Gauss's Law - Insulating Material

[forestmine]

Gauss's Law -- Insulating Material

Homework Statement

A slab of insulating material of uniform thickness d, lying between -{d}/{2} to +{d}/{2} along the x axis, extends infinitely in the y and z directions. The slab has a uniform charge density rho. The electric field is zero in the middle of the slab, at x=0.

What is E_out, the magnitude of the electric field outside the slab?

As implied by the fact that E_out is not given as a function of x, this magnitude is constant everywhere outside the slab, not just at the surface.

Homework Equations

Flux = E*A

ρ=q/V

The Attempt at a Solution

I'm finding myself really confused about this problem. If I'm visualizing it correctly, if we use Gaussian surfaces, then the charge enclosed by the surface is a line charge which extends infinitely along x=0. Since the electric field is 0 at that point, the angle the field makes with the surface of the slab is simply 90 degrees. But already right there I'm confused...if the electric field is 0 at that point, can a charge exist at that point?

At that point, when asked, "What is q, the charge enclosed by the Gaussian surface," I get q=ρAd/2 coming from the fact that the charge density in this case equals the charge over a volume.

From here, given a charge, I don't understand how to get to the electric field, presumably through calculating the flux.

I'm really lost, and I'm having a hard time connecting these concepts.

I'm not looking for an answer...I just don't think I'm thinking of any of this correctly. Any help would be greatly appreciated! Thank you!

 

[SammyS]

Homework Statement

A slab of insulating material of uniform thickness d, lying between -{d}/{2} to +{d}/{2} along the x axis, extends infinitely in the y and z directions. The slab has a uniform charge density rho. The electric field is zero in the middle of the slab, at x=0.

What is E_out, the magnitude of the electric field outside the slab?

As implied by the fact that E_out is not given as a function of x, this magnitude is constant everywhere outside the slab, not just at the surface.

Homework Equations

Flux = E*A

ρ=q/V

The Attempt at a Solution

I'm finding myself really confused about this problem. If I'm visualizing it correctly, if we use Gaussian surfaces, then the charge enclosed by the surface is a line charge which extends infinitely along x=0. Since the electric field is 0 at that point, the angle the field makes with the surface of the slab is simply 90 degrees. But already right there I'm confused...if the electric field is 0 at that point, can a charge exist at that point?

At that point, when asked, "What is q, the charge enclosed by the Gaussian surface," I get q=ρAd/2 coming from the fact that the charge density in this case equals the charge over a volume.

From here, given a charge, I don't understand how to get to the electric field, presumably through calculating the flux.

I'm really lost, and I'm having a hard time connecting these concepts.

I'm not looking for an answer...I just don't think I'm thinking of any of this correctly. Any help would be greatly appreciated! Thank you!

This is not a line charge.

This is a thick slab of thickness, d, with infinite width & length.

 

[forestmine]

In a previous part of the problem, we determine that the angle the field makes with the surface of the slab is 90 degrees. And if the field is 0 at x=0, I don't understand what other kind of charge this could be?

 

[SammyS]

In a previous part of the problem, we determine that the angle the field makes with the surface of the slab is 90 degrees. And if the field is 0 at x=0, I don't understand what other kind of charge this could be?

It's zero at x=0 because, that's the center of the slab in the x direction. (It's infinite in the y & z directions.)

 

[forestmine]

In that case, it's a field, but there is no charge? I don't understand at all.

 

[SammyS]

In that case, it's a field, but there is no charge? I don't understand at all.

What ever could possibly lead you this conclusion?Let's describe the slab in somewhat different terms.

One surface of the slab is at x = d/2. This surface, call it the front surface, is a plane which is parallel to the yz-plane. The other surface, call it the rear surface, of the slab is at x = -d/2. This surface is also parallel to the yz-plane.

Mid way between the front and back surfaces, at x = 0, is the yz-plane.

 

[Tsunoyukami]

It might help if you were to consider the plane that divides the surface into two (the yz-plane) at x = 0. If you think about this material as having a uniform charge density that means it has an equal amount of charge on either side of this imaginary central "wall". Because there is an equal amount of charge on either side the electric field at x = 0 along this line msut be zero because the charge on one side of the wall is pulling it one way (say right) while the charge on the other side of the wall is pulling it in the other direction (say left) with an equal magnitude. This means that the electric field at a central point between the two "walls" (boundaries of the surfaces) must be 0.

 

[forestmine]

Ok, I think I'm with you guys up until that point, then. In terms of it being a gaussian surface, however, I don't understand where the enclosed charge lies.

For that matter, where do I go from here in order to determine E_out in terms of rho, d, and epsilon. I'm thinking of the field in terms of E=kq/r^2. Q in this case would be Apd/2. If I plug that into the field equation, I get E=(Apd/2)/4(pi)epsilon(d/2)^2

 

[SammyS]

Ok, I think I'm with you guys up until that point, then.
Up until what point ?

...In terms of it being a gaussian surface, however, I don't understand where the enclosed charge lies.
There's no enclosed charge until a closed surface is defined.

For that matter, where do I go from here in order to determine E_out in terms of rho, d, and epsilon. I'm thinking of the field in terms of E=kq/r^2.
Don't try doing this with Coulomb's Law.

Use Gauss's Law.

Q in this case would be Apd/2. If I plug that into the field equation, I get E=(Apd/2)/4(pi)epsilon(d/2)^2

Comments above.

Furthermore,

I suggest Gauss's Law, using a cylinder whose sides are parallel to the x-axis. One end of the cylinder at x=0. The other end at any value of x that's greater than d/2 .​

From symmetry you can argue that any E-field is parallel to the x-axis.

 

[forestmine]

I've given this more thought, and I realized I'm still not following.

The problem presents a cube, with ends at x=d/2, and x=-d/2. If there are no charges present that we ought to be concerned with, is it safe to assume (for purely conceptual reasons) that there lies a charge somewhere on either side of d/2 and -d/2 that is causing the electric fields? That being said, I still can't conceptualize why the electric field is 0 at x=0, and suddenly constant on either side of that point.

 

[SammyS]

I've given this more thought, and I realized I'm still not following.

The problem presents a cube, with ends at x=d/2, and x=-d/2.

Have you been instructed to use such a cube? This is the first time I've seen mention of a cube in this thread.

I think it makes more sense to use a cube with one end at x=0, and the other at x = d/2 or beyond that. … or the cylinder I suggested previously.

The idea that E=0 at x=0 comes mainly by considering the symmetry of the situation. Symmetry also suggests that the electric field is in the x direction.

If there are no charges present that we ought to be concerned with, is it safe to assume (for purely conceptual reasons) that there lies a charge somewhere on either side of d/2 and -d/2 that is causing the electric fields?

The charge density, ρ, of the slab is uniform, so there is charge present for -d/2 ≤ x ≤ d/2 . That charge is all that's producing the E field.

How much charge is contained in any of the above cubes, or cylinder?

That being said, I still can't conceptualize why the electric field is 0 at x=0, and suddenly constant on either side of that point.

]The charge density, ρ, of the slab is uniform, so there is charge present for -d/2 ≤ x ≤ d/2 . So there is as much charge to the left of x=0 as there is to the right of x=0 . Thus the E-field at x=0 is 0.

The field is constant outside of the slab, because the there is no charge outside of the slab, and because the slab is infinite.

 

[forestmine]

I got the answer, and I think I've come to terms with it conceptually.

The cube is a suggestion later in the problem, and like you said, you're instructed to place one end at x=0 and the other at x=d/2. By doing so, you've now established a gaussian surface, and so we can solve for a charge, which in this case, given the charge density, comes to q=A(rho)d/2. Since electric flux = EA = q/(epsilon), solving for E gives us q/A(epsilon) and substituting A from the previous equation gives us the correct answer.

I think my hang up was the charge, and where its location was. But if I'm understanding correctly, when we're dealing with electric fields and no given charge, we ought to assume a Gaussian surface in order to find said charge which supplies the given field.

Thank you so much for the help! And especially for your patience. :)

 

[forestmine]

I got the answer, and I think I've come to terms with it conceptually.

The cube is a suggestion later in the problem, and like you said, you're instructed to place one end at x=0 and the other at x=d/2. By doing so, you've now established a gaussian surface, and so we can solve for a charge, which in this case, given the charge density, comes to q=A(rho)d/2. Since electric flux = EA = q/(epsilon), solving for E gives us q/A(epsilon) and substituting A from the previous equation gives us the correct answer.

I think my hang up was the charge, and where its location was. But if I'm understanding correctly, when we're dealing with electric fields and no given charge, we ought to assume a Gaussian surface in order to find said charge which supplies the given field.

Thank you so much for the help! And especially for your patience. :)

 

[SammyS]

I got the answer, and I think I've come to terms with it conceptually.

The cube is a suggestion later in the problem, and like you said, you're instructed to place one end at x=0 and the other at x=d/2. By doing so, you've now established a gaussian surface, and so we can solve for a charge, which in this case, given the charge density, comes to q=A(rho)d/2. Since electric flux = EA = q/(epsilon), solving for E gives us q/A(epsilon) and substituting A from the previous equation gives us the correct answer.

I think my hang up was the charge, and where its location was. But if I'm understanding correctly, when we're dealing with electric fields and no given charge, we ought to assume a Gaussian surface in order to find said charge which supplies the given field.

Thank you so much for the help! And especially for your patience. :)

You're welcome.

I'm glad to see that you didn't give up on the problem.