# Homework Help: Electric Field Over A Slab

1. Sep 13, 2010

### xxbigelxx

1. The problem statement, all variables and given/known data

Find the electric field produced by a uniform density plane of thickness d. This slab is very large in the x and y planes, extends from –d/2 to d/2 in z direction. Use Gauss’s Law. Note there are 3 regions.

2. Relevant equations

3. The attempt at a solution

I need help from the beginning. I think the E above and below the slab is sigma/(2 epsilon-naught). I am not sure if this is right, and what else to do. Thank you

2. Sep 13, 2010

### stevenb

As a hint, remember that electric field has a direction. In the answer you give above and below the slab, what is the direction? Doesn't symmetry require that if the field is pointing away from the slab on one side, it needs to be facing away from the slab on the other side too, and vice versa? If the fields are opposing (symmetrically) on each side of the slab, then what happens inside the slab?

3. Sep 13, 2010

### xxbigelxx

Well above the slab, the lines will be pointing upwards. Vice versa for the bottom. I'm not sure what to do inside the slab, that's my biggest problem.

4. Sep 13, 2010

### stevenb

OK, how did you get the answer outside the slab? With Gauss's Law, I assume? If so, just use Gauss's Law inside the slab as well. Make a pillbox surface with thickness in the range of 0 to d. As you go inside, the charge contained in the bounded surface will decrease, and the field will decrease as well.

Just set up Gauss's law and trust it.

5. Sep 13, 2010

### xxbigelxx

Right, I understand that conceptually, but I still don't completely understand how to work it out numerically. dA and Qenclosed are what are giving me trouble. I don't know what to write for the area of the pillbox inside of the slab. I know that 'd' has to be used somehow, but I am struggling on figuring out how.

6. Sep 13, 2010

### stevenb

Ah, OK, I see. Yes, d does not matter for that area part because the area on the sides of the pillbox is orthogonal to the D (and E in this case) vector. Where the d comes in is for calculating the enclosed charge. The slab has uniform charge density, so the total charge enclosed is:

$$Q=\rho A t$$

where Q is the charge, $$\rho$$ is the volume charge density, A is the area of one flat side of the pillbox, and t is the thickness of the pillbox. It's hard to explain without a picture. If it's still not clear, let me know and I'll post a picture.

7. Sep 13, 2010

### xxbigelxx

No I think understand that. So would E for that part be equal to rho*d/epsilon-naught?

8. Sep 13, 2010

### stevenb

No, the electric field will be a function of z within the slab. Another hint is that it will be zero at z=0. This makes sense from symmetry. There is also the boudary condition for the normal component of electric field, but remember that there is no surface charge density at the surface of the slab, since it has uniform volume charge density. A conductive disc would have all of it's charge on the surface, which would be a different problem. You don't really need the boundary condition to solve, but it's a good way to check your answer. So you know that the field is zero at z=0 and the field at the surface of the slab is continuous (since epsilon does not change at the interface).

This brings up another point. In your solution outside, you use sigma which is surface charge density. This solution comes from an infinitely thin slab. Remember that $$\sigma=\rho d$$ in the case of a finite thickness.

9. Sep 13, 2010

### xxbigelxx

Ok I get what you said in the second paragraph. Thank you.

For the first paragraph, I understand now what you are saying. However, I am not sure how to alter Gauss Law so E is a function of z. Would the value of Qenc be what changes? Would it become A*z*rho? This is where I think I am having most trouble.

10. Sep 13, 2010

### stevenb

Yes, that is basically it, but maybe it's better to use the absolute value of z or a different letter like t, at first. Later, you can change the |z| or t into z when you express the field. Personally, I would use a variable like t to represent the thickness of the pillbox, and then later try to correlate that with the coordinate z. This is more correct because the field has direction that changes based on the sign of z, but the charge enclosed is either positive or negative depending on the sign of rho. It's not too critical as long as you are correctly visualizing what the charge is and the direction of the field on each side of the xy-plane. The critical thing is that you get the final sign right to correctly specify the direction of the field.

EDIT: No wait, there is a problem there. There should be a factor of 2 introduced because the z=t/2. Sorry about that.

Last edited: Sep 13, 2010
11. Sep 13, 2010

### xxbigelxx

Ok this is what I have so far. How's it look? Thanks again.

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12. Sep 13, 2010

### stevenb

Hmmm. I may be misunderstanding your diagram and notation, but it looks like your solution is for the region outside the slab, not inside, as stated on your page.

The Gaussian (pillbox) surface needs to be inside the slab to find the field inside.

I think you may be unsure on how to make the correct pillbox. I'll draw a diagram and post it.

13. Sep 13, 2010

### xxbigelxx

The top half is for outside the slab, and the bottom is for inside. I hope that makes it more clear.

14. Sep 13, 2010

### stevenb

Ah, OK, that is the problem. That is actually a creative way to approach the problem which I never thought of before. Usually, one creates a pillbox symmetrically around the xy-plane, but you use a half pillbox with one face at xy-plane.

The reason why this works is that you can use symmetry to argue that the field must be zero at the xy-plane. Then you only need one surface of the pillbox. Thank you, I learned something new.

Anyway with either approach, the entire pillbox must be inside the slab to find the field inside the slab. Your approach finds the field outside, but just be careful about the sign of the field. Now let the upper surface of the pillbox go into the slab, and you will get a z dependence of enclosed charge.

15. Sep 13, 2010

### xxbigelxx

Hmm so would this be it? And Z goes to d/2.

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16. Sep 13, 2010

### stevenb

Well, I mostly agree with this, but I'm confused a little. Your diagram seems to show a pillbox that is symmetric about the xy-plane, but the analysis seems to be valid for a half-pillbox that rests on the xy-plane. I would just say to be sure that you are being consistent.

I really like your idea to use the half-pillbox. Honestly, I've never seen this in a book and never thought to do it this way myself. Maybe it's commonly done, but I've just never seen it. It seems perfectly valid to me though.

Your answer ends up being $$E_z=z\rho/\epsilon_o$$ for $$-d/2<z<d/2$$.

You can see that this meets the criteria to be $$E_z=0$$ at z=0, and $$E_z=d\rho/(2\epsilon_o)$$ at z=d/2, and $$E_z=-d\rho/(2\epsilon_o)$$ at z=-d/2.

So, basically I agree with you.

17. Sep 13, 2010

### xxbigelxx

Ok so I see that for inside the surface. Now would my final answer just state Ez=(what you have above) for inside, and rho*t/2epsilon-naught for outside? Is that the final form?

18. Sep 13, 2010

### stevenb

Oops, be careful here. There are two issues with what you said in this part.

1. The thickness is d and not t. The variable t is not part of the problem, but something you and I introduced for the pillbox (Gaussian surface).

2. Be careful of the sign of the field. A positive charged slab has field lines that are directed away from the slab. This means that on the negative z side of the xy-plane, the field is in the negative direction. The equation inside the slab automatically assigns the correct sign because z is explicitly in the formula (odd powers are OK, but not even powers of z).

19. Sep 13, 2010

### xxbigelxx

Ohhh right, your first point was a silly mistake on my part.

For the second point, I should just break it up into rho*d/2epsilon-naught for the top and -rho*d/2epsilon-naught for the bottom, right?

20. Sep 13, 2010

### stevenb

I believe this is correct. Please double check everything, I'm just checking this out by scribbling on a napkin, and I'm known to make silly mistakes even when I'm trying hard.

21. Sep 13, 2010

### xxbigelxx

I will try my best to double check with someone in my class tomorrow. I will upload my final work tomorrow to see what ya think. Thanks a lot for all your help, and hopefully we can wrap this up tomorrow!

22. Sep 13, 2010

### Mindscrape

23. Sep 14, 2010

### xxbigelxx

Ok I think I have finally got this. I will scan it as soon as I get to my apartment (couple hours), and upload it for you to see if you agree. Thanks again.

24. Sep 14, 2010

### xxbigelxx

Also, thanks Mindscrape. They did help.