Electric Flux and Gauss's Law

Gauss's law is the most powerful tool in electrostatics. It lets you compute electric fields with breathtaking simplicity — whenever the problem has the right symmetry.

Concept

Electric flux through a surface measures the number of field lines crossing it. For a small surface element $d\vec A$ in a field $\vec E$: $d\Phi_E = \vec E\cdot d\vec A = E\,dA\cos\theta,$ where $\theta$ is the angle between $\vec E$ and the outward normal. For an entire surface: $\Phi_E = \int_S \vec E\cdot d\vec A.$

For a closed surface (a "Gaussian surface"), this is written $\oint_S \vec E\cdot d\vec A$.

Gauss's law: $\oint_S \vec E\cdot d\vec A = \frac{Q_\text{enclosed}}{\varepsilon_0}.$

The flux out of any closed surface equals the total charge inside divided by $\varepsilon_0$. Charges outside contribute zero net flux (their incoming lines equal their outgoing lines).

Derivation

Consider a point charge $q$ enclosed by a sphere of radius $r$. By symmetry $\vec E$ is radial with magnitude $kq/r^2$, parallel to $d\vec A$ everywhere on the sphere: $\Phi = \oint E\,dA = \frac{kq}{r^2}\cdot 4\pi r^2 = 4\pi kq = \frac{q}{\varepsilon_0}.$ The $r$ dependence cancels — the flux depends only on the enclosed charge.

For an arbitrary closed surface, deform it from the sphere: any point that moves outward subtends the same solid angle from $q$, so the flux through that patch is unchanged. Charges outside contribute equal positive and negative flux as their field lines enter and leave the surface — net zero.

Worked Example

A point charge $+5\,\mu C$ sits at the centre of a cube of side $10\,cm$. Find the flux through one face.

Total flux through the cube: $\Phi_\text{total} = q/\varepsilon_0 = 5\times 10^{-6} / 8.854\times 10^{-12} \approx 5.65\times 10^5\,N\cdot m^2/C$.

By symmetry, flux through each of the six faces is $\Phi_\text{total}/6 \approx 9.42\times 10^4\,N\cdot m^2/C$.

If instead the charge were at a corner of the cube, you could imagine the corner shared by 8 cubes (octant symmetry): each cube would receive $1/8$ of $q/\varepsilon_0$.

Common Confusions

• Flux depends only on enclosed charge. Charges outside the Gaussian surface produce zero net flux through it, even though they may produce a strong field at every point of the surface.
• Gauss's law is always true, but only useful when symmetry lets you pull $E$ out of the integral.
• Direction of $d\vec A$: always outward for a closed surface.
• Flux is a scalar. Its sign indicates whether lines leave (positive) or enter (negative).

Key Takeaways

• $\Phi_E = \int \vec E\cdot d\vec A$.
• Gauss: $\oint \vec E\cdot d\vec A = Q_\text{enc}/\varepsilon_0$.
• Flux is independent of the shape of the closed surface.
• External charges contribute zero net flux.
• Use symmetry (spherical, cylindrical, planar) to extract $E$.