# Homework Help: Conducting slab between two plates

1. Sep 26, 2009

### TwoTruths

1. The problem statement, all variables and given/known data
An uncharged conducting slab is placed between two infinitesimally thin insulating plates with surface charge densities $$\sigma_1$$ and $$\sigma_2$$, respectively. Describe what fields are induced in the conducting slab and their magnitudes.

2. Relevant equations
None?

3. The attempt at a solution
This is the answer I was given with no explanation. I would like someone to explain this, please.

" When the conducting slab is inserted in these respective ﬁelds, the induced charges on its surfaces, say $$\sigma_i$$ on the top surface and $$-\sigma_i$$ on the bottom surface are such that the ﬁeld these charges produce inside the slab (outside the induced ﬁeld is zero)

$$E_i = \frac{-\sigma_i}{\epsilon_0}$$"

My two questions: why is the induced field outside = 0? Why is the induced field inside like above?

2. Sep 26, 2009

### kuruman

What is the electric field in the space between the slabs before the conducting slab is inserted? If you can answer this question, then you need to realize that the charge distribution on the conductor faces needs to be such that this filed is cancelled exactly once the conductor is inserted.

3. Sep 26, 2009

### TwoTruths

I know these must be true, and I've tried to solve the problem myself with these ideas in mind. Could you explain further why these ideas dictate that the induced field outside is 0, and the induced field inside is the negative charge distribution / electric constant? I understand why the two faces of the slab are the same magnitude and how to calculate the field between the two plates without the slab there (Gauss's Law).

4. Sep 26, 2009

### kuruman

Consider the region of space in which the conductor is gong to be placed

1. The electric field in that space is initially not zero.
2. When you put the conductor in, the electric field becomes zero because the electric filed in a conductor is always zero.
3. What kind of charge distribution do you need on the faces of the conductor so that the field that was initially there becomes zero?

Note that outside the conducting slab the field is not zero. It is whatever it was before the slab was put in place.

5. Sep 26, 2009

### TwoTruths

I will show my steps and perhaps I can make my questions more clear. I understand they are quite vague right now. My apologies.

First, the space that is going to be occupied by the conductor has an electric field $$E = \frac{\sigma_0-\sigma_1}{2\epsilon_0}$$. When the conductor occupies this space, the induced field inside must exactly cancel this. However, I do not understand how to calculate this induced inside electric field. Assume the two surfaces of the conductor have induced charge $$\sigma_i$$ and $$-\sigma_i$$. Could you take me through the steps to find $$E_\textbr{induced inside}$$?

Last edited: Sep 26, 2009
6. Sep 26, 2009

### kuruman

OK. First things first.

1. Do you understand that the electric field inside the conducting slab has to be zero?
2. Do you understand that insulating plate 1 creates an electric field that is σ/(2ε0) everywhere in space in the region outside it?
3. Now consider a region of space to the right of an insulating plate carrying charge density σ. Do you understand that to cancel the electric field beyond some distance x to the right of of the insulating plate, you need to place charge density -σ in a plane at x?

I will stop here and wait for your answers. Let me know if there is a "No" to any of the questions and which one it is.

7. Sep 26, 2009

### TwoTruths

1. Yes, as evidenced by "When the conductor occupies this space, the induced field inside must exactly cancel this."

2. Yes, as evidenced by the equation (fixed) that I found for the electric field between two plates.

3. No. You have not described the orientation of the plate you are thinking about. If the plate is oriented left to right, I don't understand what you're talking about there. If the plate is oriented up to down, then yes, and the equation to describe the electric field beyond that new plate is $$E = \frac{\sigma_0 + \sigma_1}{2\epsilon_0}$$.

What I expect the induced field inside the conductor (not the net field) to be is $$E = \frac{\sigma}{\epsilon_0}$$. This is described by the equation above in #3. However, my friend says that the induced field inside is the negative of what I expect. Is it that he's accounting for the fact that the induced field must exactly cancel the original field in his calculation of the induced field, rather than somewhere else? Or is it that the induced field is always induced opposite from the original field, so he places the negative in there? Or is there something else I'm missing?

8. Sep 27, 2009

### kuruman

OK, let's agree that we start with two parallel insulating plates separated by some distance. They are oriented in a horizontal plane. The top plate carries charge density σ1 and the bottom plate σ2. Region I is above the top plate, Region II is between the plates and Region III below the bottom plate. "Up" is positive and "down" is negative for the field directions.

In the absence of the conducting slab the electric field in the three regions is

EI=(σ12)/(2ε0)
EII=(-σ12)/(2ε0)
EI=(-σ12)/(2ε0)

I now place the conducting slab in Region II with its faces parallel to the plates. In all regions where there is vacuum, the electric field is the same as before. In the space occupied by the conductor in Region II the electric field is zero.

I construct a cylindrical Gaussian "pillbox" such that one face is in Region I and the other inside the conductor.

The electric flux through the top face is
Φ1=EIA=(σ12)A/(2ε0)
The electric flux through the bottom face is zero because the electric filed is zero there. The total electric flux is
ΦE=EIA+0=(σ12)A/(2ε0)
Let σtop=the surface charge density on the top face of the conducting slab.
The charged enclosed by the Gaussian surface is
qencl1A+σtopA

Gauss' Law says
ΦE=qencl0 or
12)A/(2ε0)=(σ1A+σtopA)/ε0

Solve to get
σtop=(-σ12)/2

If you construct another pillbox with one face in Region III and the other in the conductor and follow the same procedure, you will get
σbottom=(σ12)/2

This should make it clear, I hope.