3.2.1 Electric flux density and electric flux

The electric flux density D

We start from the one result of Section 3.1 on which everything in this section is built. A point charge $q$ located at the origin produces, at a point P at distance $r$ from it, the electric field

where $a_r$ is the unit vector pointing radially away from the charge and ${\epsilon }_0=10^{-9}/(36\pi )\approx 8.842\times 10^{-12}F/m$ is the permittivity of free space, the value used throughout this book, Eq. (1.4.2). Three features of this field are all we shall need: it is radial, it has the same magnitude at every point of a sphere centred on the charge, and it falls off as $1/r^2$.

The magnitude of $E$ depends on two things: the charge $q$, and the permittivity ${\epsilon }_0$ of the space in which the charge sits. It is useful to separate the two by introducing a second field vector that depends on the charge alone.

Equation (3.2.3) has a striking interpretation. The denominator $4\pi r^2$ is the area of the sphere of radius $r$ centred on the charge, so

It is as if the charge $q$ sent out a fixed amount of "flux", equal to $q$ coulombs, spread uniformly over every concentric sphere (Figure 3.2.1): the larger the sphere, the thinner the flux is spread, and $|D|$ is the amount of flux per unit area. This is why $D$ is called a flux density and why its unit is the coulomb per square metre. The permittivity ${\epsilon }_0$ no longer appears: $D$ is fixed by the charge and the geometry, while $E=D/{\epsilon }_0$ also carries the information about the medium.

In free space, $D$ and $E$ differ only by the constant factor ${\epsilon }_0$ and carry the same information. The distinction becomes important in Section 3.4, where material media are introduced: there $D$ keeps the property we are about to discover, namely that its flux counts the (free) charge, while $E$ is modified by the material.

Flux: how much field passes through a surface

The word flux comes from the Latin for "flow". Imagine rain falling straight down, and hold a circular net (a hoop with a mesh) in it (Figure 3.2.2). How much rain passes through the net per second?

• If the net is held horizontally, facing the rain, every drop that falls on its area $S$ goes through: the amount is (rain per square metre) $\times S$.
• If the net is held vertically, edge-on to the rain, the drops slide past and nothing goes through.
• If the net is tilted by an angle $\theta$, look at its shadow on the ground: the drops that pass through the net are exactly those that would have fallen on its shadow. The shadow of a tilted disk is an ellipse of area $S\cos \theta$, so the amount is (rain per square metre) $\times S\cos \theta$.

The angle $\theta$ is measured between the rain and the normal $\hat{n}$ of the net, the unit vector perpendicular to its plane. "Rain per square metre times $S\cos \theta$" is precisely the scalar product of a "rain vector" with the vector $S\hat{n}$.

For the electric field the "rain" is the vector $D$, and the surface need not be flat or the field uniform: we cut the surface into small elements $dS$, each practically flat, with its own normal $\hat{n}$, and add up the contributions $D\cdot \hat{n}dS$. It is convenient to combine the area of an element and its normal into one vector, the vector surface element $dS=\hat{n}dS$, familiar from the surface integrals of Section 2.3.

The unit of flux is the coulomb: $D$ is in $C/{\mathrm{m}}^2$ and $dS$ in ${\mathrm{m}}^2$. Because the integrand is the scalar product $D\cdot \hat{n}dS=DdS\cos \theta$, only the component of $D$ along the normal contributes. Figure 3.2.3 shows the three typical situations for a flat surface of area $S$ in a uniform field. When the normal is parallel to $D$, every line crosses the surface and the flux is $DS$. When the normal is perpendicular to $D$, the lines slide along the surface and the flux is zero. In general, with $\theta$ the angle between $\hat{n}$ and $D$,

Equivalently, $S\cos \theta$ is the projection of the surface onto a plane perpendicular to the field, the "shadow" it casts in the stream of field lines.

For a closed surface, such as the skin of a balloon, there is a natural choice of normal: the one pointing outward. With it, a field line that leaves the enclosed region contributes positively ($D\cdot \hat{n}>0$), and a line that enters contributes negatively ($D\cdot \hat{n}<0$). The outward flux is therefore a balance: what flows out minus what flows in.