What is the electric field in a thin slab with uniform charge distribution?

[SherlockOhms]

Homework Statement

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.]

Homework Equations

E = σ/2ε₀.

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.

 

[Doc Al]

So in vector form the electric field will be negative in the - x direction and positive in the +x direction.

Yes. Assuming the charge is positive.

Use the hint provided to choose a Gaussian surface.

 

[SherlockOhms]

Yes. Assuming the charge is positive.

Use the hint provided to choose a Gaussian surface.

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?

 

[Doc Al]

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.

Use Gauss's law! (When you are within the slab, the total charge within your pillbox will be a function of x.)

Any suggestions on how to make E = σ/2ε₀ a function of the distance form the centre,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.

I think the Electric field at the centre is 0, is that correct?

Right! So that's one surface of your Gaussian pill box.

 

[SherlockOhms]

Thanks. So, this is what I've derived...
Beginning with these equations:
ρ = Q / V.
q_en / 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!

 

[Doc Al]

Thanks. So, this is what I've derived...
Beginning with these equations:
ρ = Q / V.
q_en / 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.

Looks good to me.

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.

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

 

[SherlockOhms]

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

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!

 

[Doc Al]

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?

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?

 

[SherlockOhms]

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?

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!

 

[SherlockOhms]

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

 

[SherlockOhms]

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

 

[Doc Al]

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!

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?

 

[SherlockOhms]

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?

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

 

[Doc Al]

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

Good!

 

[SherlockOhms]

So, x doesn't even come into the equation for part b?
Thanks. Sorry that you had to spell it out for me!

 

[SherlockOhms]

Am I right in saying that so?

 

[Doc Al]

So, x doesn't even come into the equation for part b?

Correct.

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