I wrote this lecture, because I watched many of you work problems using \(\vec{D}\) and I realized we needed to talk about the raison d’etre of \(\vec{D}\).

\(\vec{D}\) was invented to make your life easier, believe it or not. It’s not a physical quantity, but it’s a convenient non-physical quantity. You can’t compute anything you can measure with it. For example, for potential you still need \(\vec{E}\). For Work or Energy you need \(V\) which means you need \(\vec{E}\). For Force you need \(\vec{E}\).

Some things that make sense to me

The Displacement vector is not a physical thing. You couldn’t measure it. If you divide it by the permittivity it’s the field that would be there if you only had free charge and no bound charge. So the only time it’s physical (and you could measure it) is when there’s no dielectric, in which case \(\vec{D} = \epsilon_0\vec{E}\).

So then, you have to be careful about when you use it. There’s a Gauss’ Law for displacement, as we have seen:

\[\nabla\cdot\vec{D} = \rho_f\]

which is generally REALLY useful for starting problems involving dielectric. If you start with that formula (or its integral version) you never get in trouble putting \(\epsilon\)’s or \(\epsilon_0\)’s in the wrong places.

Going back to our roots

In order to use \(\vec{D}\) appropriately I think it’s useful to remember where we started. We started with:

\[\rho = \rho_b + \rho_f\]

So if we want to calculate the field \(\vec{E}\) everywhere, we need to use the total charge.

Then we wrote down our familiar Gauss’ law for this total charge:

\[\epsilon_0\nabla \cdot \vec{E} = \rho = \rho_b + \rho_f = −\nabla \cdot \vec{P} + \rho_f\]

So \(\vec{E}\) is the total field, not just the portion generated by polarization.

Then we took the divergence term on the right over to the left, which is sort of like subtracting the effect of the bound charge from both sides:

\[\nabla \cdot (\epsilon_0 \vec{E} + \vec{P}) = \rho_f\]

so it makes sense that we get something that only depends on free charge. But, mind you, that was the step where we created an unphysical quantity on the left that we are about to call “electric displacement.”

We define \(\vec{D}\) the electric displacement as:

\[\vec{D} = \epsilon_0 \vec{E} + \vec{P}\]

and it is the TOTAL electric field (times \(\epsilon_0\)), PLUS the polarization.

Adding the polarization is sort of like adding back in the field that we lost to the polarization. That’s confusing. Remember, the polarization will (almost) always serve do undo the electric field a little (or a lot), so \(\epsilon_0 \vec{E} + \vec{P}\) will almost always be larger than \(\epsilon_0 \vec{E}\) by itself. And \(\vec{E}\) is the total field, not \(\vec{D}\). (I’m saying “almost” because when we start to deal with dielectric functions, that are functions of the frequency of the electromagnetic wave, you can get negative dielectrics).

I think this equation is confusing because it looks like \(\vec{D}\) is the field of the free charges plus the field of the polarization, but that’s not what it is. \(\vec{E}\) is the total field, the field of the bound and the free charges, and when you add \(\vec{P}\) you’re replacing the field lost to the bound charges.

So now Gauss’s law is just

\[\nabla \cdot \vec{D} = \rho_f\]

and the integral version is

\[\oint \vec{D} \cdot d\vec{a} = Q_{f,{\rm enc}}\]

That’s why it’s convenient, because now we don’t have to worry about the infinite cycle of the polarization contributing to the field, and thereby changing the polarization, and changing the contribution of the polarization to the field…

\[\vec{D} = \epsilon_0(1 + \chi_e)\vec{E} = \epsilon\vec{E}\]

where \(\chi_e\) is called the electric susceptibility and \(\epsilon\) is the permittivity.

\(\chi_e\) is always positive, so \(\epsilon\) is always larger than 1, so \(\vec{D}\) is always larger than \(\epsilon_0\vec{E}\) which makes sense because \(\vec{D}\) is the electric field times \(\epsilon_0\) that we would have if we didn’t have any bound charge.