Gauss's Law

Solving for the electric field of a single particle is simple, but solving for the electric field with space is more challenging. To do this, we must enclose the space in a Gaussian shape and then use Gauss’s Law.

Gauss’s Law states that the electric flux through a closed surface $\propto$ the charge enclosed within that surface.

$${\varphi }_E=\oint \vec{E}d\vec{A}=\frac{Q_{enc}}{{ϵ}_0}$$

Uniform Charge Density §

Electric Field Outside of Object §

To demonstrate Gauss’s Law in this situation, we take a cylinder with radius $R$, height $L$, and charge $Q$.

$$\begin{array}{rl}\varphi =\oint \vec{E}d\vec{A}& =\frac{Q_{enc}}{{ϵ}_0}\\ ⟹\vec{E}\cdot 2\pi rL& =\frac{Q}{{ϵ}_0}\\ ⟹\vec{E}& =\frac{Q}{2\pi rL{ϵ}_0}\end{array}$$

This holds true for both conductors and non-conductors.

Electric Field Inside Object §

Conductors §

Conductors have an internal electric field of 0.

Non-Conductors §

For this, we need the density of the charge, $\rho$.

$$\begin{array}{rl}\vec{E}\cdot 2\pi rL& =\frac{Q_{enc}}{{ϵ}_0}\\ \rho & =\frac{Q_{enc}}{V_{enc}}=\frac{Q}{V}\\ ⟹\frac{Q}{\pi R^{2}L}& =\frac{Q_{enc}}{\pi r^{2}L}\\ ⟹Q_{enc}& =\frac{Q{r}^2}{R^2}\\ ⟹\vec{E}& =\frac{Qr}{2\pi R^{2}L{ϵ}_0}\end{array}$$

Nonuniform Density §

Electric Field Outside Object §

Conductors and Non-Conductors §

Gauss’s Law states that

$$\vec{E}2\pi rL=\frac{Q_{enc}}{{ϵ}_0}$$

Given the same cylinder dimensions and $\rho =-Cr$, we solve for $Q_{enc}$.

$$\begin{array}{rl}{Q}_{enc}& ={\int }_0^{L}dz\cdot {\int }_0^R-Cr\cdot rdr{\int }_0^{2\pi }d\theta \\ & =2\pi L{\left[\frac{-C{r}^3}{r}\right]}_0^R\\ & =-\frac{2C\pi L{R}^3}{3}\end{array}$$

Plugging back into Gauss’s Law:

$$\begin{array}{rl}{\vec{E}}_{out}\cdot 2\pi rL& =-\frac{2C\pi L{R}^3}{3{ϵ}_0}\\ ⟹{\vec{E}}_{out}& =\frac{-C{R}^3}{3r{ϵ}_0}\hat{i}\end{array}$$