip? So say it is L in the x-axis and infinite in z. But now you have y from 0 to L? And in what sense are you wanting to "represent" the surface?

Sorry, i think i was a little confusing in choosing the words. Let's imagine it like this. You see the wall, the $y-axis$ is to the right and left of this wall, the $x-axis$ is up and down and the $z-axis$ is behind and in front of you. The surface is placed so that it is infinite up and down and of length $L$ from $y=0$ to $y=L$. Is now clearer?

 

[haruspex]

Sorry, i think i was a little confusing in choosing the words. Let's imagine it like this. You see the wall, the $y-axis$ is to the right and left of this wall, the $x-axis$ is up and down and the $z-axis$ is behind and in front of you. The surface is placed so that it is infinite up and down and of length $L$ from $y=0$ to $y=L$. Is now clearer?

Ok, but then what do you mean by the required transformation into a linear density? Where is this in the picture, and in what sense does it relresent the original charge? The fields are obviously not the same.

 

[diredragon]

So with the set up coordinate system you rotate it so that the x-axis looks to you and you see only the z ad the y. The surface looks like a line on the y-axis you right? So given the surface charge density can you transform it into a linear charge density along the y-axis bu cutting the surface into small elements of $dy$ (lines along the surface). This is the problem.

 

[haruspex]

So with the set up coordinate system you rotate it so that the x-axis looks to you and you see only the z ad the y. The surface looks like a line on the y-axis you right? So given the surface charge density can you transform it into a linear charge density along the y-axis bu cutting the surface into small elements of $dy$ (lines along the surface). This is the problem.

Ok. I think you are asking whether the field in the YZ plane can be simulated by a uniform charge along the Y axis.
The answer is no.
Close to the plane, the field is approximately constant (as for an infinite plane sheet). For a line of charge, the field gets very much stronger as you approach the line.