Define electric flux. Applying Gauss's law and serive the expression for electric intensity due to an infinite long straight charged wire. (Assume that the electric field everywhere radial and depends only on the radial distance r of the point from the wire.)

Explanation

Electric flux : The number of electric lines of force passing perpendicular to the area is known as electric flux `(phi)`. Electric flux `phi=vec(E).vec(A)`. So flux is a scalar.
Expression for E due to an infinite long straight charged wire :
(1) Consider an infinitely long thin straight wire with uniform linear charge density `'lambda'`.
(2) Linear charge density `lambda=("change q")/("length l") implies q=lambda l rarr (1)`
(3) construct a coaxial cylindrical gaussion surface of length `'l'` and radius 'r'. Due to symmetry we will assume that electric field is radial i.e., normal to the conducting wire.
(4) The flat surfaces AB and CD are `bot^(r)` to the wire. Select small area `ds_(1)` and `ds_(2)` on the surface as AB and CD. They are `bot^(r)` to `vec(E)`. So flux coming out through them is zero.
Since flux `phi=oint vec(E).d vec(s)=Eds cos 90^(@)=0`
(5) So flux coming out through the cylindrical surface ABCD is taken into account.
(6) From Gauss's law
`underset(S)(oint)vec(E).dvec(s)=int E.ds=Es=E(2pi rl)` ...(2)
From Gausses theorem, `underset(S)(oint)vec(E).dvec(s)=q/epsi_(0)` ...(3)
From (2) and (3), `E(2pi rl)=Q/epsi_(0)=(lambda //)/epsi_(0)[ :' Q=lambda //]`
`:. E=(lambda l)/(2pi epsi_(0) rl)=1/(2pi epsi_(0)) lambda/r`
(8) Therefore electric intensity due to an infinitely long consucting wire `E=lambda/(2pi epsi_(0) r)`.