Figure shows a section of an infinite rod of charge having linear charge density λ which is constant for all points on the line. Find electric field E at a distance r from the line.

$\frac{\lambda}{2 \pi \varepsilon_0 r}$

From symmetry, $\vec{E}$ due to a uniform linear charge can only be radially directed. As a Gaussian surface, we can choose a circular cylinder of radius r and length l, closed at each end by plane caps normal to the axis.

$\varepsilon_0 \oint \vec{E} . d \vec{s}=q_{in}$

$\varepsilon_0\left|\int \vec{E} . d \vec{s}+\int \vec{E} . d \vec{s}\right|=q_{in}$

Cylindrical Plane Surface

$\varepsilon_0 E(2 \pi rl)+\int E . ds . \cos 90^{\circ}=\lambda l$

$E=\frac{\lambda l}{\varepsilon_o 2 \pi r l}=\frac{\lambda}{2 \pi \varepsilon_0 r}$

The direction of $\vec{E}$ is radially outward for a line of positive charge.