1. Mar 12, 2006

### mewmew

For this problem I am giving the following:

An infinite slab of charge parallel to the yz plane whose density is given by:

p(x)= t, -b<x<b;
0, |x|>b;

Where t and b are constants.

And I am to find the electric field.

I am pretty confused on how to do this problem. I know that the electric field is in the x direction, and that is also where the normal to the surface is directed. So I thought it would be:

$$2 v E =\frac{q}{\epsilon}$$ where the q = $$\frac{p}{v}$$

Where v is the volume, similar to how you do an infinite sheet but I don't get the right answer. But then q would be 0 for outside of b and that is not right so I guess my method is completely wrong.

Last edited: Mar 12, 2006
2. Mar 12, 2006

### lightgrav

The big deal about Gauss is that the E-field piercing the surface
is written in terms of the charge enclosed.

A cylinder that cuts all the way through the slab encloses
the same amount of charge as it would if the Q were just a thin sheet.
So the E-field that pierces the end caps is the same as for a thin sheet...
I would write this as E dot A = (4 pi k) Q_enclosed , using k = 1/(4pi epsilon)
Q enclosed = rho * volume ... = t * (A h) ... what volume contains charge?

This isn't true for the interior of the slab, because the endcap's Area is the same,
but the cylinder encloses less charge.

3. Mar 12, 2006

### mewmew

Thanks, but I think I need to do it a different way. We are using Gauss' law in its basic form of $$\int\int_s E . n ds=\frac{q}{\epsilon}$$ I guess I am confused because for a infinite thin sheet you get $$\frac{\sigma}{2 \epsilon}$$ and it would seem outside of the slab this is what the field would be, but it isn't, it is $$\frac{\rho b}{\epsilon}$$

By looking on another website I think to have found it, I did the following integral:

$$\int_b^b \rho dx$$ which seems to give me the right answer but I don't really understand it. The bottom limit is -b but I couldn't get latex to do it right

Last edited: Mar 12, 2006
4. Mar 12, 2006

### lightgrav

the integral of the charge density IS the enclosed charge.
sigma is SURFACE charge density, which is rho * thickness.

If you think that your equation for Gauss is different from mine,
maybe you don't recognize an Area when they write it as ds instead of dA.

How much charge is enclosed by a cylinder that extends from -b to +b?
What's hard to understand?

5. Mar 13, 2006

### mewmew

Ya, I figured it out it just took me a while to remember all the stuff I learned last year. It wasn't that I thought what you had wasn't Gauss' law, I just wasn't sure how to put it into the form I needed. Thanks for the help.

Last edited: Mar 13, 2006