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# 1.7: Electric Field D

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

We have been assuming that all “experiments” described have been carried out in a vacuum or (which is almost the same thing) in air. But what if the point charge, the infinite rod and the infinite charged sheet of Section 1.6 are all immersed in some medium whose permittivity is not $$\epsilon_0$$, but is instead $$\epsilon$$? In that case, the formulas for the field become

$\nonumber E=\frac{Q}{4\pi\epsilon r^2},\quad \frac{\lambda}{2\pi\epsilon r},\quad \frac{\sigma}{2\epsilon}$

There is an $$\epsilon$$ in the denominator of each of these expressions. When dealing with media with a permittivity other than $$\epsilon_0$$ it is often convenient to describe the electric field by another vector, $$\textbf{D}$$, defined simply by

$\textbf{D}=\epsilon \textbf{E}\label{1.7.1}$

In that case the above formulas for the field become just

$\nonumber D=\frac{Q}{4\pi r^2},\quad \frac{\lambda}{2\pi r},\quad \frac{\sigma}{2}$

The dimensions of $$D$$ are Q L-2, and the SI units are C m-2.

This may seem to be rather trivial, but it does turn out to be more important than it may seem at the moment.

Equation \ref{1.7.1} would seem to imply that the electric field vectors $$\textbf{E}$$ and $$\textbf{D}$$ are just vectors in the same direction, differing in magnitude only by the scalar quantity $$\epsilon$$. This is indeed the case in vacuo or in any isotropic medium – but it is more complicated in an anisotropic medium such as, for example, an orthorhombic crystal. This is a crystal shaped like a rectangular parallelepiped. If such a crystal is placed in an electric field, the magnitude of the permittivity depends on whether the field is applied in the $$x$$- , the $$y$$- or the $$z$$-direction. For a given magnitude of $$E$$, the resulting magnitude of $$D$$ will be different in these three situations. And, if the field $$\textbf{E}$$ is not applied parallel to one of the crystallographic axes, the resulting vector $$\textbf{D}$$ will not be parallel to $$\textbf{E}$$. The permittivity in Equation \ref{1.7.1} is a tensor with nine components, and, when applied to $$\textbf{E}$$ it changes its direction as well as its magnitude.

However, we shan’t dwell on that just yet, and, unless specified otherwise, we shall always assume that we are dealing with a vacuum (in which case $$\textbf{D}$$ = $$\epsilon_0$$$$\textbf{E}$$) or an isotropic medium (in which case $$\textbf{D}$$ = $$\epsilon$$$$\textbf{E}$$). In either case the permittivity is a scalar quantity and $$\textbf{D}$$ and $$\textbf{E}$$ are in the same direction.