Gauss's law also handles two other shapes with strong symmetry: a very long line (or cylinder) of charge and a very large flat sheet. The trick is choosing a Gaussian surface whose pieces either catch the field head-on or not at all.
Why it works
Line: by symmetry the field points straight out from the line and depends only on the distance $r$. Wrap the line in a Gaussian cylinder of radius $r$ and length $\ell$. The flat ends get no flux (the field skims them); the curved side gets $E\cdot 2\pi r\ell$. It encloses $\lambda\ell$:
$$E\,(2\pi r\ell) = \frac{\lambda\ell}{\varepsilon_0} \Rightarrow E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r}.$$
Sheet: the field points straight away from the sheet on both sides. Use a pillbox through the sheet with caps of area $A$. Only the caps get flux, $2EA$, and it encloses $\sigma A$:
$$2EA = \frac{\sigma A}{\varepsilon_0} \Rightarrow E = \frac{\sigma}{2\varepsilon_0}.$$
Key idea
Long line: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$, falling as $1/r$. Large sheet: $E = \dfrac{\sigma}{2\varepsilon_0}$, the same at every distance. Inside a long uniform cylinder of radius $R$: $E = \dfrac{\rho r}{2\varepsilon_0}$.
Worked examples
Worked example: A long wire
A long wire carries $5.0\ \text{nC/m}$. Find the field $4.0\ \text{cm}$ away.
- $$E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{5.0 \times 10^{-9}}{2\pi(8.85 \times 10^{-12})(0.040)} = 2250\ \text{N/C}$$
- This matches the close-up limit $2k\lambda/d$ of the rod integral, with far less work.
Answer: $E \approx 2250\ \text{N/C}$
Worked example: A large sheet
A large sheet carries $4.0\ \text{nC/m}^2$. Find the field $1.0\ \text{cm}$ and $5.0\ \text{cm}$ from it.
- $E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{4.0 \times 10^{-9}}{2(8.85 \times 10^{-12})} = 226\ \text{N/C}$.
- The distance never entered, so it is the same at both points (as long as we are close compared with the sheet's size).
Answer: $226\ \text{N/C}$ at both
Worked example: Two sheets: a capacitor
Two large parallel sheets carry $+4.0$ and $-4.0\ \text{nC/m}^2$. Find the field between them and outside them.
- Each sheet makes $226\ \text{N/C}$ on both sides: away from the positive sheet, toward the negative one.
- Between the sheets both point the same way: $E = \sigma/\varepsilon_0 = 452\ \text{N/C}$.
- Outside, they point opposite ways and cancel: $E = 0$. This is the uniform field of a parallel-plate capacitor.
Answer: $452\ \text{N/C}$ between, $0$ outside
Common mistake
"A sheet's field gets weaker as you move away." For a sheet that is large compared with your distance, it does not. As you back away, each piece of the sheet is farther, but more of the sheet's pieces push in your direction rather than sideways, and the two effects cancel exactly. Check the formula: $\sigma/(2\varepsilon_0)$ has no distance in it. (A real sheet of size $D$ behaves this way only while you are much closer than $D$; far away it looks like a point.)