Gauss' law for a physical capacitor with finite thickness plates

[etotheipi]

In derivations of capacitance it is standard to consider two oppositely charged, infinitely thin sheets. If we construct a Gaussian cylinder across one sheet, we obtain $E_{1} = \frac{\sigma}{2\epsilon_{0}}$ for one sheet, and then we can superpose this field with that from the other at an arbitrary point between the plates to obtain the result $E = \frac{\sigma}{\epsilon_{0}}$.

However, a (slightly...) better model of a capacitor might be one with plates which are still infinite, but now have some thickness. One plate has a charge $Q$ and the other $-Q$. Since the electric field is zero inside the conducting plates, the electric field strength at the surface of the plate is now just $\frac{\sigma}{\epsilon_{0}}$ (as is the case with any other conductor) since we only have to count flux through one end of the cylinder which contains a small section of the surface. Likewise, the electric field strength at any point outside the other plate due to the other plate also has to be $\frac{\sigma}{\epsilon_{0}}$. When we superpose the two electric fields, the total electric field strength between the plates is $\frac{2\sigma}{\epsilon_{0}}$,... but that can't be right!

I can't however see where the second approach fails. The plates are still infinite so due to the symmetry of the infinite plane the field must still always be orthogonal to the sheet, and so the field strength at an arbitrary point on that side of the sheet must be the same as that at the surface.

The only resolution I can think of is that the field strength inside the plate is non-zero, though this goes against standard theory for conductors (since then the charges would just move until an equipotential volume was again obtained). Any guidance would be appreciated, thank you!

 

[Dale]

The thickness of the plates is not really relevant. The charge is a thin surface charge even for a thick plate. The derivation will be the same as for the thin plate.

 

[kuruman]

When we superpose the two electric fields, the total electric field strength between the plates is 2σϵ0\frac{2\sigma}{\epsilon_{0}},... but that can't be right!

It's not right. What do you mean by "when we superimpose"? What counts for the flux is only the charge that is enclosed by the Gaussian surface. So if one face of the Gaussian pillbox is through the conductor, the flux through the other face is (a) $EA$ if the other face is between the plates; (b) zero if it is in the other plate or beyond it on the other side.

 

[etotheipi]

The thickness of the plates is not really relevant. The charge is a thin surface charge even for a thick plate. The derivation will be the same as for the thin plate.

Right, in that case we've got a layer of surface charge on the inner face of the conductor and nothing on the outer face, so I suppose that means that we do have an electric field within the conductor?

It's not right. What do you mean by "when we superimpose"? What counts for the flux is only the charge that is enclosed by the Gaussian surface. So if one face of the Gaussian pillbox is through the conductor, the flux through the other face is (a) $EA$ if the other face is between the plates; (b) zero if it is in the other plate or beyond it on the other side.

I guess that works if we consider the two sheets together, the way I did it was to consider a single sheet in isolation and then use the principle of superposition to find the electric field at some point between them.

 

[hutchphd]

For a finitely thick plate the charge distributes on two faces. For the same total Q the charge density σ will be cut in half relative to "thin" plate. That simple I think.

 

[kuruman]

I guess that works if we consider the two sheets together, the way I did it was to consider a single sheet in isolation and then use the principle of superposition to find the electric field at some point between them.

It would be a good exercise to consider the following problem: Start with two infinite conducting sheets infinitely far apart carrying surface charge densities $\sigma_1$ and $\sigma_2$. Bring them close to each other to form a parallel plate capacitor. Find the surface charge distribution on the two surfaces of each plate.

 

[etotheipi]

It would be a good exercise to consider the following problem: Start with two infinite conducting sheets infinitely far apart carrying surface charge densities $\sigma_1$ and $\sigma_2$. Bring them close to each other to form a parallel plate capacitor. Find the surface charge distribution on the two surfaces of each plate.

I've scribbled a few things down and it seems to me that in the infinitely far apart case the surface charge is distributed evenly on both faces, whilst when brought close together the whole surface charge is on the inner face.

I think my error is that we can't consider the plate with thickness in isolation and use the principle of superposition, since the fact its surface charge is all on one side is precisely due to the presence of the other plate.

If we take the other plate away, it just becomes another regular conductor and goes back to $\frac{\sigma}{\epsilon_{0}}$.

Contrast this with the infinitely thin sheet, it's charge distribution doesn't change even if we bring another plate close to it so we can apply the principle of superposition without trouble.

 

[kuruman]

$\dots$ whilst when brought close together the whole surface charge is on the inner face.

Rethink this part. If the Gaussian pillbox encloses both plates and sticks out in vacuum, what is the electric flux through the faces?

 

[Dale]

so I suppose that means that we do have an electric field within the conductor?

No, the conductor is still a conductor. No e-field.

 

[Dale]

For a finitely thick plate the charge distributes on two faces. For the same total Q the charge density σ will be cut in half relative to "thin" plate. That simple I think.

I don’t think this is correct. All of the charge should be on the inner surface.

 

[etotheipi]

This is what I came up with,

I did take a bit of a shortcut, for the long pillbox all that I can say is that the net flux is zero, however due to the symmetry of the situation I can pretty much say that the electric field outside must be zero. It then follows that, since the E field within the conductor is zero as well, all of the charge must be on the inner face.

 

[etotheipi]

I think to use the principle of superposition we need to be a little sneaky. If we assemble the full capacitor, there will be no field within either of the conductors. Suppose then we went and "glued" all of the charges to their current positions, and then removed one of the plates.

Now there would indeed be an E field inside the one remaining conductor, however the E field at its surface would be $\frac{\sigma}{2\epsilon_{0}}$ just as in the infinitely thin case.

The important part is that when we use the principle of superposition, we need to consider the charges as is. If I just remove one of the plates and don't "glue" the other one, then evidently once I remove the other one the charge will redistribute across both faces and the outside field will change.

 

[mitochan]

Say ideal zero thickness plate has charge density $\sigma$, I suppose more real conductor plate with thickness has charge density of
$\frac{\sigma}{2}+\delta$ on inside surface and $\frac{\sigma}{2}-\delta$ on outside surface. $\delta$ is necessary to cancel electric field in the conductor plate that comes from charges on the opposite plate.

EDIT : I am thinking above said is wrong for INFINITE plate conductors. For finite plate some line of electric field go around the condenser and come to outside surface. For infinite plate no lines come to outside surface.

 

[hutchphd]

I don’t think this is correct. All of the charge should be on the inner surface

Indeed. And thanks also to @etotheipi for the Gaussian pillbox drawing!

 

[kuruman]

It helps to solve the most general case then look at special cases. We have three regions in space, L (left), M (middle) and R (right).

Place conductor 1 to the left of conductor 2. The field in each region is independent of position. This means that by symmetry if we swap the plates, the fields in L and R must not change, i.e. $E_L=E_R$. Then, for a pillbox with faces in L and R,$$E_R A+E_LA=\frac{(\sigma_1+\sigma_2)A}{\epsilon_0}=2 E_RA ~\rightarrow~E_R=E_L=\frac{(\sigma_1+\sigma_2)}{2\epsilon_0}.$$For a pillbox with one face in L and one inside the conductor 1 there is only flux through one face so$$E_LA=\frac{(\sigma_1+\sigma_2)A}{2\epsilon_0}=\frac{\sigma_{1L} A}{\epsilon_0}~\rightarrow~\sigma_{1L} =\frac{(\sigma_1+\sigma_2)}{2}.$$Since $\sigma_{1L}+\sigma_{1R}=\sigma_1$, it follows that $\sigma_{1R} =\dfrac{(\sigma_1-\sigma_2)}{2}.$

The charge densities on conductor 2 are found simply by swapping indices 1 and 2 in the above expressions. Note that when the plates carry equal and opposite charges ($\sigma_1=-\sigma_2=\sigma$), all the charge is on the inside plates as it should be in the case of a parallel plate capacitor charged by a battery. For a pillbox with one face inside conductor 1 and one in the middle region M,$$E_MA=\frac{\sigma_{1R}}{\epsilon_0}A=\frac{\sigma-(-\sigma)}{2\epsilon_0}A~\rightarrow~E_M=\frac{\sigma}{\epsilon_0}.$$Note that there is superposition in the middle but also a factor of 2 in the denominator. In general though, if you start with two uncharged plates and you put total charge $Q_1$ on one and $Q_2$ on the other, the distributions will be as above.