Knockout NEET May 2021 (One Month)
An exhaustive E-learning program for the complete preparation of NEET..
₹ 14000/-
₹ 6999/-
Buy Now
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, the electric field at P due to this length dx is
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.
In the question linear charge density =E
therefor