Gauss’s Law

Definition

$\oint_{S} \vec{E} \cdot \hat{n} d a = \frac{Q}{ε_{0}}$ . $\hat{n}$ pointing out of the surface (positively oriented).

Verifying with Coulomb’s Law

Single Point Charge

$\frac{q}{4 π ε_{0}} \oint_{S} \frac{\hat{r} \cdot \hat{n}}{r^{2}} d r = \frac{q}{4 π ε_{0}} \oint_{S} \frac{\cos θ d r}{r^{2}} = \frac{q}{4 π ε_{0}} \oint_{S} \frac{d a_{⊥}}{r^{2}} = \frac{q}{4 π ε_{0}} \oint_{S} d Ω = \frac{q}{ε_{0}}$ Note, this surface is closed so the integral resolves to the total solid angle.

Note, $ℓ = r θ$ so $A = r^{2} Ω$

Two Point Charges

Using linearity of Electric field, we see that Coulomb’s Law proves Gauss’s Law.

Verifying with Divergence Theorem

$\oint_{S} \vec{E} \cdot \vec{n} d a = \int_{V} \nabla \cdot \vec{E} d V = \frac{1}{ε_{0}} \int_{V} ρ d V$ .