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

In summary, Gauss' law is not applicable to this problem because of the non-uniform charge density. Coulomb's law is applicable, but it is complicated to integrate because of the field directions.
  • #1
automat
3
0

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 [tex]-\rho_{0}[/tex] to [tex]\rho_{0}[/tex], 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 [tex]\rho(z)=\rho_{0}z/L[/tex]. 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",
[tex]\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'[/tex]
A bit of clarification on the vectors: [tex] \textbf r [/tex] is the vector from the origin to the field point of interest. [tex]{\textbf r}{\bf '}[/tex] is the vector from the origin to the location of the volume element (the charge). And [tex]\bf\cursive{r}[/tex] is the vector from the volume element (charge) to the field point, [tex]{\bf\cursive{r}} = {\textbf r} - {\textbf r}{\bf '}[/tex].
Note the odd formatting:[tex]\cursive{r}\neq \tau[/tex]. Rather, [tex]\tau'[/tex] 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,
[tex]{\textbf E}={\sigma \over 2 \epsilon_{0}}\hat{\textbf n}[/tex].
From here, it should be possible to integrate from z=-L to +L, where the surface charge density, [tex]\sigma[/tex], 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?
 
Physics news on Phys.org
  • #2
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 [tex]\rho (z) dz[/tex] 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...
 
Last edited:
  • #3
dynamicsolo said:
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?
 
  • #4
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... :) ]
 
Last edited:
  • #5
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.
 
  • #6
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.
 

1. What is an electric field?

An electric field is a physical quantity that describes the influence that an electric charge has on other charges in its vicinity. It is represented by a vector and is measured in units of force per unit charge (newtons per coulomb).

2. What is a charged insulating slab?

A charged insulating slab is a large, flat object made of an insulating material (such as plastic or rubber) that has a net electric charge. This charge is evenly distributed throughout the slab's surface and does not move freely like in a conductor.

3. How does the electric field within a charged insulating slab differ from that of a conductor?

In a conductor, the electric field is zero inside the material due to the presence of free electrons that can easily move and neutralize any external electric field. In a charged insulating slab, the electric field is non-zero and varies throughout the material, as the charges are unable to move freely.

4. What factors affect the strength of the electric field within a charged insulating slab?

The strength of the electric field within a charged insulating slab depends on the amount and distribution of charge on the slab, as well as the distance from the slab's surface. It also depends on the permittivity of the material, which describes how easily the material can be polarized by an electric field.

5. How is the electric field within a charged insulating slab calculated?

The electric field within a charged insulating slab can be calculated using Coulomb's law, which states that the electric field at a point is equal to the force per unit charge at that point. This can also be calculated using the electric potential, which is a measure of the electric potential energy per unit charge at a point.

Similar threads

  • Advanced Physics Homework Help
Replies
13
Views
2K
Replies
11
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
26
Views
4K
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Replies
1
Views
1K
Replies
5
Views
844
  • Advanced Physics Homework Help
Replies
16
Views
3K
Back
Top