Obtain the formula for the electric field due to a long thin wire of uniform linear charge density `lambda` without using Gauss's law. [Hint. use Coulomb's law directly and evaluate the necessary integral].

Explanation

In Fig, AB is a long thin wire of uniform linear charge density, `lambda`. We have to obtain formula for electric field due to this wire at any point P at a perpendicular distnace `PC = r` form the wire.
Consider a small element of length, dx of the wire with center 0, such that `OC = x`
Charge on the element, `q = lambda dx`
`:.` Electric intensity at p due to the element,
`dE = (1)/(4pi in_(0)) (lambda dx)/(OP^(2)) = (lambda dx)/(4pi in_(0) (r^(2) + x^(2)))`
`vec(dE)` is along OPD. If `/_OPC = 0, vec(dE)` can be resolved into two recantugular components : dE cos `theta` along `PE _|_ AB` and dE `sin theta` along `PF || BA`. The parallel components component will be cancelled by the parallel component of the field due to charge on a simillar element dx of wire on the other half. The radial component gets added.
`:.` effective component of electric intensity due to the charge element, `vec(dE) = dE = dE cos theta`
`dE = (lambda dx cos theta)/(4pi in_(0) (r^(2) + x^(2)))` ....(i)
From `Delta OCP, x = r tan theta :. dx = r sec^(2) theta d theta`
Now, `r^(2) + x^(2) = r^(2) + r^(2) tan^(2) theta = r^(2) (1+ tan^(2) theta) = r^(2) sec^(2) theta`
From (i), `dE' = (lambda r sec^(2) theta d theta)/(4 pi in_(0) r^(2) sec^(2) theta) cos theta = (lambda)/(4pi in_(0) r) cos theta d theta`
As wire has infinite length, its ends A and B are infinite distance apart. Therefore, `theta` varies from `-pi//2 to + pi//2` :. Electric intensity at P due to the whole wire,
`E' int_(-pi//2)^(pi//2) (lambda)/(4pi in_(0) r) cos theta d theta = (lambda)/(4pi in_(0) r) [sin theta]_(-pi//2)^(pi//2) = (lambda)/(2pi in_(0) r)`