AcademyGauss's Law
Academy
Symmetry Applications of Gauss's Law
Level 1 - Physics topic page in Gauss's Law.
Principle
Gauss's law solves fields directly only when symmetry makes the flux integral simple.
Notation
\(\vec E\)
electric field
\(\mathrm{N\,C^{-1}}\)
\(r,R\)
field radius and source radius
\(\mathrm{m}\)
\(q_{\mathrm{enc}}\)
charge enclosed by the Gaussian surface
\(\mathrm{C}\)
\(\lambda\)
linear charge density
\(\mathrm{C\,m^{-1}}\)
\(\sigma\)
surface charge density
\(\mathrm{C\,m^{-2}}\)
\(\epsilon_0\)
permittivity of free space
\(\mathrm{C^{2}\,N^{-1}\,m^{-2}}\)
Method
Derivation 1: Spherical symmetry
For a spherically symmetric charge distribution, the field is radial and has one magnitude at a fixed radius.
Gaussian sphere
\[A=4\pi r^2\]
Flux
\[\oint\vec E\cdot d\vec A=E(4\pi r^2)\]
Gauss's law
\[E(4\pi r^2)=\frac{q_{\mathrm{enc}}}{\epsilon_0}\]
Spherical field
\[E=\frac{q_{\mathrm{enc}}}{4\pi\epsilon_0r^2}\]
Derivation 2: Cylindrical symmetry
For a long uniform line charge, choose a coaxial cylinder. The curved side contributes; the end caps do not.
Curved area
\[A_{\mathrm{side}}=2\pi r\ell\]
Enclosed charge
\[q_{\mathrm{enc}}=\lambda\ell\]
Gauss's law
\[E(2\pi r\ell)=\frac{\lambda\ell}{\epsilon_0}\]
Line-charge field
\[E=\frac{\lambda}{2\pi\epsilon_0r}\]
Derivation 3: Planar symmetry
For an infinite charged sheet, use a pillbox that crosses the sheet. Equal flux leaves through both caps.
Pillbox caps
\[\Phi_E=EA+EA=2EA\]
Enclosed charge
\[q_{\mathrm{enc}}=\sigma A\]
Gauss's law
\[2EA=\frac{\sigma A}{\epsilon_0}\]
Sheet field
\[E=\frac{\sigma}{2\epsilon_0}\]
Rules
Use these only when the stated symmetry assumptions apply.
Spherical symmetry
\[E=\frac{q_{\mathrm{enc}}}{4\pi\epsilon_0r^2}\]
Outside sphere
\[E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\quad(r\ge R)\]
Uniform solid sphere
\[E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}\quad(r<R)\]
Line charge
\[E=\frac{\lambda}{2\pi\epsilon_0r}\]
Infinite sheet
\[E=\frac{\sigma}{2\epsilon_0}\]
Examples
Question
Find the field
\[0.40\,\mathrm{m}\]
from a long line with \[\lambda=5.0\,\mathrm{nC\,m^{-1}}\]
Answer
Use a coaxial cylinder.
\[E=\frac{\lambda}{2\pi\epsilon_0r}=\frac{5.0\times10^{-9}}{2\pi(8.85\times10^{-12})(0.40)}=2.25\times10^2\,\mathrm{N\,C^{-1}}\]
Checks
- First identify symmetry; then choose the Gaussian surface.
- Pull \(E\) out of the integral only where it is constant.
- End caps of a line-charge cylinder have zero flux because the field is radial.
- Infinite-sheet field does not decrease with distance in the ideal model.