Gauss's law — integral form, choosing Gaussian surfaces

WHAT is Gauss's Law?

Electric flux $\Phi_E = \oint \vec E\cdot d\vec A$ measures how much field "flows through" the surface. The dot product picks out the component of $\vec E$ along the outward normal — sideways field contributes nothing.

WHY is it true? (Derive from Coulomb's law)

We build it from scratch, step by step.

Step 1 — Flux of a single point charge through a sphere. Put charge $q$ at the centre of a sphere of radius $r$. By Coulomb's law, $\vec E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat r.$ Why this step? We start with the only field law we already trust (Coulomb), in the simplest symmetric case.

On the sphere $\vec E \parallel d\vec A$ (both radial), and $|\vec E|$ is constant. So $\Phi = \oint E\,dA = E\oint dA = \frac{q}{4\pi\varepsilon_0 r^2}\cdot(4\pi r^2)=\frac{q}{\varepsilon_0}.$ Why this step? The $r^2$ in Coulomb's denominator cancels the $r^2$ in the sphere's area. The radius vanished — flux is the same for any sphere. That cancellation is the secret heart of Gauss's law.

Step 2 — Any closed shape, charge inside. A field line from $q$ that exits a tilted patch crosses an area $dA$, but only the perpendicular bit $dA\cos\theta$ catches lines, while the patch is further away by exactly the factor that grows $r^2$. The two effects cancel, so solid angle is what counts: $d\Phi = \frac{q}{4\pi\varepsilon_0}\,d\Omega,\qquad \oint d\Omega = 4\pi \ \Rightarrow\ \Phi=\frac{q}{\varepsilon_0}.$ Why this step? It shows the sphere was not special — any enclosing surface gives $q/\varepsilon_0$.

Step 3 — Charge outside. A field line entering one side of the surface must leave the other side: it pierces an even number of times, contributing $+$ then $-$. Net $\Phi = 0$ for outside charges. Why this step? This proves outside charges don't enter $Q_{\text{enc}}$.

Step 4 — Superposition. Real fields add: $\vec E = \sum \vec E_i$. Flux is linear, so $\Phi = \sum_i \frac{q_i^{\text{(inside)}}}{\varepsilon_0} = \frac{Q_{\text{enc}}}{\varepsilon_0}.\qquad\blacksquare$

HOW to use it: choosing the Gaussian surface

The law is always true but only solvable when you can pull $E$ out of the integral. Choose a surface so that on each face either:

1. $\vec E$ is constant in magnitude and parallel to $d\vec A$ ($\theta=0$), so $\int E\,dA = EA$, or
2. $\vec E\perp d\vec A$ ($\theta=90°$), so the flux is zero.

Worked examples

Example 1 — Field of a point charge

Find $\vec E$ at distance $r$ from charge $q$.

• Surface: sphere radius $r$. Why? Spherical symmetry → $E$ depends only on $r$ and points radially.
• On sphere: $\Phi = E\,(4\pi r^2)$. Why? $E$ constant & parallel to normal.
• Gauss: $E\,4\pi r^2 = q/\varepsilon_0 \Rightarrow E = \dfrac{q}{4\pi\varepsilon_0 r^2}$. We recovered Coulomb — consistency check passed.

Example 2 — Infinite line charge, linear density $\lambda$

• Surface: cylinder radius $r$, length $L$, coaxial.
• Curved face: $\vec E$ radial, $\parallel d\vec A$, constant → flux $= E\,(2\pi r L)$.
• Flat end caps: $\vec E \perp d\vec A$ → flux $=0$. Why? Field points radially, caps face along axis.
• $Q_{\text{enc}} = \lambda L$. $E\,(2\pi r L) = \frac{\lambda L}{\varepsilon_0}\ \Rightarrow\ \boxed{E=\frac{\lambda}{2\pi\varepsilon_0 r}}$ Why the length $L$ cancels? The trapped charge and the curved area both scale with $L$.

Example 3 — Infinite charged plane, surface density $\sigma$

• Surface: pillbox, cross-section area $A$, straddling the sheet symmetrically.
• Two flat faces: $\vec E\parallel d\vec A$, total flux $= 2EA$ (field exits both sides).
• Side wall: $\vec E\perp d\vec A$ → 0.
• $Q_{\text{enc}}=\sigma A$. $2EA=\frac{\sigma A}{\varepsilon_0}\ \Rightarrow\ \boxed{E=\frac{\sigma}{2\varepsilon_0}}$ Why the factor 2? Flux escapes through both faces. Why $E$ independent of distance? Infinite plane = constant solid-angle coverage; field is uniform.

Example 4 — Uniformly charged solid sphere, radius $R$, total $Q$, inside ($r<R$)

• Surface: sphere radius $r<R$.
• $Q_{\text{enc}} = Q\dfrac{r^3}{R^3}$ (charge scales with enclosed volume). $E\,4\pi r^2=\frac{Q r^3/R^3}{\varepsilon_0}\Rightarrow \boxed{E=\frac{Qr}{4\pi\varepsilon_0 R^3}}$ Why linear in $r$ inside? Less charge is enclosed as you go deeper; $E\to0$ at the centre.

Common mistakes (steel-manned)

Flashcards

Connections

• Coulomb's law — Gauss's law is derived from it; they're equivalent for statics.
• Electric flux — the quantity Gauss's law constrains.
• Electric field of conductors — field inside = 0, surface field $\sigma/\varepsilon_0$ via a pillbox.
• Divergence theorem — converts integral form to the differential form $\nabla\cdot\vec E=\rho/\varepsilon_0$.
• Maxwell's equations — Gauss's law is the first of the four.
• Symmetry in physics — why surface choice works.

Concept Map

Hinglish (regional understanding)