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)`.
