1A · Gauss Law & Flux — Proof + Solved PYQs

This page covers Gauss's law in electrostatics. We define electric flux, state the theorem, and prove it step-by-step. The proof is the most repeated question in Section 1 (asked 4 times across 2018–2023), so the full derivation is given here.

1. Electric flux

Flux counts how many field lines cross a surface. Take a small patch $d\vec{S}$ on the surface — its arrow is the outward normal. The flux through the patch is $d\Phi = \vec{E}\cdot d\vec{S} = E\,dS\cos\theta$. Adding all patches gives the total flux through the surface:

The circle on the integral $\oint$ means the surface is closed. Think of the closed surface as a net that catches field lines.

2. Gauss's theorem — statement

For any closed surface in electrostatics, the total outward flux equals the enclosed charge divided by $\varepsilon_0$:

3. Gauss's theorem — proof

We use Coulomb's result: the field of a point charge $Q$ at distance $r$ is $\vec{E} = \dfrac{Q}{4\pi\varepsilon_0 r^2}\hat{r}$ (radially outward).

Fig 1.1: Gaussian sphere of radius $r$ centred on $+Q$. On the surface, $\vec{E}$ is everywhere parallel to $d\vec{S}$, so $\vec{E}\cdot d\vec{S} = E\,dS$. Integrating gives $\Phi = E\cdot 4\pi r^2 = Q/\varepsilon_0$.

3.1 Proof for a spherical surface

Take a sphere of radius $r$ centred on the charge $Q$. Every point on the sphere is at distance $r$ from $Q$.

1. On the sphere, $\vec{E}\parallel d\vec{S}$, so $\vec{E}\cdot d\vec{S} = E\,dS$ (reason: $\cos 0° = 1$).
2. $E$ is constant on the sphere (reason: every point is at the same distance $r$ from $Q$).
3. Take $E$ outside the integral: $\Phi = E\oint dS = E\cdot 4\pi r^2$ (reason: area of sphere is $4\pi r^2$).
4. Substitute $E$: $\Phi = \dfrac{Q}{4\pi\varepsilon_0 r^2}\cdot 4\pi r^2 = \dfrac{Q}{\varepsilon_0}$ (reason: the $r^2$ cancels — area grows as $r^2$, field falls as $1/r^2$).

This proves Gauss's law for one charge inside a sphere.

3.2 Proof for any closed surface

Now consider a closed surface of any shape around the same charge $Q$. A small patch $dS$ at distance $r$ from $Q$, with its normal making angle $\theta$ with $\hat{r}$, subtends a solid angle $d\Omega = dS\cos\theta/r^2$. The flux through it is:

1. Flux through patch: $d\Phi = E\cos\theta\,dS = \dfrac{Q}{4\pi\varepsilon_0 r^2}\cdot \dfrac{\cos\theta\,dS}{1}$ (reason: $E = Q/4\pi\varepsilon_0 r^2$).
2. Rewrite as $d\Phi = \dfrac{Q}{4\pi\varepsilon_0}\cdot \dfrac{dS\cos\theta}{r^2} = \dfrac{Q}{4\pi\varepsilon_0}\,d\Omega$ (reason: definition of solid angle).
3. Integrate over the closed surface: $\Phi = \dfrac{Q}{4\pi\varepsilon_0}\oint d\Omega = \dfrac{Q}{4\pi\varepsilon_0}\cdot 4\pi = \dfrac{Q}{\varepsilon_0}$ (reason: closed surface subtends $4\pi$ sr).
4. The shape does not change the answer — only the enclosed charge matters.

3.3 Proof for many charges (superposition)

If there are $n$ charges $Q_1, Q_2, \dots, Q_n$ inside the surface, the total field is the sum $\vec{E} = \sum_i \vec{E}_i$ by superposition.

1. Total flux: $\Phi = \oint \vec{E}\cdot d\vec{S} = \oint \sum_i \vec{E}_i\cdot d\vec{S} = \sum_i \oint \vec{E}_i\cdot d\vec{S}$ (reason: linearity of the integral).
2. Each $\oint \vec{E}_i\cdot d\vec{S} = Q_i/\varepsilon_0$ by the single-charge proof above.
3. Add: $\Phi = \sum_i Q_i/\varepsilon_0 = Q_{\text{in}}/\varepsilon_0$ (reason: sum of enclosed charges is $Q_{\text{in}}$).
4. Charges outside contribute zero net flux: every line that enters the closed surface also leaves it.

4. Differential form of Gauss's law

Shrink the closed surface to a point. Use the divergence theorem $\oint \vec{E}\cdot d\vec{S} = \int (\nabla\cdot\vec{E})\,dV$ and write $Q_{\text{in}} = \int \rho\,dV$. Both sides of Gauss's law become volume integrals, so the integrands must be equal at every point:

Solved PYQs

Reference. Brij Lal & Subrahmanyam, Electricity & Magnetism Ch-2 (flux, Gauss theorem, differential form).