$$ \oint_S\vec{E}\cdot \text{d} \vec{A}=\frac{Q}{\epsilon_0} $$

• $\oint$ is a closed integral, in this case over a closed surface, $S$
• $\vec{E}$ is the electric field
• $\text{d} \vec{A}$ is an infinitesimal area element
• $\vec E\cdot\text{d}\vec A$ is the dot product
• $Q$ is the total charge encloses within the surface $S$
• $\epsilon_0$ is the permittivity of free space

Area vector

• $\text{d} \vec{A}$ is the vector perpendicular to the infinitesimal area

Example: Cartesian coordinates:

• $\text{d}A=\text{d}x\,\text{d}y$
• $\text{d}\vec A=\text{d}x\,\text{d}y \,\hat{k}$ where $\hat k$ is in the direction perpendicular to the area

Example of an area vector for cartesian coordinates

Example of an area vector for cartesian coordinates

Gauss’ law for a point charge

$$ \oint_S\vec{E}\cdot \text{d} \vec{A}=\oint_SE\, \text{d} A=E\oint_S\, \text{d} A=\boxed{E\,4\pi r^2}=\frac{q}{4\pi\epsilon_0 r^2}4\pi r^2=\boxed{\frac{q}{\epsilon_0}} $$

• $S$ is the spherical shell at a radius $r$
• At all points of the surface, $E$ and $\text d \vec A$ are parallel $\Rightarrow \vec E \cdot \text d \vec A = E \, \text d A$
• The electric field strength is the same at all points on the surface, which is why it was moved out of the integral
• $E=\frac{q}{4\pi\epsilon_0 r^2}$
• We see for Coulomb’s law is consistent with Gauss’ law

In red is the spherical gaussian surface $\text d \vec A$, the orange and yellow vectors are the Electric field $\vec E$, and the center sphere is the charge $q$

In red is the spherical gaussian surface $\text d \vec A$, the orange and yellow vectors are the Electric field $\vec E$, and the center sphere is the charge $q$

Generalised use of gauss’s law

• Gauss’ law is applicable for any charge distribution and any close surface
• To use Gauss’ law its best to exploit symmetries such as:
  • Uniform sphere of charge
  • Infinite line of charge
  • Infinite plane of charge

Examples:

1) Infinite line charge:

$$ \oint_S\vec{E}\cdot \text{d} \vec{A}=\frac{Q}{\epsilon_0}=\int_{\text{top}}\vec{E}\cdot \text{d} \vec{A}+\int_{\text{bot}}\vec{E}\cdot \text{d} \vec{A}+\int_{\text{side}}\vec{E}\cdot \text{d} \vec{A} $$

• By symmetry we can see $\vec E$ is purely radial
• For top and bottom $\vec E \perp \text d \vec A$ so no contribution as the dot product $=0$
• For sides $\vec E \parallel \text d \vec A \Rightarrow \vec E \cdot \text d \vec A = E \,\text dA$
• By symmetry $|E|$ is uniform thus: $\int_\text{side} E \text d A = E \int_\text{side} \text d A$
• $\int_\text{side} \text dA= 2 \pi r l$
• $Q=\lambda l$ where $\lambda$ is the charge unit length

$$ \oint_S\vec{E}\cdot \text{d} \vec{A}=\frac{\lambda l}{\epsilon_0}=E2\pi rl\\\Rightarrow \vec E = \frac{\lambda}{2\pi r \epsilon_0} \hat{r} $$

2) infinite plane charge:

$$ \oint_S\vec{E}\cdot \text{d} \vec{A}=\frac{Q}{\epsilon_0}=\int_{\text{top}}\vec{E}\cdot \text{d} \vec{A}+\int_{\text{bot}}\vec{E}\cdot \text{d} \vec{A}+\int_{\text{side}}\vec{E}\cdot \text{d} \vec{A} $$