Determine electric field intensity near an infinitely long straight uniformly charge wire.
Or.
Using Gauss' law expression for electric field intensity at a point situated at a distance 'r' from an infinitely long, uniformly charged straight wire.

Explanation

Gauss law states that total eneric flux over the closed surface S in vacuum is `(1)/(epsi_(0))` times the total charge contained inside surface S.
`phi_(E)=ointvec(E).vec(dS)=(q)/(epsi_(0))`
Electric field intensity due to a line charge.
COnsider a thin charged rod with uniform linear charge density `lamda`. we wish to find an expression for electric intensity at point P at a perpendicular distance r, from the rod. consider a right circular cylinder of radius r and length l with the infinite long line of charge as its axis.
In surface I and III, `vecE and vec(dS)` are `bot` to each other. in case of surface II, `vecE and vec(dS)` are parallel to each other and hence `theta=0^(@)`

From Gauss.s theorem, we have
`ointvec(E).vec(dS)=(q)/(epsi_(0))`
or `oint_(I)vec(E).vec(dS)+oint_(II)vec(E).vec(dS)+oint_(III)vec(E).vec(dS)=(q)/(epsi_(0))`
or `int_(I)EdScos90^(@)+int_(II)EdScos0^(@)+int_(III)EdScos90^(@)`
`=(q)/(epsi_(0)0`
or `0+Eint_(II)dS+0=(q)/(epsi_(0))`
or `E(2pirl)=(q)/(epsi_(0))`
or `E=(q)/(2pirlepsi_(0))`
But `q=lamdal`
`therefore E=(lamdal)/(2pirlepsi_(0))`
or `E=(lamda)/(2pirepsi_(0))`