Electric field within a very large, charged, insulating slab?

[automat]

Homework Statement

We have a charged, insulating slab of thickness 2L in the z direction, and very large (read infinite) in the x and y directions. The slab contains a charge per unit volume that varies linearly from -\rho_{0} to \rho_{0}, from one side of the slab to the other. Specifically, taking the slab to extend from -L to +L in the z direction, the charge densiy is \rho(z)=\rho_{0}z/L. Find the electric field everywhere inside the slab (magnitude and direction), and the potential difference between the two edges of the slab.

Homework Equations

Despite the planar symmetry, it seems that Gauss' law is not applicable due to the non-uniform charge density. That leaves me with "Coulomb's law",
\textbf{E}(\textbf{r})={1\over 4\pi\epsilon_{0}} \int {\rho(\textbf{r}\bf{'})\over \cursive{r}^{2}}{\bf\hat\cursive{r}}d\tau'
A bit of clarification on the vectors: \textbf r is the vector from the origin to the field point of interest. {\textbf r}{\bf '} is the vector from the origin to the location of the volume element (the charge). And \bf\cursive{r} is the vector from the volume element (charge) to the field point, {\bf\cursive{r}} = {\textbf r} - {\textbf r}{\bf '}.
Note the odd formatting:\cursive{r}\neq \tau. Rather, \tau' is the volume element. For reference, this is all from Chapter 2 of Griffiths' Intro to Electrodynamics.

The Attempt at a Solution

So far all my attempts have been in vain. I've managed to come up with E=0 in a number of different ways... all obviously wrong. My latest attempt has been to take "slices" of the slab in the x-y plane. An "infinite" charged plane has, using Gauss' law,
{\textbf E}={\sigma \over 2 \epsilon_{0}}\hat{\textbf n}.
From here, it should be possible to integrate from z=-L to +L, where the surface charge density, \sigma, would be vary as a function of z as in the volume charge density above. Unfortunately, I haven't been able to work it out.

Any ideas?

 

[dynamicsolo]

Re-EDIT: Now I'm sorry I erased what I had here and second-guessed myself. You do want to integrate the layers of charge \rho (z) dz as infinite sheets of charge, but take care about the field directions. The overall direction of the field in the slab is "downward", i.e., toward z = -L .

There will be some cancellation, but there should be none at z = 0.

My apologies for zig-zagging on this -- it's been a long week for me...

 

[automat]

If you integrate from z = -L to z = +L , you will get the correct answer of E = 0 at z = 0.

Since the problem asks for the field everywhere, you want to find a function for the electric field, E(z). So pick an arbitrary z_0 and perform the integration of the field from z = -L up to z = z_0 and separately from z = z_0 up to z = +L, then add the results. You should find that a certain amount of cancellation takes place, but there is total cancellation only for z_0 = 0. (The integration of infinite uniformly charged sheets is the right idea, but you have to apply it more carefully.)

Thanks for the input, dynamicsolo. This was also my opinion at first glance. Taking a step back, looking at the system in the most basic terms, we have a "slab" of positive charge in the region z>0 adjacent to a "slab" of negative charge in the region z<0. As in a parallel plate capacitor (although likely not as simple), there will be an electric field oriented in the -z direction at z=0. Outside the slab, it would seem that the electric field would be zero due to the fact that the electric field of a "sheet" of charge is independent of distance from the sheet (because a test charge "sees" a larger area of the surface of charge?). The regions of positive and negative charge (despite their non-uniform charge density) should cancel outside the slab.

Does this make sense, or am I misunderstanding the question?

 

[dynamicsolo]

I re-withdrew my posts: you are probably right about the field being maximal at z = 0. Here's another way to think about the charge arrangement: it is like being inside a nested set of infinitely many parallel-plate capacitors, with the charges on the plates getting weaker as the plates get closer to the mid-level, where you have zero charge and zero separation. The field in the slab will be "downward" toward z = -L and the infinite sum should converge.

This will also help with the second question. The field is zero, so the potential inside the slab is a constant, not necessarily zero. The nested capacitor approach may help with this problem.

Sorry about the needless confusion -- wrestling with physical situations can be that way sometimes (especially when you're sleep-deprived). [We generally strive to cause only necessary confusion... :) ]

 

[automat]

Argh, having typed the last response and ruminated on what I said, I think I have the solution. At any given z_0, the net electric field is due to the regions of the slab -L < z < -z_0 and z_0 < z < L. Any charge within the region |z|<|z_0| will cancel, giving the strongest electric field for z_0=0. I'll have to work it out on paper, but I think I finally see a light at the end of the tunnel.

 

[dynamicsolo]

The approach using nested capacitors also gives the same answer, since the field is zero outside the "capacitors" with separation less than z_0. The result for the slab's interior field does peak at
z = 0 and falls to zero at the surfaces of the slab, matching the exterior field there.

The nested capacitors also lead in a straightforward way to the result for the second question concerning the potential difference between the two surfaces.