State Gauss' law in electrostatic.. A thin straight infinitely long conducting wire of linear charge density ` lambda ` is enclosed by a cylinder surface of radius 'r' and length ''l' ,its axis coinciding with the length of the wire . Qbtain the expressionn for the electric field, indicating its direction at a point on the surface of the cylinder.

Explanation

Consider an infinitely long straight charged wire of linear charge desnsity `lambda`. To find electric field at a point P situated at a distance r from the wire by using Guass. law consider a cylinder of length l and radius r as the Gaussian surface.
From symmetry consideration electric field at each point of its curved surface is `vecE` and is point outwards normally . Therefore, electric flux over the curved surface.
` int vecE hatn ds = E 2 pi r l`

On the side face 1 and 2 of the cylinder normal drawn on the surface is perpendicular to electric field E and hence these surfaces do not contribute towards the total electric flux.
`therefore ` Net electric flux over the entire Gaussian surface `phi_E = E. 2pi r l`
By Gauss law electric flux `phi_E = 1/(epsi_0)` (charge enclosed)`= (lambda l)/(epsi_0)`
Comparing (i) and (ii) , we have
`E. 2 pi rl = (lambdal)/(epsi_0)`
`implies E = (lambda)/(2 pi epsi_0 r)`
As `E prop 1/r` , hence E -r shown in fig.