Fields and forces with dielectrics
We will now prove some rather general theorems for electrostatics in situations where dielectrics are present. We have seen that the capacitance of a parallel-plate capacitor is increased by a definite factor if it is filled with a dielectric. We can show that this is true for a capacitor of any shape, provided the entire region in the neighborhood of the two conductors is filled with a uniform linear dielectric.  Without the dielectric, the equations to be solved are \begin{equation*} \FLPdiv{\FLPE_0}=\frac{\rho_{\text{free}}}{\epsO}\quad \text{and} \quad \FLPcurl{\FLPE_0}=\FLPzero. \end{equation*}  With the dielectric present, the first of these equations is modified; we have instead the equations \begin{equation} \label{Eq:II:10:26} \FLPdiv{(\kappa\FLPE)}=\frac{\rho_{\text{free}}}{\epsO}\quad \text{and} \quad \FLPcurl{\FLPE}=\FLPzero. \end{equation}  Now since we are taking $\kappa$ to be everywhere the same, the last two equations can be written as \begin{equation} \label{Eq:II:10:27} \FLPdiv{(\kappa\FLPE)}=\frac{\rho_{\text{free}}}{\epsO}\quad \text{and} \quad \FLPcurl{(\kappa\FLPE)}=\FLPzero. \end{equation}  We therefore have the same equations for $\kappa\FLPE$ as for $\FLPE_0$, so they have the solution $\kappa\FLPE=\FLPE_0$. In other words, the field is everywhere smaller, by the factor $1/\kappa$, than in the case without the dielectric. Since the voltage difference is a line integral of the field, the voltage is reduced by this same factor. Since the charge on the electrodes of the capacitor has been taken the same in both cases, Eq. (10.2) tells us that the capacitance, in the case of an everywhere uniform dielectric, is increased by the factor $\kappa$. Let us now ask what the force would be between two charged conductors in a dielectric. We consider a liquid dielectric that is homogeneous everywhere. We have seen earlier that one way to obtain the force is to differentiate the energy with respect to the appropriate distance. If the conductors have equal and opposite charges, the energy $U=Q^2/2C$, where $C$ is their capacitance. Using the principle of virtual work, any component is given by a differentiation; for example, \begin{equation} \label{Eq:II:10:28} F_x=-\ddp{U}{x}=-\frac{Q^2}{2}\,\ddp{}{x}\biggl(\frac{1}{C}\biggr). \end{equation}  Since the dielectric increases the capacity by a factor $\kappa$, all forces will be reduced by this same factor. One point should be emphasized. What we have said is true only if the dielectric is a liquid. Any motion of conductors that are embedded in a solid dielectric changes the mechanical stress conditions of the dielectric and alters its electrical properties, as well as causing some mechanical energy change in the dielectric. Moving the conductors in a liquid does not change the liquid. The liquid moves to a new place but its electrical characteristics are not changed. Many older books on electricity start with the “fundamental” law that the force between two charges is \begin{equation} \label{Eq:II:10:29} F=\frac{q_1q_2}{4\pi\epsO\kappa r^2}, \end{equation}  a point of view which is thoroughly unsatisfactory. For one thing, it is not true in general; it is true only for a world filled with a liquid. Secondly, it depends on the fact that $\kappa$ is a constant, which is only approximately true for most real materials. It is much better to start with Coulomb’s law for charges in a vacuum, which is always right (for stationary charges). What does happen in a solid? This is a very difficult problem which has not been solved, because it is, in a sense, indeterminate. If you put charges inside a dielectric solid, there are many kinds of pressures and strains. You cannot deal with virtual work without including also the mechanical energy required to compress the solid, and it is a difficult matter, generally speaking, to make a unique distinction between the electrical forces and the mechanical forces due to the solid material itself. Fortunately, no one ever really needs to know the answer to the question proposed. He may sometimes want to know how much strain there is going to be in a solid, and that can be worked out. But it is much more complicated than the simple result we got for liquids. A surprisingly complicated problem in the theory of dielectrics is the following: Why does a charged object pick up little pieces of dielectric? If you comb your hair on a dry day, the comb readily picks up small scraps of paper. If you thought casually about it, you probably assumed the comb had one charge on it and the paper had the opposite charge on it. But the paper is initially electrically neutral.  It hasn’t any net charge, but it is attracted anyway. It is true that sometimes the paper will come up to the comb and then fly away, repelled immediately after it touches the comb. The reason is, of course, that when the paper touches the comb, it picks up some negative charges and then the like charges repel. But that doesn’t answer the original question. Why did the paper come toward the comb in the first place? The answer has to do with the polarization of a dielectric when it is placed in an electric field. There are polarization charges of both signs, which are attracted and repelled by the comb. There is a net attraction, however, because the field nearer the comb is stronger than the field farther away—the comb is not an infinite sheet. Its charge is localized. A neutral piece of paper will not be attracted to either plate inside the parallel plates of a capacitor. The variation of the field is an essential part of the attraction mechanism. As illustrated in Fig. 10–8, a dielectric is always drawn from a region of weak field toward a region of stronger field. In fact, one can prove that for small objects the force is proportional to the gradient of the square of the electric field. Why does it depend on the square of the field? Because the induced polarization charges are proportional to the fields, and for given charges the forces are proportional to the field. However, as we have just indicated, there will be a net force only if the square of the field is changing from point to point. So the force is proportional to the gradient of the square of the field. The constant of proportionality involves, among other things, the dielectric constant of the object, and it also depends upon the size and shape of the object. There is a related problem in which the force on a dielectric can be worked out quite accurately. If we have a parallel-plate capacitor with a dielectric slab only partially inserted, as shown in Fig. 10–9, there will be a force driving the sheet in.  A detailed examination of the force is quite complicated; it is related to nonuniformities in the field near the edges of the dielectric and the plates. However, if we do not look at the details, but merely use the principle of conservation of energy, we can easily calculate the force. We can find the force from the formula we derived earlier. Equation (10.28) is equivalent to \begin{equation} \label{Eq:II:10:30} F_x=-\ddp{U}{x}=+\frac{V^2}{2}\,\ddp{C}{x}. \end{equation}  We need only find out how the capacitance varies with the position of the dielectric slab. Let’s suppose that the total length of the plates is $L$, that the width of the plates is $W$, that the plate separation and dielectric thickness are $d$, and that the distance to which the dielectric has been inserted is $x$. The capacitance is the ratio of the total free charge on the plates to the voltage between the plates. We have seen above that for a given voltage $V$ the surface charge density of free charge is $\kappa\epsO V/d$.  So the total charge on the plates is \begin{equation*} Q=\frac{\kappa\epsO V}{d}\,xW+\frac{\epsO V}{d}\,(L-x)W, \end{equation*}  from which we get the capacitance: \begin{equation} \label{Eq:II:10:31} C=\frac{\epsO W}{d}\,(\kappa x+L-x). \end{equation}  Using (10.30), we have \begin{equation} \label{Eq:II:10:32} F_x=\frac{V^2}{2}\,\frac{\epsO W}{d}\,(\kappa-1). \end{equation}  Now this equation is not particularly useful for anything unless you happen to need to know the force in such circumstances. We only wished to show that the theory of energy can often be used to avoid enormous complications in determining the forces on dielectric materials—as there would be in the present case. Our discussion of the theory of dielectrics has dealt only with electrical phenomena, accepting the fact that the material has a polarization which is proportional to the electric field. Why there is such a proportionality is perhaps of greater interest to physics. Once we understand the origin of the dielectric constants from an atomic point of view, we can use electrical measurements of the dielectric constants in varying circumstances to obtain detailed information about atomic or molecular structure. This aspect will be treated in part in the next chapter.