# Strange thing happens with the surface polarisation charge density

1. May 15, 2014

### Pritam

This confusion is about the following problem:

Suppose the entire region below the plane z=0 is filled with uniform linear dielectric material of susceptibility $χ_e$. Calculate the force on a point charge q situated a distance d above the origin.

Now, this problem is done in Griffiths' Electrodynamics book( Example 4.8). Its clear how to do this problem. But problem comes from a point of surface charge density $\sigma_b$ on the surface of the dielectric.

Let me clear myself.

We need to know the surface bound charge. The sign of the charge will be negative. There will be no volume charge as the dielectric is considered to be linear and homogeneous. Now, $$\sigma_b=P.n=P_z=\epsilon_0 \chi_e E_z$$
Where $E_z$ is the z component of the total field just inside the dielectric at z=0. This field is due to in part to q and in part to the bound charge itself. The z component of the field due to the bound charge is $-\sigma_b/2 \epsilon_0$. From here we can get the total electric field $E_z$ and then the bound charge. Now using method of image we can get the potential at any point anywhere may be for z>0 or z<0.

But my confusion arises from the 2 in the denominator of $\sigma_b/2 \epsilon_0$.This 2 comes when the perpendicular component of the electric field above and below the surface is eual in magnitude. So that using Gauss's law we can have a 2 in the denominator. But in this case, the field due to the bound surface charge is not the same in the dielectric and in the vacuum because of different permittivity ($\epsilon$ and $\epsilon_0$ respectively). So we cant write a 2 in the denominator. Rather I think this surface charge will not be equally divided.

I hope I have made myself clear. I better request the reader to go through the Example 4.8 in Griffiths's Electrodynamics book.

Last edited: May 15, 2014
2. May 15, 2014

### Meir Achuz

If you look at the original derivation of the bound surface charge (which may not be that clear in Griffiths),
you will see that the field due to bound surface charge affects E just like free surface charge. It is only after this step in the derivation that permittivity is introduced. Thus the effect on E is just like that of free surface charge, hence the same factor of 1/2.

3. May 15, 2014

### Pritam

Yes I am getting that! But not really satisfied! Ok I understood this bound charge is working as a free charge! But why not to consider permittivity is not clear to me! I mean is there any exact physical reason behind this?

4. May 15, 2014

### Meir Achuz

Permittivity and the D field are introduced as mathematical constructs to simplify the effect of the induced dipoles.
The effect of the dipoles (in this case as a surface charge) can be implemented with no mention of epsilon or D.

5. May 16, 2014

### Pritam

I am really confused over this actually! Please dont mind.

When I started this and encountered this fact of 2 I went for to know why this 2 comes. There I saw the derivation and find that this derivation is done over a charged surface considered to be a charged sheet having two surfaces!!!! Now this was a point! So I started to consider a charged surface as actually "two charged surfaces" so that the total surface charge density $\sigma_b$ is distributed in those two surfaces in such a case that the upper surface contributes to the electric field above only and the lower surface contributes to the electric field below only! Now then I concluded, if the electric field is same in magnitude above and below, I can say that "two surfaces" carries same charge density! Otherwise How can they produce the same field?!!! Is there any misconception over this kind of way of thinking?

Now Please tell me, if my above clarification is valid, why I cant think about this same way and say as the electric field is different so charge density should also be different!

Ok! everything will be ok for me if I consider $D$ instead of $E$ then this whole clarification is valid for any case! Now I can say $D$ is same in both above and below (In the dielectric).

I think I am right! OR I am doing some great blunder with my way of understanding! Please I request you to consider my place and to think the way I am thinking so that I can be sure about my so far progress.
Nevertheless I was really happy to get your reply as that indeed was specifically telling a solution of the confusion.Thanks a lot.

6. May 18, 2014

### Meir Achuz

There is only one surface in this case, not two. The two surfaces would be for a conducting sheet of finite thickness, which is not this case. The one surface is the surface of the dielectric, with a surface charge \sigma. The factor of 1/2 enters because the E field can leave the gaussian pillbox at each face.