Gauss's law
(E2.2.1) E2.2 · Gauss's Law →The outward flux through a closed surface equals the enclosed charge divided by ε₀.
| charge enclosed by the surface | C | |
| permittivity of free space | 8.85×10⁻¹² C²/(N·m²) | |
| outward area element | m² | |
| electric field | N/C |
Always true. Useful for finding E only where symmetry lets E leave the integral.
Always true but only useful for finding E directly where symmetry lets E be pulled out of the integral — without that symmetry, the law gives the total flux but not the field itself.
Take the easiest case first: a sphere of radius r centred on a point charge. The field there is
E is perpendicular to the sphere everywhere and has the same size, so the dot product and integral collapse:
Substitute E. Every r cancels — the answer does not depend on the size of the sphere:
Any other closed shape around the same charge is crossed by exactly the same field lines, so it gives the same flux. And a charge outside sends every line in one side and out the other, contributing zero. Superposition then gives the general law:
Independent of the size of the cube; the six faces are equivalent by symmetry. The same result by direct integration is laborious.
The field on the surface is not zero; entering and leaving flux cancel.
The law is always true, but only <i>useful</i> when symmetry lets you pull E out of the integral. Without symmetry you know the flux but cannot extract the field.