AcademyGauss's Law
Academy
Gauss's Law
Level 1 - Physics topic page in Gauss's Law.
Principle
Gauss's law links the net electric flux through a closed surface to the charge enclosed.
Notation
\(\Phi_E\)
net electric flux through a closed surface
\(\mathrm{N\,m^{2}\,C^{-1}}\)
\(\vec E\)
electric field
\(\mathrm{N\,C^{-1}}\)
\(d\vec A\)
outward area vector
\(\mathrm{m^{2}}\)
\(q_{\mathrm{enc}}\)
net charge enclosed by the Gaussian surface
\(\mathrm{C}\)
\(\epsilon_0\)
permittivity of free space
\(\mathrm{C^{2}\,N^{-1}\,m^{-2}}\)
\(S\)
closed Gaussian surface
1
Method
Derivation 1: Check a point charge with a sphere
For a point charge at the center of a sphere, the electric field is radial and has the same magnitude everywhere on the sphere.
Point-charge field
\[E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\]
Spherical area
\[A=4\pi r^2\]
Flux product
\[\Phi_E=EA=\frac{q}{\epsilon_0}\]
Derivation 2: State the closed-surface law
The result depends only on the net charge inside the closed surface. Charges outside can reshape the local field, but their field lines enter and leave the surface in equal net amount.
Closed flux
\[\Phi_E=\oint_S\vec E\cdot d\vec A\]
Gauss's law
\[\oint_S\vec E\cdot d\vec A=\frac{q_{\mathrm{enc}}}{\epsilon_0}\]
No enclosed charge
\[q_{\mathrm{enc}}=0\Rightarrow\Phi_E=0\]
Derivation 3: Use the law only after choosing the surface
A Gaussian surface is imaginary. It helps find \(\vec E\) only when symmetry makes the flux integral simple.
Choose surface
\[S\ \text{matches the field symmetry}\]
Simplify flux
\[\oint_S\vec E\cdot d\vec A=EA\]
Solve field
\[E=\frac{q_{\mathrm{enc}}}{\epsilon_0 A}\]
Rules
These are the compact forms of Gauss's law.
Gauss's law
\[\oint_S\vec E\cdot d\vec A=\frac{q_{\mathrm{enc}}}{\epsilon_0}\]
Net flux
\[\Phi_E=\frac{q_{\mathrm{enc}}}{\epsilon_0}\]
Zero enclosed charge
\[q_{\mathrm{enc}}=0\Rightarrow\Phi_E=0\]
Symmetric field step
\[EA=\frac{q_{\mathrm{enc}}}{\epsilon_0}\]
Examples
Question
A closed surface encloses
\[+4.0\,\mathrm{nC}\]
Find the net electric flux.Answer
\[\Phi_E=\frac{q_{\mathrm{enc}}}{\epsilon_0}=\frac{4.0\times10^{-9}}{8.85\times10^{-12}}=4.5\times10^2\,\mathrm{N\,m^2\,C^{-1}}\]
Checks
- Gauss's law uses net enclosed charge, not nearby charge.
- The surface must be closed for \(q_\{\\mathrm\{enc\}}/\\epsilon_0\).
- Zero net flux does not imply zero field everywhere on the surface.
- Gauss's law gives flux always, but field magnitude only when symmetry simplifies the integral.