Using Gauss’ law, derive an expression for the electric field at a point near an infinitely long straight uniformly charged wire.

Explanation

Considering the whole line charge to be made of symmetrically situated length elements, each of length dI, it is clear from the figure shown that, the resultant field will be directed perpendicular to the line charge as shown-

To find magnitude
Now considering a cylindrical gaussian surface of radius r and length l , passing through point p , as shown

Flux through the Gaussian Surface= Flux through the curved Cylindrical part `Phi= E xx 2 pi r l ` .....(1) , But according to Gauss’ law `[Phi= (Sigma q)/( epsilon_(0))]`
net electric flux emerging from cylindrical surface is `[Phi = (lambda l ) / ( epsilon_(0))]` ....(2) [ The surface includes charge equal to `lambda l` ]
Now , eliminating`ul(Phi)` from equation (1) & (2), we get
`Exx 2 pi r l = (lambda)/(epsilon_(0))` or `E=(lambda)/(2 pi epsilon_(0)r ) = ( 2 k lambda)/(r)`
And in vector form `vec(E)=(2 k lambda)/(r^(2)) vec(r)`