Teardown: Maxwell's equations

Arthur J. Glazar -August 19, 2012

A VOLUME THAT ENCLOSES NO CHARGE

Let's now take a look at Figure 6 to help us visualize what we've just described.

FIGURE 6: The cube represents a small volume (ΔX ΔY ΔZ) that we wish to investigate

In Figure 6, the cube represents a small volume (ΔX ΔY ΔZ) that we wish to investigate¹⁷. This imaginary cube is immersed in a uniform electric field that is directed from left to right. In Figure 6 the uniform electric field is designated E, and the uniform Electric Flux Density Vector is designated (εE). Both are represented by the same arrow.

The area of the left face of the cube is A1, and is equal to ΔX • ΔZ. Similarly, the area of the right face of the cube is A2 and it is also equal to ΔX • ΔZ.

Cube face A1 receives a total flux of (εE) (A1) [ampere-seconds]. Cube face A2 receives a total flux (εE) (A2) [ampere-seconds]. Being imaginary, the cube cannot influence either the direction or magnitude of (εE) so that the flux entering through A1 must be equal to the flux leaving at A2. All of the other cube faces are parallel to (εE) and therefore do not intercept flux either entering or leaving the cube. By convention, flux entering a volume is considered positive (+) and flux leaving a volume is considered negative (-). Therefore the sum of the flux entering and the flux leaving the cube is zero. Or we can say that the net flux leaving the cube is zero.

Now let's consider the above description in terms of equation (III'). The constant amplitude and direction of (εE) means that there is no spatial rate of change around the cube, so that in equation (III') all three partial derivatives are zero:

∂(εE) ∂(εE) ∂(εE)

----- = 0, --------- = 0, and --------- = 0

∂x ∂Y ∂Z

and the equation reduces to 0 = ρ. The meaning of this seemingly trivial result is that the volume charge density contained within the cube is zero, or in other words, the cube contains no electrical charge if div(εE) = 0 (and vice-versa). For this example we chose a uniform electric field and a cube for easy visualization. The result would be the same for any external electric field and a closed volume of any shape.¹⁸

A VOLUME THAT ENCLOSES A CHARGE

Since we've stipulated that shape doesn't matter, we can switch to a spherical volume to simplify our next topic, which is to investigate a volume that does contain an electrical charge as shown in Figure 7. We know that an externally-produced field (Figure 6) contributes nothing to div(εE), so it has been removed from further consideration.

Figure 7: The figure shows a point charge, q, placed at the center (x=0, y=0, z=0) of a small sphere.

Figure 7 shows a point charge, q, placed at the center (x=0, y=0, z=0) of a small sphere. Physics textbooks tell us that the electric field produced by a point charge is

E = ------------ u_R [volts per meter]

4 π ε R²

where q is the charge magnitude in coulombs, R is the radial distance from the charge to a target point, in meters, and u_R is a "unit vector" (its magnitude or amplitude is 1) which indicates that the E field is directed along the radial line from source to the target point. Other symbols are as defined before. The field produced by a stationary point charge varies in both direction and magnitude. First, the field is radial, and therefore changes continuously in direction around the charge. In Figure 7 just one E vector is shown although there are an infinite number radiating in all directions. Second, the magnitude of the field varies inversely with the square of distance from the charge.

The Flux Density Vector is obtained from this equation simply by multiplying both sides by ε:

(εE) = ------------ u_R [ampere-seconds] / [square meter]

4 π R²

or [coulombs] / [square meter]. Now consider this: at a distance R from the charge, the magnitude of flux density is

(εE) = ---------- [coulombs] per [square meter].

4 π R²

And also at distance R, the total surface area, A, of the sphere is 4π R² [square meters]. The total flux thru the surface of the sphere is obtained by multiplying the flux density, (εE), by the total surface area, A, which is simply q [coulombs]. That is,

Total flux = (εE) A = ---------- 4 π R^{2 =}q

4 π R²

From geometry we know that the volume,Ѵ , of any sphere is equal to 4/3 π R³,where R is its radius.So that in our example, we can say that the total flux per unit volume is

q q

----- = ------------ [coulombs] / [cubic meter].

Ѵ 4/3 π R³

The same result would be obtained by calculating the divergence of (εE) for the same point charge¹⁹. The bottom line is this: at a given point in space, if div(εE) is not zero, then a charge must exist at that point, and the volume density of that charge is div(εE). That is the meaning of equation (III).

We chose a sphere with q at its center because all of the flux density vectors pass through the sphere's surface in a direction perpendicular to (or normal to) the surface. In that way we avoided having to resolve the vectors into their x, y and z components; that is, the vectors in our special example automatically met the requirements for perpendicularity.

MAGNETIC MONOPOLES (NOT)

In the electrostatic world there exists a fundamental entity called "electric charge", which bears the name and unit "coulomb". But there is no equivalent entity (i.e., a "magnetic charge") in the electromagnetic world. This asymmetry has always bothered physicists and experimenters. But like it or not, the divergence equations are forced to take this asymmetry into account.

And so, to analyze a magnetic field, we could modify Figure 6 by just re-labeling the field as "H or (μH)", and make the same statements that we did for the electric field. But we cannot press Figure 7 into service for magnetic fields because there is no magnetic equivalent to a "point charge" that we could insert at 0,0,0. It might be argued that a small bar magnet, or one pole of a horseshoe magnet would qualify. But careful analysis will always show that any magnetic flux vector leaving a closed volume around the magnet would have an offsetting vector entering the volume. The conclusion must be that the net outward flux i.e., div(μH)) is always zero. And that's the meaning of Equation (IV).

VOLTAGE DENSITY REDUX

There is one piece of unfinished business to attend to before concluding our guide. We promised to revisit the interpretation of "voltage density" in the section "NEW VECTORS CREATED."

Recall that square brackets indicate units (not quantities), and [μH] is the magnetic flux density. Then we can write,

[volt] [seconds]

[μH] = -----------------------

[square meter]

dividing both sides by [seconds] gives us:

[μH] [volts]

------------- = --------------------

[seconds] [square meter]

In words, this says that the time-rate-of-change of Magnetic Flux Density is equal to [volts per square meter], which seems to describe a "voltage density".

From there we can take another step and multiply both sides of the last equation by [square meters], which leaves [volts] isolated on the right side of the equation:

[μH]

------------- X [square meters] = [volts]

[seconds]

This is equivalent to Faraday's Law of induction which states that a voltage, proportional to the rate-of-change of magnetic flux density, is induced in an area bounding the flux. You will find this in textbooks written as

∂φ

V = - -------

∂t

where V is the induced voltage, and φ is the magnetic flux density.

CONCLUSION

If for no other reason, Maxwell's equations are worth knowing about just because of their place in the history of science and technology. The late physicist Richard P. Feynman characterized Maxwell's Equations as "the most significant event of the 19th century".