State ‘Gauss law’ in electrostatics. Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density λ cm-1.

Gauss’s law in electrostatics: It states that “the total electric flux over the surface S in vacuum is $\frac{1}{{\mathrm{\epsilon }}_0}$ times the total charge (q).

$\mathrm{contained} \mathrm{in} \mathrm{side} \mathrm{S} \therefore {\mathrm{\varphi }}_{\mathrm{E}}=\underset{\mathrm{S}}{\oint }\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{S}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$

Electric field due to an infinitely long straight wire: Consider an infinitely long straight-line charge having linear charge density X to determine its electric field at distance r. Consider a cylindrical Gaussian surface of radius r and length l coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface S1 and is directed radially outward.
Total flux through the cylindrical surface,

$$\begin{gathered} \oint \overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{S}}=\underset{{\mathrm{s}}_1}{\oint }\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{{\mathrm{S}}_1}+\underset{{\mathrm{s}}_2}{\oint }\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{{\mathrm{S}}_2}+\underset{{\mathrm{s}}_3}{\oint \overrightarrow{\mathrm{E}}}.\mathrm{d}\overrightarrow{{\mathrm{S}}_3} \\ =\underset{{\mathrm{s}}_1}{\oint }{\mathrm{EdS}}_1 \cos 0°+\underset{{\mathrm{s}}_2}{\oint }{\mathrm{EdS}}_2\cos 90°+\underset{{\mathrm{s}}_3}{\oint }{\mathrm{EdS}}_3\cos 90° \\ =\mathrm{E}\oint {\mathrm{dS}}_1=\mathrm{E}\times 2\mathrm{\pi rl} \\ \mathrm{As} \mathrm{\lambda } \mathrm{is} \mathrm{the} \mathrm{charge} \mathrm{per} \mathrm{unit} \mathrm{length} \mathrm{and} \mathrm{l} \mathrm{is} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{wire}, \mathrm{so} \mathrm{charge} \mathrm{enclosed} \mathrm{is}, \\ \mathrm{q}=\mathrm{\lambda l} \\ \mathrm{By} \mathrm{Gauss}\text{'}\mathrm{s} \mathrm{theorem} : \\ \oint \overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{{\mathrm{S}}_1}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0} \Rightarrow \mathrm{E}\times 2 \mathrm{\pi rl}=\frac{\mathrm{\lambda l}}{{\mathrm{\epsilon }}_0} \\ \therefore \mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}} \end{gathered}$$