A Capacitor With A Dielectric But Also Free Space in Between

[rtareen]

Lets go through the example problem until we get to the part I don't understand. Figure 25-17 can be used as a reference to all questions. From part (a) to part (b) we eventually find the charge q on one plate (and by default the charge -q on the other). No problem there. The battery is then removed so we know the plate charges will remain constant throughout the problem.

Part (c) is where i start getting confused. We use Gauss' law to find the field between the plate and the dielectric and the result is the same as if the dielectric was not there. Doesn't the induced charge on the dielectric contribute the field in between? We know the result is true due to Gauss; law saying so but is there any explanation? My guess is that we are treating faces where the induced charges are as infinite planes and since the field of infinite planes don't depend on distance they cancel out outside the of dielectric? Is that right? Please explain. It doesn't make sense to treat such a small dielectric as an infinite plane.

In part (d) we use Eq. 25-36 but that equation was derived for a dielectric that that fills the space in between and is in contact with the plates. Why is it the same? Also, we can only use Gauss' Law to find the field that is constant at all points on the Gaussian surface. Is the field the same at both ends of Gaussian Surface II? It is clearly not in the area in between. Intuitively it is not either because the field in the upper part of Gaussian Surface II should be weaker due to there being less positive charge to attract the induced negative charge.

I might be missing something fundamental that is not allowing me to fully understand what is going on here.

 

[etotheipi]

You're probably familiar with Gauss law in the form$$\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}$$That's always true, but it's not always the easiest to work with. The charge density is a sum of bound charges (e.g. those in the dielectric) and free charges (e.g. those in the capacitor plates), $\rho = \rho_{\text{bound}} + \rho_{\text{free}}$. In your problem, you know the free charges ($q_{\text{free}} = \pm q$ on either plate), but as of yet we have no way of dealing with the charges in the dielectric.

There is a quantity called polarisation which is related to the total electric field as $\vec{P} = \varepsilon_0 \chi \vec{E}$, where $\chi$ is the susceptibility. The electric displacement field is $\vec{D} = \varepsilon_0 \vec{E} + \vec{P}$, so $$\vec{D} = (1+\chi)\varepsilon_0 \vec{E}$$The factor $(1+\chi)$ is called the relative permittivity $\varepsilon_r$ (or alternatively dielectric constant, like in your book). And we define $\varepsilon=\varepsilon_r \varepsilon_0$, so that $\vec{D} = \varepsilon \vec{E}$. Also, take the divergence of $\vec{D} = \varepsilon_0 \vec{E} + \vec{P}$, and use that the bound charge satisfies $\rho_{\text{bound}} = -\nabla \cdot \vec{P}$, to get$$\nabla \cdot \vec{D} = \varepsilon_0 \nabla \cdot \vec{E} + \nabla \cdot \vec{P} = \rho - \rho_{\text{bound}} = \rho_{\text{free}}$$You'll notice that this is the same form as the original equation, and can be integrated,$$\int_{\Sigma} \vec{D} \cdot d\vec{S} = {q}_{\text{free}} \quad \left( = \varepsilon_0 \int_{\Sigma} \varepsilon_r\vec{E} \cdot d\vec{S} \right)$$That means, it's possible to define a Gaussian surface and use this version of the equation, counting only the free charge (on the capacitor plates) and ignoring the bound charge (in the dielectric). That's what's being used in (c) and (d)!

 

[rtareen]

Im not familiar with Gauss' law in that form. I only know the integral form. So i couldn't understand most of your explanation. But thanks for replying. I liked your post!

 

[etotheipi]

Thanks! But in that case I should clarify. I can state it in a more direct way; you essentially have two versions of Gauss' law,$$\int_{\Sigma} \vec{E} \cdot d\vec{S} = \frac{q_{\text{tot}}}{\varepsilon_0}$$where $\vec{E}$ is the usual electric field, and $q_{\text{tot}}$ is the total charge enclosed (bound, free, whatever) inside the surface $\Sigma$. There is another form, which looks like$$\int_{\Sigma} \vec{D} \cdot d\vec{S} = q_{\text{free}}$$here $\vec{D} = \varepsilon_0 \varepsilon_r \vec{E} = \varepsilon \vec{E}$ is the displacement field, defined in the last post. This version can be stated then equivalently as$$ \int_{\Sigma} \varepsilon_0 \varepsilon_r \vec{E} \cdot d\vec{S} = q_{\text{free}}$$This is a great version to use in some cases when you have dielectrics present, since you can completely ignore the bound charges in the dielectric and only worry about the free charges in the capacitor plates, for example! Your book uses the symbol $\kappa$ intead of the symbol $\varepsilon_r$, but they're exactly the same quantity.

You should check the example in your book to see that, when they calculate the field inside the dielectric, they use a Gaussian surface which includes part of the dielectric and part of the plates, but they only consider the free charges, i.e. the $-q$ inside the plate.

Hopefully that clears things up

 

[rtareen]

Thanks! But in that case I should clarify. I can state it in a more direct way; you essentially have two versions of Gauss' law,$$\int_{\Sigma} \vec{E} \cdot d\vec{S} = \frac{q_{\text{tot}}}{\varepsilon_0}$$where $\vec{E}$ is the usual electric field, and $q_{\text{tot}}$ is the total charge enclosed (bound, free, whatever) inside the surface $\Sigma$. There is another form, which looks like$$\int_{\Sigma} \vec{D} \cdot d\vec{S} = q_{\text{free}}$$here $\vec{D} = \varepsilon_0 \varepsilon_r \vec{E} = \varepsilon \vec{E}$ is the displacement field, defined in the last post. This version can be stated then equivalently as$$\varepsilon_0 \varepsilon_r \int_{\Sigma} \vec{E} \cdot d\vec{S} = q_{\text{free}}$$This is a great version to use in some cases when you have dielectrics present, since you can completely ignore the bound charges in the dielectric and only worry about the free charges in the capacitor plates, for example! Your book uses the symbol $\kappa$ intead of the symbol $\varepsilon_r$, but they're exactly the same quantity.

You should check the example in your book to see that, when they calculate the field inside the dielectric, they use a Gaussian surface which includes part of the dielectric and part of the plates, but they only consider the free charges, i.e. the $-q$ inside the plate.

Hopefully that clears things up

But why does this equation still work even when there is space between the capacitor plates and the dielectric? I could show you the book's whole derivation, and when they derived this they used it for the case where the dielectric fills up the entire space between the plates. I can show you the figure. So why doesn't it matter?

 

[etotheipi]

It doesn't matter if there's space between them. They have drawn a cuboidal Gaussian surface $\Sigma$ that encloses the lower part of the dielectric and the upper part of the negative plate. The free charge enclosed by the surface is $-q$, because the whole charge on the lower plate is on the surface nearer the dielectric, and there's no free charge inside the dielectric.

The second half of the problem is to calculate the flux of the $\vec{D}$ field through the Gaussian surface. We assume the field is perfectly uniform, so no flux out the sides of $\Sigma$. Also, the lower plate is a conductor, so no field in there either. So the only part of the surface we need to worry about is the upper face through the dielectric. The flux through that part of surface, which we'll just call $\Sigma_1$, is just$$\Phi_D = \int_{\Sigma_1} \vec{D} \cdot d\vec{S} = \varepsilon_0 \varepsilon_r \int_{\Sigma_1} \vec{E} \cdot d\vec{S} = -\varepsilon_0 \varepsilon_r E A$$because $\vec{E}$ is antiparallel to $d\vec{S}$ inside the dielectric, taking normals to point out of $\Sigma$. So now you just equate$$\Phi_D = -q$$