Derive an expression for the electric field at a point due to an infinitely long thin charged straight wire using Gauss Law.

Explanation

Let the Gaussian surface be taken as a cylinder of length 1 and radius r. Electric field lines drawn from a point on the wire to every point on the rim of a circular plane make an angle of zero with respect to the normal drawn to the surface. Electric flux through the Gaussian surface
= E(SA), SA = surface area 1
= E(2`pi`rl) If X represents linear charge density of the thin wire, then total charge on the wire = `lambda` l. From Gauss Law, total outward electric flux through a closed surface =`(1)/(epsilon_(0))`
vectorally `vecE=(lambda)/(2 pi epsilon_(0)r)hatn` where `hatn` is the radial unit vector in the plane normal to the wire passing throught the point