Gauss's law in integral form

Put a point charge q at the centre of an imaginary sphere of radius r. Coulomb's law gives a field of magnitude q/4πε₀r² pointing radially outward, and on a sphere centred on the charge the outward area element dA is radial too, so E ⋅ dA is simply E dA on every patch. E has the same magnitude everywhere on that sphere, so it comes straight out of the integral: Φ = E × 4πr² = (q/4πε₀r²)(4πr²) = q/ε₀. Watch what just happened. The r² in the denominator of Coulomb's law was cancelled by the r² in the area of the sphere, and the radius vanished from the answer. Take q = 3.00 nC. At r = 0.100 m the field is 2.70 × 10³ N C⁻¹ across an area of 0.1257 m², giving 339 N m² C⁻¹; at r = 0.200 m the field has fallen to 674 N C⁻¹ but the area has grown to 0.5027 m², and the product is 339 N m² C⁻¹ again. Only an inverse-square force law does this. Under a 1/r³ law the flux would leak away with distance and there would be no Gauss's law to write.

Nothing in that argument obviously survives if the surface is a cube or a crumpled bag. Solid angle shows that it does. Draw a narrow cone of solid angle dΩ outward from the charge. Wherever the cone meets the closed surface, at slant distance r and with the surface tilted at angle θ to the cone's axis, the patch it cuts has area dA = r² dΩ / cos θ. The flux through that patch is E dA cos θ = (q/4πε₀r²)(r² dΩ) = (q/4πε₀) dΩ, and both r and θ have dropped out. Every cone contributes the same amount however far away or however steeply tilted the surface happens to be there, because pushing the surface outward weakens the field by exactly the factor by which it enlarges the patch. Now add the cones. The solid angle all around an interior point is 4π steradians, so Φ = (q/4πε₀)(4π) = q/ε₀ for any closed surface whatever that encloses q. Shape has been eliminated from the problem.

Now move the charge outside the closed surface and run the same cones. A cone from an external charge either misses the surface entirely or pierces it an even number of times — twice for a convex surface, once going in and once coming out. On the entry patch the field points into the volume, so E ⋅ dA is negative; on the exit patch it points out and E ⋅ dA is positive. Both patches subtend the same solid angle dΩ, so both have magnitude (q/4πε₀) dΩ, and they cancel exactly. Sum over every cone and an outside charge contributes zero net flux, however close it sits and however large it is. Be careful about what has just been proved. Nothing here says the external charge produces no field on the surface. A +10.0 nC charge parked 1.00 mm outside makes a field of about 9 × 10⁷ N C⁻¹ at the nearest patch, far larger than anything inside is likely to make there. Its flux in and its flux out merely happen to be equal.

Real problems have many charges, and superposition finishes the job. The total field on the surface is the vector sum of the field of every charge, inside and out, and the surface integral is linear in the field, so the total flux is the sum of the fluxes each charge would produce on its own. Each interior charge contributes qᵢ/ε₀ and each exterior charge contributes zero, so adding them gives ∮E ⋅ dA = Q_enc/ε₀ with Q_enc the algebraic sum of the enclosed charges, signs included. Suppose a surface encloses +12.0 nC and −4.50 nC while a −20 nC charge sits somewhere outside. Then Q_enc = +7.50 nC and Φ = 7.50 × 10⁻⁹ C ÷ 8.854 × 10⁻¹² C² N⁻¹ m⁻² = +847 N m² C⁻¹, the positive sign meaning net outward. When the charge is smeared rather than pointlike the same sum becomes an integral, Q_enc = ∫ρ dV over the volume the surface bounds — and only over that volume.

ε₀ is not a new piece of physics; it is Coulomb's constant wearing different clothes. SI fixes the unit of charge through the ampere, so the force law needs an experimental constant out front: F = k q₁q₂/r² with k = 8.988 × 10⁹ N m² C⁻². Writing k as 1/4πε₀ moves the 4π off the force law and parks it beside the 4π that the surface area of a sphere was always going to supply, which is exactly why Gauss's law emerges with no stray numerical factor. That gives ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻², the same number as 8.854 × 10⁻¹² F m⁻¹ when capacitance arrives later. Check the units of the law itself: C ÷ (C² N⁻¹ m⁻²) = N m² C⁻¹, which is the unit of flux and equals V m. So the flux out of a closed surface is readable as a measurement of charge: one coulomb inside registers 1.13 × 10¹¹ N m² C⁻¹ of outward flux.

Gauss's law holds for every closed surface you can imagine drawing, around any arrangement of charges, static or moving — it is the first of Maxwell's equations, and no experiment has caught it failing. But it is one scalar equation, and an electric field is three functions of three coordinates. One number per surface cannot possibly determine all of that. What the law pins down is an average: the outward component of E summed over the whole surface, not its value anywhere on it. Draw a 3 cm sphere around one of two equal charges held 10 cm apart and the flux through it is exactly q/ε₀, yet the field on that sphere is lopsided and the law will not tell you its value at a single point. Gauss's law delivers E only when you can argue in advance, from symmetry, that |E| is constant over the surface and meets dA at a fixed angle, so that ∮E ⋅ dA collapses to E × A and can be solved for E. Hunting for those surfaces is the next topic's work.