Divergence and Gauss's Law
Shrink a Gaussian surface down around a single point and Gauss's law becomes a statement about that point: the divergence of , its outflow per unit volume, equals the charge density there over . Field lines start on positive charge, end on negative charge, and never begin or end in empty space.
What the symbols mean
- Divergence of E
Net outflow of field per unit volume at a point: ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z.
- Charge density
Charge per cubic meter at that point, in C/m³.
What the symbols mean
- Divergence theorem
The total outflow through the boundary equals the divergence added up over the volume.
- Gauss's law
The charge inside, divided by ε₀.
What the symbols mean
- Spherical radial component
The field's strength along r̂, depending only on the distance r from a center.
- Cylindrical radial component
The field's strength straight away from an axis, depending only on the distance s from it.
Explanation
Divergence is simply a measure of how much a field spreads out from a single point: how much more comes out of a tiny box around it than goes in. The name fits, since field lines diverge, or fan apart, from a source. It's useful because it turns Gauss's law, a statement about whole bags, into a statement about each point in space, which is the form computers and differential equations need. The formula looks dense, but it's just two pieces: (how much field gushes out of a tiny box around the point, per unit volume) and (how much charge is packed in there, divided by the usual constant). It's the same law as before, just zoomed in; bear with it.
From a bag to a point. Gauss's law in integral form talks about a whole closed surface. Shrink that surface around one point , so it encloses a tiny volume . The flux out is about , because divergence is outflow per unit volume (Curl and Divergence). Gauss's law says the same flux is . Divide by :
The proof, with the divergence theorem. For any region with boundary , the divergence theorem and Gauss's law give two expressions for the same flux, so
If the integrand were positive at some point, it would stay positive on a small ball around that point (it's continuous), and the integral over that ball would be positive. The same goes for negative. So the integrand is zero everywhere. Running the argument backward recovers the integral form, so the two forms say exactly the same thing.
Reading it. Positive divergence marks positive charge, a source of field lines; negative divergence marks negative charge, a sink; in empty space . That doesn't mean the field is zero there, only that every field line entering a small box also leaves it.
Radial fields. For , take a thin spherical shell between and . Its net outward flux is . Divide by the shell's volume, :
A thin can around an axis gives the same way. Notice it's not just . A field can weaken outward and still have zero divergence, because it's spreading over more area, and the (or ) keeps track of that spreading.
The point charge. For , is constant, so everywhere except at . The inverse-square law is exactly the falloff that keeps the divergence zero in empty space. All of the charge's divergence is packed into the single point where it sits; physicists write its density as , a spike whose integral is .
Two directions of travel. Given , differentiating gives at once. Given , finding means integrating, which needs symmetry, or Poisson's equation for the potential, which you'll meet in the next unit. This equation is the first of Maxwell's Equations, where the course ends up; its magnetic twin, , says there are no magnetic charges.
A common mistake. It's tempting to think zero divergence means zero field, since "nothing is coming out." It doesn't: a point charge's field is strong all around it, yet its divergence is zero everywhere except at the charge, because every line entering a small box also leaves. The other usual slips are dropping the , using the Cartesian formula on spherical components, and reading at a point as the total charge (that takes a volume integral).
See it
Worked examples
In a region of space, N/C, with and in meters. Find the charge density at m.
- 1
N/(C·m).
Inside a sphere of radius , the field is . Find the charge density , and check it against the total charge from the integral form.
- 1
, so , giving : zero at the center, growing outward.
Practice
Type your answer and check it. Solve them all to complete the lesson.
Inside a long, uniformly charged cylinder, with kN/(C·m). Find the charge density, in nC/m³, to 3 significant figures.
Which of these could be the electrostatic field in a region with no charge? (Constants are dropped.)
In a region, N/C, with coordinates in meters. Find the charge density at m, in nC/m³, to 3 significant figures.
Inside a sphere of radius cm, the field is , with kN/C. Find the charge density 5.0 cm from the center, in μC/m³, to 3 significant figures.
In a silicon diode, the p side next to the junction holds fixed negative ions at a density of m⁻³ (each with charge ) out to a depth of μm, and the field is zero beyond that depth. In silicon, is replaced by (the capacitors lesson explains why). Use the one-dimensional Gauss's law to find the field strength at the junction, in MV/m, to 3 significant figures.
The field N/C (with in meters) fills a region. Find the total charge inside the cube m, in pC, to 3 significant figures. Then check your answer with the flux through the cube's faces.
On a fair-weather day the atmosphere's field points straight down, with strength 120 N/C at the ground and 80 N/C at an altitude of 1.0 km. Assuming it changes steadily in between, find the charge density in that layer of air, and from it the number of excess elementary charges per cubic meter, in millions, to 3 significant figures.
The hardest one here. Get it, and you've got this lesson.
A long electron beam has total charge nC per meter, spread out so that its field at distance from the axis is , with mm. Find the charge density , check that it adds up to per meter, and give on the axis in μC/m³, to 3 significant figures.
Where it's used
Semiconductor engineers solve , usually in the form of Poisson's equation, millions of times to design transistors: device simulators chop a chip into tiny cells and demand that the flux out of each cell match the charge inside, exactly the box in the first picture. The same cell-by-cell bookkeeping runs the field solvers used for antennas, high-voltage insulators and particle accelerators. Atmospheric scientists use the one-dimensional version to measure the charge floating in the air from how the fair-weather field changes with height.