# Q 1.30: Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law.[Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

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

## Related Chapters

### Preparation Products

##### Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
##### Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
##### Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-