Obtain the

Let AB be a long thin wire of uniform linear charge density λ.

Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.

The charge on a small length dx on the line AB is q which is given as q = λdx.

So, according to Coulomb’s law, electric field at P due to this length dx is

where, ϵ₀ is the vacuum permittivity of the medium

But

⇒

This electric field at P can be resolved into two components as dEcosθ and dEsinθ. When the entire length AB is considered, then the dEsinθ components add up to zero due to symmetry. Hence, there is only dEcosθ component.

So, the net electric field at P due to dx is

dE' = dE cosθ

⇒ ………………….(1)

In ΔPOC,

⇒ x = h tan θ

Differentiating both sides w.r.t. θ,

⇒ dx = h sec²θ dθ …………………….(2)

Also, h² + x² = h² + h²tan²θ

⇒ h² + x² = h²(1+ tan²θ)

⇒ h² + x² = h² sec²θ ………….(3)

(Using the trigonometric identity, 1+ tan²θ = sec²θ)

Using equations (2) and (3) in equation (1),

⇒,

The wire extends from to since it is very long.

Integrating both sides,

⇒

⇒

⇒

⇒

This is the net electric field due to a long wire with linear charge density λ at a distance h from it.

NOTE: In the given case, it has been assumed that the length of the wire tends to infinity. In case of wires of finite lengths, the sin(θ) components cancel out only along the perpendicular bisector of the wire. At any other point, the net electric field will have both sin(θ) and cos(θ) components.