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

Take a long thin wire XY of uniform linear charge density λ.

Consider a point A at a perpendicular distance l from the mid-point O of the wire.

Let E be the electric field at point A due to the wire, XY.

Consider a small length element dx on the wire section with OZ = x Let q be the charge on this piece.

Therefore q =λdx

Electric field due to the piece,

dE = 1 /4πε₀ . λdx/(AZ)²

However , AZ = √ (l²+ x ²)

Therefore dE = 1 /4πε₀ . λdx/(l²+ x ²)
The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθ is the parallel component. When the whole wire is considered, the component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point A. Hence, effective electric field at point A due to the element dx is dE₁.
Therefore dE₁ = 1 /4πε₀ . λdx cosθ/(l²+ x ²)—–Eq-1

In  ΔAZO, tan θ =x/l ⇒ x = l.tan θ —————————–Eq-2

From Equation-2 we obtain

dx/dθ = l sec²θ ⇒ dx= l sec²θ dθ ——————————Eq-3

From Equation-2 we have

x² + l² = l² tan² θ +l²= l² ( tan² θ +1)= l² sec²θ———-Eq-4

Putting equations-3 & 4  in Equation-1 ,we obtain

dE₁ = 1 /4πε₀ . λ (l sec²θ dθ ) cosθ/(l² sec²θ)

=1 /4πε₀ .λcosθ dθ /l—————————-Eq-5

The wire is so long that θ tends from -π/2 to π/2

By integrating Eq-5 ,we obtain the value of field E₁ as,    

⌠ ^π/2 dE ₁ = ⌠ ^π/2 1 /4πε_{0 .} λcosθ dθ /l

_⁻ π/2⌡             _⁻ π/2⌡

⇒ E₁ =1 /4πε_{0 .} λ/l  [sinθ _{⁻ π/2} ] ^π/2

⇒ E₁ = 1 /4πε_{0 .} λ/l  [sinθ _{⁻ π/2} ] ^π/2

⇒ E₁ = 1 /4πε_{0 .} λ/l x 2  ⇒  E₁ =  λ/2πε₀ l

Therefore .the electric field due to long wire is   λ/2πε₀ l