Use Gauss's law to obtain the expression for electric field due to a thin infinitely long straight uniformly charged wire.

Flux through the Gaussian surface \[\begin{aligned} & =\text { flux through the curved cylindrical part of the surface } \\ & =\mathrm{E} \times 2 \pi \mathrm{rl} \end{aligned} \] The surface includes charges equal to \(\displaystyle \lambda 1\), Gauss's law then gives \[\begin{aligned} \mathrm{E} \times 2 \pi r l & =\frac{\lambda l}{\epsilon_{0}} \\ \mathrm{E} & =\frac{\lambda}{2 \pi \epsilon_{0} r} \end{aligned} \]