Electric Flux and Gauss’s Theorem — Notes

• Area vector $\Delta\vec S$: magnitude = area, direction normal to the surface (outward for a closed surface).
• Uniform field, plane area: $\Delta\Phi = \vec E\cdot\Delta\vec S = E\Delta S\cos\theta$ (θ between E and the normal): maximum at 0°, zero at 90°, negative (−EΔS) at 180°.
• General: $\Phi = \oint \vec E\cdot d\vec S$ (surface integral). Flux is a scalar; unit N m² C⁻¹ (= V m).

• Statement: the total electric flux through any closed surface is $1/\varepsilon_0$ times the net charge enclosed: $\oint \vec E\cdot d\vec S = q_\text{enc}/\varepsilon_0$.
• Charges outside contribute zero net flux (they still affect E at the surface). Flux is independent of the shape and size of the surface.
• Charge at the centre of a cube: $q/(6\varepsilon_0)$ through each face.
• Gaussian surface: an imaginary closed surface chosen so that over each part E is either normal and of constant size, or parallel to the surface; it must not pass through a point charge.
• Gauss’s theorem and Coulomb’s law are equivalent for static charges.

Coaxial cylinder (radius r, length l): flux only through the curved surface, $E \cdot 2\pi rl = \lambda l/\varepsilon_0$, so $E = \dfrac{\lambda}{2\pi\varepsilon_0 r} = \dfrac{2k\lambda}{r}$, radially outward for λ > 0; $E \propto 1/r$.

Pillbox of cross-section A through the sheet: flux through the two ends only, $2EA = \sigma A/\varepsilon_0$, so $E = \dfrac{\sigma}{2\varepsilon_0}$, independent of distance, normal to the sheet. Two parallel sheets $+\sigma$ and $-\sigma$: $E = \sigma/\varepsilon_0$ between them (from + to −), $E = 0$ outside.

• Outside ($r > R$): $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ — as if all the charge were at the centre.
• On the surface ($r = R$): $E = \dfrac{q}{4\pi\varepsilon_0R^2} = \dfrac{\sigma}{\varepsilon_0}$, with $\sigma = q/4\pi R^2$.
• Inside ($r < R$): $q_\text{enc} = 0$, so $E = 0$ (electrostatic shielding).
• Graph of E against r: zero up to R, a jump to its maximum at R, then $\propto 1/r^2$.