Dielectric in an electric field

A charge \( +Q \) placed in free space sets up an electric field around itself. This electric field is described by the Coulomb’s law of electrostatics. But a charge in a free space is quite useless: in real life we are surrounded by things of all sorts, and if we really want to make any use of our knowledge of electrostatics, we'll need to learn to apply it in presence of other materials. Motivated by the need to make our knowledge useful, let’s make this system a bit more interesting, and close to reality, by introducing a dielectric (insulator) material around the charge, as shown in the figure below. The region \(B\) is dielectric, whereas regions \(A\) and \(C\) are empty. The questions that naturally arise are:

a) How do the dielectric and the charge \(+Q\) interact?
b) Do properties of the dielectric subjected to the electric field from \(+Q\) remain the same or change?
b) Does the presence of dielectric affect the way the electric field is setup in space?

Let us try to answer these questions using logic and some fundamental facts we already know. In this post, we will try to answer question (a) qualitatively. We will delve into quantitative analysis once we have a good qualitative grasp on key characteristics of this system.

We know that this dielectric consists of atoms as its building block. So, the natural way to approach the problem would be to think about how an atom would behave when placed in an electric field. Since an atom is composed of a positive nucleus and a cloud of negative electrons around that nucleus, our problem is really reduced to figuring how these fundamental charges within the atom interact when placed in the electric field of the charge \(+Q\). But don’t we know the answer to this already? The like charges repel, and the opposite attract.

Applying this to the atom, we find that the positive nucleus in the atom is repelled away from \(+Q\) whereas the negative electron cloud is attracted toward \(+Q\). So, effectively the atom has been distorted or stretched a bit along the direction of the field. It is not hard to see that the stretched atom shown in the figure above is a bit more negative in the region on the right compared to the region on the left. (\(Z\) is the atomic number, \(e\) is the magnitude of electronic charge, \(+Ze\) is the nuclear charge, and \(-Ze\) is the charge on electron cloud.) Thus, the atom now has a positive and a negative pole at opposite ends. We say that the atom has been polarized and has a dipole (two poles). To analyze anything in science, we first need to find a way to quantify it, and a dipole is no exception. So, let’s see how a dipole is quantified.

Clearly, the behavior of a dipole will depend on magnitude of charge that is separated (i.e. how big \(\pm\delta\) are), and the distance between them. Therefore, the simplest way to define dipole moment is would be \(\vec{p} = \delta \vec{d}\); \(\vec{d}\) is the position vector of \(+\delta\) with respect to \(-\delta\). It follows from the definition that direction of \(\vec{p}\) is from \(-\delta\) to \(+\delta\) as opposed to the direction of \(\vec{E}\) which always runs from \(+ve\) charge to \(-ve\) charge. This is important to remember as it is common to get confused between the relative directions of \(\vec{p}\) and \(\vec{E}\) in a dipole.

Now that we've figured out qualitatively how an atom is affected in the presence of an electric field, let's try to extend this understanding to include the whole of dielectric. Since the dielectric consists of many such atoms and since all of them are subjected to the electric field from \(+Q\), we may safely assert that all of the atoms in it are polarized. The electron clouds on all atoms will be displaced toward \(+Q\) and the nuclei away from \(+Q\). Therefore, we can say the dielectric itself has now been polarized. This is a general conclusion and applies to all dielectrics. It is not hard to see that the side of the dielectric closest to the charge \(+Q\) will be negative whereas the side farthest will be positive. So, in essence, the surface of the dielectric closes to \(+Q\) has become negative and the surface farthest positive. The overall system can be represented in the following way (the green arrows represent atomic dipoles within the dielectric):

To further analyze the system, we need to find a way to quantify this collection of dipoles. It is important to realize at this point that although the atomic dipoles are discrete, they are extremely small compared to the dimensions of objects we use in real life. So small that we may treat a collection of them as a continuous distribution rather than a distribution of discrete particles. At macro scales, this assumption will not compromise the validity of our theory. On microscopic scales, however, we can't make this assumption and need to resort to a different methodology to treat this system. Therefore, for a macroscopic system, distribution of dipoles can be described using polarization \(\vec{P}(r)\) defined as the dipole moment per unit volume or density of dipole moment. Obviously, it is a function of position \(\vec{r}\) in space because the density of dipoles may be different in different locations within the dielectric. Therefore, the dipole moment \(d\textbf{p(r)}\) in the volume \(dV\) would be

$$ d\vec{p}(\vec{r}) = \vec{P}(\vec{r}) \, dV$$ And the total dipole moment of a polarized dielectric object would be $$ \vec{p}(\vec{r}) = \int d\vec{p}(\vec{r}) = \int \vec{P}(\vec{r}) \, dV$$

Finally, we have figured out how the charge \(+Q\) affects the dielectric: presence of a charge or an electric field polarizes a dielectric. A dielectric is polarized when subjected to an electric field. Following examples that illustrate surface charge induced due to electric field from a charge \(+Q\) placed in the vicinity.

Do properties of the dielectric subjected to the electric field from \(+Q\) remain the same or change? Of course the properties change. Now there is an electric field associated with dielectric whereas earlier there was none. If you were to place a small charge near it, the charge would move under the influence of the electric field from the dielectric. One question still remains: How does all of this affect the electric field distribution in space? This is the subject of our next inquiry. But why in the world should anyone care about polarization, dielectrics, or anything we've talked about in this section? Of what use is any of this? I think it's a good time to take a break and look at an interesting family of dielectrics known as piezoelectrics.

NEXT: A brief tangent: Piezoelectric Dielectrics