A very long uniformly thread oriented along the axis of a a circle of raius `R` rests on its centre with one of the ends. The charge of the thread per unit length is equla to `lambda`. Find the flux of the vector `E` across the circle area.

Explanation

From the solution of 3.14 field strength at a perpendicular distance `r lt R` from its left end
`vec(E) (r ) = (lambda)/(4pi epsilon_(0) r) (-vec(i)) + (lambda)/(4pi epsilon_(0) r) (hat(e_(r )))`
Here `hat(e_(r))` is a unit vector along radial direction.
Let us consider an elemental surface, `dS = dy dz = sz (r d theta)` a figure. Thus fluix fo `vec(E) (r )` over the element `vec(dS)` is given by
`d Phi = vec(E). vec(dS) = [(lambda)/(4pi epsilon_(0) r) (vec(-i)) + (lambda)/(4pi epsilon_(0) r) (hat(e_(r )))]. dr (r d theta) vec(i)`
`= - (lambda)/(4pi epsilon_(0)) dr d theta ("as " vec(e_(r)) _|_ vec(i))`
The sought flux, `varphi = - (lambda)/(45pi epsilon_(0)) int_(0)^(R) dr int_(0)^(2pi) d theta = - (lambda R)/(2 epsilon_(0))`,
If we have taken `vec(ds) uarr uarr (- vec(i))`, then `varphi` were `(lambda R)/(2 epsilon_(0))`
Hence `|varphi| = (lambda R)/(2 epsilon_(0))`