EM-06 · Electrostatics field

Gauss's Law

The net electric flux through a closed surface is proportional only to the charge inside it. Grow the Gaussian surface or move the charge outside, and see for yourself that the flux is independent of the surface size.

Grow the surface, the flux stays

Grow the Gaussian sphere wrapping a point charge. On the surface the field weakens as 1/R² while the area grows as R². What happens to the net flux? And what if you move the charge outside the surface?

Gaussian surface radius RR = 1.50

Drag to orbit. Slider for surface radius R, buttons for charge inside/outside.

Net flux through the closed surface

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

E shrinks as 1/R² and 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 we defined the net flux ∮E·dA through a closed surface. The Gauss law says this value is proportional only to the charge Q inside the surface: ∮E·dA = Q_enc/ε₀. The vacuum permittivity ε₀ is the constant. Whatever the surface, wrapping the same charge gives the same flux.

Observe∮E·dA = Q_encε₀

Net flux is the enclosed charge over ε₀.

Independent of the surface shape and size

Double the sphere around a point charge and the field on the surface drops to a quarter while the area quadruples. Their product, the flux, is unchanged. A dented surface or a cube — same thing. As long as it wraps the same charge, the net flux is always Q/ε₀. Flux equals the number of field lines piercing the surface, and that number is set by the charge.

Choose∮E·dA = ?

Only the charge inside the surface counts.

Outside charges do not count

When a charge sits outside the surface, its field lines enter on one side and leave on the other. As many go in as come out, so the net flux is zero. That is why the Gauss law counts only the charge trapped inside. However large a charge sits outside, it contributes nothing to the net flux — though the field E at each point on the surface is still affected by it.

With symmetry, E without integrating

When symmetry is good, the Gauss law becomes a shortcut. Around a point charge or a sphere, place a concentric Gaussian sphere: E is constant over the surface and perpendicular to it, so the integral collapses to E·(4πr²). That gives E = Q/(4πε₀r²) — exactly the Coulomb field of EM-03. For an infinite line you pick a cylinder, for an infinite plane a box. Choosing the surface that matches the symmetry, learned in EM-01, is the key.

Fill inE · 4πr² = ?

With symmetry, E is constant on the surface and the integral becomes a product.

On your ownE = ? r̂

The Gauss law gives back the Coulomb field of a point charge.

Back to the first screen

On the first screen, growing the Gaussian surface left the net flux untouched: as the field on it weakened by 1/R², the area grew by R², cancelling exactly. Move the charge outside and the flux dropped to zero — every field line that entered also left. In the end, what sets the net flux through a closed surface is not its size or shape, but only the charge trapped inside.

The Gauss 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 is independent of the surface's shape and size and of outside charges. With good symmetry, you pull E out of the surface integral and find the field without integrating.

What comes next

The Gauss law is written here as a surface integral (the integral form). The divergence theorem in the next unit (EM-07) converts this surface integral into a volume integral, pulling out the pointwise differential form ∇·E = ρ/ε₀. That is the first of Maxwell’s equations. The integral form serves symmetric problems; the differential form, the local structure of the field.