# Electric field of an infinite charged plate surrounded by different dielectric medium

1. Oct 7, 2011

### Barloud

Hi everybody,

I am getting confused by a very simple problem: lets take an infinite sheet electrically charged with a surface density σ. Because the sheet is infinite, we can first determine that the electrical field on both side contains only components normal to the sheet. If the plate is surrounded on both side by vacuum, we know from the symmetry principle that the components of the electrical field on both sides of the sheet have the same magnitude and opposite directions. From Gauss law, we can then determine that the magnitude is σ/2ε0.

Now, lets consider the case where the sheet is surrounded on one side by dielectric material 1 with relative permittivity ε1 and on the other by material 2 with relative permittivity ε2. I want to determine the characteristics of the electrical field E1 and E2 on both sides of the sheet.

My first reasoning approach was that, compared to the vacuum case, the electrical field is simply reduced by a factor ε1 in material 1 and by a factor ε2 in material 2 due to the induced polarization, so that we would get E1=σ/2ε0ε1 and E2=σ/2ε0ε2. Well, it cannot be true as it violates the continuity condition on the electric displacement (D2=D1+σ → ε2 E2=ε1E1+σ → σ=0), but it confuses me.

My second approach is to consider first the continuity condition of the electric displacement which tells that ε2E2=ε1E1+σ. But then I need a second equation to determine E1 and E2. Using some numerical software, I observed that E1=E2 which leads to E1=E2=σ/(ε0ε1+ε0ε2). However, the fact that E1=E2 is really counter-intuitive to me as the problem appears to be non symmetric.

Could someone give me some arguments on why the first approach is not correct? Is the second approach correct? If yes, how to justify the assumption E1=E2?

2. Oct 7, 2011

### syph3r

Re: Electric field of an infinite charged plate surrounded by different dielectric me

I think your confusion is arising because you've forgotten to treat the fields as vector quantities.

Let's place the infinite sheet with surface charge density σ in the XY plane. Region 1, above the XY plane (z>0), has permittivity $ε_{1}$, and region 2, below the XY plane (z<0), has permittivity $ε_{2}$.

Using Gauss's law, you can find that the electric flux density in region 1 is $D_{1}=\frac{σ}{2} \hat{z}$, and in region 2 is $D_{2}=-\frac{σ}{2} \hat{z}$

Now you can find the electric field in both regions using the constitutive relation, $D=εE$, which leads to:

$E_{1}=\frac{σ}{2ε_{1}} \hat{z}$ and $E_{2}=-\frac{σ}{2ε_{2}} \hat{z}$

This is the answer you developed using your "first reasoning", which is correct. You then say that it cannot be true, because it violates the continuity condition. But using the vector expressions for the flux you will find this is not the case, since the continuity condition in this case is:

$\hat{n} \cdot (D_{1}-D{_2})=σ$

Substituting the expressions for $D_{1}$ and $D_{2}$, and $\hat{n}=\hat{z}$ gives:

$\hat{z} \cdot ((\frac{σ}{2} \hat{z}) - (-\frac{σ}{2} \hat{z})) = σ$

$\hat{z} \cdot (\frac{σ}{2} \hat{z} + \frac{σ}{2} \hat{z}) = σ$

$\hat{z} \cdot (σ \hat{z}) = σ$

$σ = σ$

So, the continuity condition holds.

3. Oct 9, 2011

### Barloud

Re: Electric field of an infinite charged plate surrounded by different dielectric me

Thanks for your reply syph3r. Indeed, a basic calculus error ...but there is still something that I do not get.

Let's define $\stackrel{}{D}$$_{1}$=$\stackrel{}{D}$$_{1z}$$\hat{z}$ and $\stackrel{}{D}$$_{2}$=$\stackrel{}{D}$$_{2z}$$\hat{z}$. By applying Gauss' law on the interface, the only derived relation that I can figure out is $\stackrel{}{D}$$_{1z}$-$\stackrel{}{D}$$_{2z}$=σ. However, you mention in your reply that Gauss' law allows determining $D_{1z}=\frac{σ}{2}$ and $D_{2z}=-\frac{σ}{2}$. Do you assume in combination with Gauss law that the magnitudes of $\stackrel{}{D}$$_{1}$ and $\stackrel{}{D}$$_{2}$ are equal to obtain this result? If so, how is it justified?

The trigger of all this story are the results of numerical simulations made with a finite elements software (Comsol). Apparently, the software assumes that the magnitude of $\stackrel{}{E}$$_{1}$ and $\stackrel{}{E}$$_{2}$ are equal rather than $\stackrel{}{D}$$_{1}$ and $\stackrel{}{D}$$_{2}$. In this case, the obtained results for the electric displacement are $D_{1z}=\frac{ε_{1}σ}{ε_{1}+ε_{2}}$ and $D_{2z}=-\frac{ε_{2}σ}{ε_{1}+ε_{2}}$. The continuity of the electric displacement is also respected with these expressions.

This makes me crazy, who's wrong who's right?

4. Oct 10, 2011

### DiracRules

Re: Electric field of an infinite charged plate surrounded by different dielectric me

I think I got the answer.

Think it this way: if you had to do this without the dielectric material, you would have applied Gauss getting $\vec{E_1}-\vec{E_2}=\frac{\sigma}{\epsilon_0}$. The other equation comes from symmetry: being in vacuum, you can "guess" that both the fields are equal in magnitude: $|\vec{E_1}|=|\vec{E_2}|$. So you get that $E_1=\frac{\sigma}{2\epsilon_0},\,\,E_2=-\frac{\sigma}{2\epsilon_0}$

Now, let's reason the same way with the dielectric: using Gauss, you get $\vec{D_1}-\vec{D_2}=\sigma$. Now, since we are using the electric displacement vector, the situation IS symmetric, because we take count of the "asymmetries" of the space domain in the equation $\vec{D}=\epsilon\vec{E}$. So, the situation is symmetric again, $|\vec{D_1}|=|\vec{D_2}|$
So, linking the two equations, we get $\vec{D_1}=-\vec{D_2}=\frac{\sigma}{2}\Rightarrow \vec{E_1}=\frac{\sigma}{2\epsilon_1},\,\vec{E_2}=-\frac{\sigma}{2\epsilon_2}$

5. Oct 10, 2011

### Born2bwire

Re: Electric field of an infinite charged plate surrounded by different dielectric me

Probably the best way to look at this problem is as a limiting case. Take for example the problem where we have a half-space with the interface in the x-y plane at z=0 and a sheet of charge at some z = h where h>0. Now we can use image theory to solve for the image charge inside the second region and then take the limit of h to 0.

So, let's assume that the top region is vacuum and the bottom is a region with permittivity of \epsilon. Now, we can solve for the image charge of a point charge, this is done in Griffiths example 4.8 on page 189 in my third edition, and we find that the image charge is

$$q_b = -\frac{\epsilon-\epsilon_0}{\epsilon+\epsilon_0}$$
Now, by extension of superposition we can therefore say that for a sheet of charge that the image charge will scale by the same factor. Now, if we take the limit of h to zero, then the resulting charge sheet will be the summation of the original charge density and the image charge density,

$$\sigma_{total} = \sigma \left( 1 + -\frac{\epsilon-\epsilon_0}{\epsilon+\epsilon_0} \right) = \frac{2\sigma\epsilon_0}{\epsilon+\epsilon_0}$$
Thus, the resulting electric field above the charge sheet is

$$E_1 = \frac{\sigma}{\epsilon+\epsilon_0}$$
Now let's find the electric field below the charge sheet. Now we know from the boundary conditions of the original problem that

$$\epsilon_0E_1+\epsilon E_2 = \sigma$$
We substitute the above in and we find that

$$E_2 = \frac{\sigma}{\epsilon+\epsilon_0}$$
This would agree with the simulation results given in the OP. Note, the above assumes that E_1 points in the positive z direction and E_2 points in the negative z direction. Thus,

$$\mathbf{E}_1 = \frac{\sigma}{\epsilon+\epsilon_0} \hat{z}$$
$$\mathbf{E}_2 = -\frac{\sigma}{\epsilon+\epsilon_0} \hat{z}$$
My intuition would be that if we assume that the upper half-space is a dielectric as opposed to a vacuum then we would substitute in the appropriate permittivity in place of the vacuum permittivity in the above equations.

EDIT: The point here being is that we have not made any additional assumptions except from what we can already gather from symmetry and boundary conditions. We did not assume that the magnitudes of the electric fields had to be equal.

Last edited: Oct 10, 2011
6. Oct 10, 2011

### DiracRules

Re: Electric field of an infinite charged plate surrounded by different dielectric me

Premising that I am not 100% sure of my reasoning in the previous post, I'd like to point out that my "educated guess" that the magnitudes of the electric displacement vectors have to be equal comes out from the symmetry of the system, once you look at it not according to E (in which case you would see the asymmetry due to the different dielectrics) but according to D (in which case there is symmetry because all the "things" linked to the dielectric are condensed in D itself)

7. Oct 10, 2011

### Born2bwire

Re: Electric field of an infinite charged plate surrounded by different dielectric me

There isn't anything in here that would suggest to me that there should be an expected symmetry about the boundary. Any idea about how the electric displacement behaves is ruined by the free charge that is impressed on the surface. All we can say is that the electric displacement will be discontinuous by the free charge density. Take for example if you have the charge sheet displaced above the boundary. We know that the displacement field is continuous below and above the sheet because there are no other free charges. But that does not mean that the displacement field across the sheet is symmetric.

Let's say we have the following geometry:

Region 1

-------------------------------- σ

Region 2

----------------------------------- Boundary

Region 3

Where regions 1 and 2 are vacuum and region 3 is \epsilon. Thus, the electric and electric displacement fields are:

$$\mathbf{E}_1 = \frac{\sigma}{\epsilon+\epsilon_0} \hat{z}$$
$$\mathbf{E}_2 = -\frac{\sigma\epsilon}{\epsilon_0 \left( \epsilon+\epsilon_0 \right) } \hat{z}$$
$$\mathbf{E}_3 = -\frac{\sigma}{\epsilon+\epsilon_0} \hat{z}$$
$$\mathbf{D}_1 = \frac{\sigma\epsilon_0}{\epsilon+\epsilon_0} \hat{z}$$
$$\mathbf{D}_2 = -\frac{\sigma\epsilon}{\epsilon+\epsilon_0} \hat{z}$$
$$\mathbf{D}_3 = -\frac{\sigma\epsilon}{\epsilon+\epsilon_0} \hat{z}$$

The displacement field is still continuous across the interface but it is not symmetric across the plate. Now we know that the general behavior of the above is as expected for a few things. First, the boundary conditions due to the plate leads us to expect a discontinous displacement field across the plate but a continuous one across the boundary. However, the plate is going to induce a polarization inside the dielectric half-space which will result in a polarization charge (boundary charge) along the surface of the half-space due to the discontinuity of the bulk dielectric. This means that we expect that the electric field to be discontinuous across both the charge sheet AND the boundary due to the free and bound charges. If we were to assume that the displacement field is symmetric about the charge sheet in this problem then it would appear that the boundary charges were not being induced since the electric field in region 1 and region 2 would be symmetric.

Last edited: Oct 10, 2011
8. Oct 11, 2011

### Barloud

Re: Electric field of an infinite charged plate surrounded by different dielectric me

Thank you all for your answers. I am happy to see that it was not such a trivial problem at the end :)

I finally got the same answer than Born2bwire by introducing the electrical potential in the problem and considering the continuity of this potential across the interface.