electric flux and Gauss theorem

Coulomb's law gives the field of a point charge, and the superposition principle lets us add the fields of several charges. Electric flux $\Phi_E$ through a surface measures how much field pierces that surface: for a small area $d\Phi_E=\vec E\cdot d\vec A$, and for a finite surface $\Phi_E=\int \vec E\cdot d\vec A$. For a uniform field and a flat area, $\Phi_E=E A\cos\theta$, where $\theta$ is the angle between the field $\vec E$ and the normal $\vec n$ to the area. Gauss's theorem states that the flux through a closed surface equals $\oint \vec E\cdot d\vec A = Q_{\text{enc}}/\varepsilon_0$; charge outside the surface does not change this flux. The sign of the flux is set by the outward normal convention: field leaving the surface gives positive flux. Symmetry of the problem — spherical, cylindrical, or planar — often reduces the integral to a simple product and lets us find the field from $Q_{\text{enc}}$.

A uniform electric field $E=200\,\mathrm{N/C}$ passes through a flat area $A=0.50\,\mathrm{m^2}$. The normal to the area makes an angle $\theta=60^\circ$ with the field vector $\vec E$. Find the electric flux through the area.

A point charge $Q=+2.0\,\mu\mathrm{C}$ is at the center of a spherical surface of radius $r=0.10\,\mathrm{m}$. Find the electric flux through this closed sphere and the magnitude of the electric field on its surface. Use $\varepsilon_0=8.854\times10^{-12}\,\mathrm{C^2/(N\,m^2)}$.

A cube of side $a=2.0\,\mathrm{m}$ lies in the first octant with one corner at the origin and edges along the $x$, $y$, and $z$ axes. In this region the field is $\vec E = (5x)\,\hat i\ \mathrm{N/C}$, where $x$ is in meters. Find the net flux through the cube and the charge inside the cube implied by Gauss's theorem. Use $\varepsilon_0=8.854\times10^{-12}\,\mathrm{C^2/(N\,m^2)}$.

Carry out a safe independent investigation in an online electrostatics simulation such as PhET 'Charges and Fields' or 'Gauss's Law', or on paper with a spreadsheet. Place a virtual point charge $+1\,\mathrm{nC}$ at the center and construct a spherical Gaussian surface of radius $r$. Record the field magnitude $E$ on the surface, compute $E\cdot 4\pi r^2$, and compare it with $Q/\varepsilon_0$. Repeat for $r=1,2,3$ arbitrary units and confirm that the flux does not depend on radius. Then add a second charge outside the sphere and verify that the flux through the sphere is unchanged. Do not use real high-voltage sources or household mains electricity.

Show that flux is the projection of area onto the field direction and illustrate why Gauss's theorem links the net flux only to the charge inside a closed surface.

A two-panel sketch. Left panel: a uniform field is drawn as parallel arrows $\vec E$, with a tilted area $A$, its normal $\vec n$, and the angle $\theta$; shading shows the projection $A\cos\theta$ through which field lines pass. Right panel: a closed spherical surface around charge $+Q$, outward normals $d\vec A$ at several points, and radial field lines; beside it write $\oint \vec E\cdot d\vec A = Q_{\text{enc}}/\varepsilon_0$.

A phone screen of area $A=0.012\,\text{m}^2$ gets a uniform surface charge density $\sigma=5.0\times10^{-6}\,\text{C/m}^2$ after being rubbed in a pocket. A Gaussian pillbox with one face just outside the screen and one just inside encloses the charged patch. The enclosed charge is $Q=\sigma A=(5.0\times10^{-6})(0.012)=6.0\times10^{-8}\,\text{C}$.

Gauss's theorem gives the flux through the closed pillbox: $\Phi_E=Q/\varepsilon_0=(6.0\times10^{-8})/(8.85\times10^{-12})\approx6.8\times10^3\,\text{N}\cdot\text{m}^2/\text{C}$. By symmetry the field just outside the thin charged sheet is $E=\sigma/(2\varepsilon_0)\approx2.8\times10^5\,\text{N/C}$. This exercises the choice of Gaussian surface and the link between flux and enclosed charge.

An insulated charging cable for an electric bus has a uniform linear charge density $\lambda=2.0\times10^{-6}\,\text{C/m}$. A cylindrical Gaussian surface of radius $r=1.5\,\text{m}$ and length $L=10\,\text{m}$ is coaxial with the cable. The enclosed charge is $Q=\lambda L=(2.0\times10^{-6})(10)=2.0\times10^{-5}\,\text{C}$.

Gauss's theorem gives $\Phi_E=Q/\varepsilon_0=(2.0\times10^{-5})/(8.85\times10^{-12})\approx2.3\times10^6\,\text{N}\cdot\text{m}^2/\text{C}$. The radial field at $r=1.5\,\text{m}$ follows as $E=\lambda/(2\pi\varepsilon_0 r)\approx2.4\times10^4\,\text{N/C}$. The example uses cylindrical symmetry and the fact that the flux through the curved surface equals $E(2\pi rL)$.

A storm cloud has a net charge $Q=-20\,\text{C}$ concentrated in a region. A spherical Gaussian surface of radius $R=1500\,\text{m}$ is centered on that region. The enclosed charge is $Q=-20\,\text{C}$; the flux is the same for every sphere that encloses this charge.

Gauss's theorem gives $\Phi_E=Q/\varepsilon_0=(-20)/(8.85\times10^{-12})\approx-2.3\times10^{12}\,\text{N}\cdot\text{m}^2/\text{C}$. With spherical symmetry the field magnitude at $R=1500\,\text{m}$ is $E=|Q|/(4\pi\varepsilon_0 R^2)\approx8.0\times10^4\,\text{N/C}$. The example shows that flux depends on enclosed charge, not on the radius of the chosen Gaussian sphere.