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Gauss's Law

Total electric flux through any closed surface equals the net enclosed charge divided by ε₀. Equivalent to Coulomb's law for static fields.
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Derivation

Proof for a spherical surface enclosing a point charge

Place charge qq at the centre of a sphere of radius rr. The field everywhere on the surface:

E⃗=q4πε0r2r^\vec{E} = \frac{q}{4\pi\varepsilon_0 r^2}\hat{r}

r^\hat{r} is parallel to dA⃗d\vec{A} at every point (both radially outward). So:

Φ=∮SE⃗⋅dA⃗=q4πε0r2∮SdA=q4πε0r2⋅4πr2=qε0\Phi = \oint_S \vec{E} \cdot d\vec{A} = \frac{q}{4\pi\varepsilon_0 r^2}\oint_S dA = \frac{q}{4\pi\varepsilon_0 r^2} \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}

The r2r^2 from the inverse-square law exactly cancels the r2r^2 from the surface area. This is not a coincidence — it is a direct consequence of the 1/r21/r^2 nature of the Coulomb force.

Extending to arbitrary surfaces

The flux through any closed surface enclosing qq equals q/ε0q/\varepsilon_0, regardless of the shape of the surface. This follows from the solid angle argument: every field line from qq that passes through the sphere also passes through any enclosing surface.

Multiple charges and superposition

By superposition, the total flux through a closed surface:

Φ=∑iqiε0=qencε0\Phi = \sum_i \frac{q_i}{\varepsilon_0} = \frac{q_{enc}}{\varepsilon_0}

where qencq_{enc} is the algebraic sum of all charges inside.

Charges outside the surface contribute zero net flux — field lines entering from outside must also exit.

∮SE⃗⋅dA⃗=qencε0\boxed{\oint_S \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}}
Remember
Gauss's law is equivalent to Coulomb's law for static fields in vacuum, but is far more powerful for problems with symmetry. It also generalises to time-varying fields (as part of Maxwell's equations) while Coulomb's law does not.