Electric field inside a slab of with constant charge density.

[theBEAST]

Homework Statement

I have attached the question in a picture. I also attached a picture a drew as my attempt at understanding the question.

Homework Equations

flux = EA = Q_encl/ε_o

The Attempt at a Solution

So I am not sure how to interpret this question. Is the charge distributed uniformly? If it is then it can be very difficult to solve as you can see in my diagram the electric field goes every where.

Instead of a cylinder I was thinking of using a sphere as my surface... Does that even make sense?


Here is an answer I found online:
http://answers.yahoo.com/question/index?qid=20100124155001AAmSOpU

Do you guys agree? I notice the solution only used the top and bottom area of the gaussian surface...

 

[BruceW]

usually, the slab is taken to be infinitely (or at least, very) wide in the x and y planes. So using symmetry, what do you think would be the electric field component in the direction of the cylinder's curved side?

 

[theBEAST]

usually, the slab is taken to be infinitely (or at least, very) wide in the x and y planes. So using symmetry, what do you think would be the electric field component in the direction of the cylinder's curved side?

It seems like it should be zero, if it isn't it's extremely difficult to solve... But I don't know why...?

 

[BruceW]

Yeah, its zero. When I was doing undergraduate physics, the explanation they offered was that due to symmetry, it will be zero. If you think about the situation, it would be hard to imagine how the electric field in this direction would be non-zero, while still keeping the symmetry of the problem.

I guess to do this problem properly, you could use rigorous mathematical arguments about symmetry groups. But that is stuff beyond most physics undergraduates, so an intuitive explanation is OK.

 

[asteriatic]

I would first clarify the question by asking for more information about the slab. Is the charge density constant throughout the slab or only on the surfaces? Is the slab a perfect rectangular prism or does it have irregular edges? These details can greatly affect the solution.

Assuming the charge density is constant throughout the slab, we can use Gauss's Law to find the electric field inside the slab. As you correctly stated, the flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (εo). In this case, we can choose a Gaussian surface that is a rectangular prism with one side inside the slab and the other two sides parallel to the faces of the slab. This way, the electric field is perpendicular to the surface and the flux can be easily calculated.

Using the equation for flux, we can rewrite it as EA = Qenclosed/εo. Since the slab has a constant charge density, we can rewrite the enclosed charge as ρV, where ρ is the charge density and V is the volume of the Gaussian surface. Therefore, we can rewrite the equation as EA = ρV/εo, and solving for the electric field, we get E = ρ/εo. This means that the electric field inside the slab is directly proportional to the charge density.

As for the solution you found online, it seems to be assuming that the charge density is only on the top and bottom surfaces of the slab. In this case, the electric field inside the slab would be zero, but there would be a non-zero electric field outside the slab. This is because the electric field inside a conductor is always zero, so the electric field would only be present on the surface where the charge is located.

In conclusion, the solution to this problem depends on the assumptions made about the charge distribution inside the slab. As a scientist, it is important to clarify these details before attempting to solve the problem.