Homework Help: Thin Slab, Gaussian Pill Box.

1. Jan 25, 2013

SherlockOhms

1. The problem statement, all variables and given/known data
4.10.4 Thin Slab
Let some charge be uniformly distributed throughout the volume of a large planar slab of plastic of thickness d. The charge density is ρ. The mid-plane of the slab is the y-z plane.
(a) What is the electric field at a distance from the mid-plane when |x| < d/2?
(b) What is the electric field at a distance from the mid-plane when |x| > d/2?
[Hint: put part of your Gaussian surface where the electric field is zero.]

2. Relevant equations
E = σ/2ε0.

3. The attempt at a solution
So in vector form the electric field will be negative in the - x direction and positive in the +x direction. This isn't a homework assignment. They're course notes with sample problems. I don't really have any ideas how to go about this so any hints would be great like.

2. Jan 25, 2013

Staff: Mentor

Yes. Assuming the charge is positive.

Use the hint provided to choose a Gaussian surface.

3. Jan 25, 2013

SherlockOhms

Thanks. Yeah, I chose the surface to be a pill box. I'm just not sure how to make the Electric field a function of x though. I've worked through one asking you to calculate the electric field everywhere in space which seems to be a lot easier. Any suggestions on how to make E = σ/2ε0 a function of the distance form the centre,x? I think the Electric field at the centre is 0, is that correct?

4. Jan 25, 2013

Staff: Mentor

Use Gauss's law! (When you are within the slab, the total charge within your pillbox will be a function of x.)
That's the formula for the field outside an infinitely large charged disk. Forget it. You'll derive your own formula using Gauss's law.
Right! So that's one surface of your Gaussian pill box.

5. Jan 25, 2013

SherlockOhms

Thanks. So, this is what I've derived.....
Beginning with these equations:
ρ = Q / V.
qen / ε0 = ∫E.dA

The upper surface of the pill box i.e. the circular part has an area A. So ρ = Q/ A(x) where x is the distance from the centre. ∫E.dA = EA. EA = ρ (A)(x) / ε0
The A's the cancel and you get E = (ρ)(x) / ε0 which is a function of x. I think this is right anyway.

Could you tell me where d/2 comes into the calculation? I don't think I fully understand |x| < d/2 and |x| > -d/2. Thanks!

6. Jan 25, 2013

Staff: Mentor

Looks good to me.

Within the slab versus outside the slab. Once you're outside the slab, the enclosed charge remains constant with increasing x.

7. Jan 25, 2013

SherlockOhms

So I just calculate the Electric field from the centre to d/s and then seperately calculate the electric field from d/2 to x? Not sure if I understood what you meant. Sorry for the hassle man!

8. Jan 25, 2013

Staff: Mentor

You have found the field from the center to d/2 (when x < d/2). Now use the same approach (Gauss's law) to find the field when x > d/2. Hint: What's the total charge enclosed within the pill box when x > d/2?

9. Jan 25, 2013

SherlockOhms

I would've said it was E = ρ(d/2) / ε0 + ρ(x - d/2) / ε0 for the total flux contained inside the pill box from x = 0 to x > d/2 but I don't know. Ugh! I'm so confused!

10. Jan 25, 2013

SherlockOhms

The above is the same as E = ρ(x) / ε0 so that's not right. That was silly.

11. Jan 25, 2013

SherlockOhms

Why isn't it just E = ρ(x) / ε0 again except with x > d/2?

12. Jan 25, 2013

Staff: Mentor

Hint: The charge only goes to d/2. So when x > d/2, what's the volume of the charge contained in your Gaussian pill box?

13. Jan 25, 2013

SherlockOhms

The volume will be (d/2)(A).
So ρ = qen / (d)(A)/2 => ρdA / 2 = enclosed charge.
So won't E = ρd/2ε0?

14. Jan 25, 2013

Staff: Mentor

Good!

15. Jan 25, 2013

SherlockOhms

16. Jan 25, 2013

SherlockOhms

Am I right in saying that so?

17. Jan 25, 2013

Staff: Mentor

Correct.

Just like in the formula for the field from an infinite sheet of charge. The field is independent of distance.

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