EM-06 · Electrostatics field

Gauss's Law

The net electric flux through a closed surface is proportional only to the charge inside it. You grow the Gaussian surface or move the charge outside, and confirm for yourself that the flux does not depend on the size of the surface.

Grow the surface and the flux stays the same

The wire-mesh ball is a Gaussian surface around a point charge. Growing it with the slider weakens the field on the surface as 1/R², but the surface area grows as R², so their product, the net flux, stays put on the indicator. Move the charge outside with the button, and the net flux disappears.

Gaussian surface radius RR = 1.50

Drag to rotate; the slider sets the surface radius R, and the buttons put the charge inside or outside.

Net flux through the closed surface

∮E·dA = Q_enc/ε₀ · only the enclosed charge counts

E shrinks as 1/R² while A grows as R², so their product, the flux, stays the same.

E 44% · A 225% · Φ 100% (invariant)

Inside · Q/ε₀ ·

What the Gauss law says

In EM-05 you defined the net flux through a closed surface, ∮E·dA. Gauss's law says this value is proportional only to the charge inside the surface; as a formula, ∮E·dA = Q_enc/ε₀. The constant is the permittivity of free space, ε₀. However you draw the surface, as long as it encloses the same charge, the same flux comes out.

Observe∮E·dA = Q_encε₀

The net flux is the enclosed charge divided by ε₀.

Independent of the surface shape and size

Doubling the radius of a ball around a point charge weakens the field on the surface to 1/4, but the area becomes 4 times larger. Their product, the flux, does not change. The same holds for a dented surface or a cube: as long as it encloses the same charge, the net flux is always Q/ε₀. If you picture flux as the number of field lines piercing the surface, that number is set by the charge, and the shape of the surface does not matter.

Choose∮E·dA = ?

Only the charge inside the surface counts.

Outside charges do not count

When the charge is outside the surface, its field lines enter on one side and leave unchanged through the other. As many leave as enter, so the net flux is 0. That is why Gauss's law counts only the charge trapped inside the surface. However large a charge sits outside, it adds nothing to the net flux. The field E at each point on the surface, however, is affected by outside charges, so when symmetry is broken the law still holds but cannot give E directly.

With symmetry, E without integrating

With good symmetry, Gauss's law becomes a shortcut. Put a concentric Gaussian sphere around a point charge or a uniformly charged ball, and the field has constant magnitude on the surface and is perpendicular to it, so the integral reduces at once to E·(4πr²). This gives E = Q/(4πε₀r²), which is exactly the Coulomb field of EM-03. For an infinite line choose a cylinder, and for an infinite plane a box-shaped surface; the key is choosing a surface that fits the symmetry, as practiced in EM-01.

Fill inE · 4πr² = ?

With symmetry, the field is constant on the surface and the integral turns into a product.

On your ownE = ? r̂

Gauss's law brings back Coulomb's point-charge field.

Back to the first screen

On the first screen, growing the Gaussian surface did not budge the net flux. As the field on the surface weakened as 1/R², the area grew as R², and the two canceled exactly. Moving the charge outside dropped the flux to 0, which means every field line that came in went back out. In the end, what sets the net flux through a closed surface is not its size or shape but only the charge trapped inside.

Gauss's law states that the net electric flux through a closed surface is proportional only to the charge inside it. ∮E·dA = Q_enc/ε₀. It does not depend on the shape and size of the surface or on outside charges. With good symmetry, E can be pulled out of the integral and the electric field found without integrating.

What comes next

Gauss's law is now written as a surface integral, the integral form. The divergence theorem in the next unit (EM-07) turns this surface integral into a volume integral and draws out the differential form that holds at a point, ∇·E = ρ/ε₀. That is the first of Maxwell's equations. The integral form serves problems with symmetry, and the differential form serves the local structure of a field.