Question

Derive an expression for the electric field at a point due to an infinitely long thin charged straight wire using Gauss law.

Answer 1

The total electric flux through a closed surface in free space is equal to \(\frac 1{\varepsilon_0}\) times the net charge enclosed by the surface.

i.e., \(\phi = \frac q{\varepsilon_0}\)

In the above fig.

AB is the infinitely long wire

E is the electric field

P is a point at a distance r from the wire

'r' is the radius of Gaussian cylinder

'l' is the length of the Gaussian cylinder

Let 'q' be the charge enclosed by the Gaussian cylinder . Let 'λ' be the linear charge density on the wire.

Flux through the end faces is zero because there are no components of electric field along the normal to the end faces.

ϕ = flux through curved surface

ϕ = E x area of curved surface (ϕ = E x area)

ϕ = E × 2πrl  .......(1)

From Gauss's theorem

\(\phi = \frac q{\varepsilon_0}\)     ......(2)

But \(\lambda = \frac ql\)

⇒ \(q = \lambda l\)

⇒ \(\phi = \frac {\lambda l}{\varepsilon_0}\)  .....(3)

On comparing (1) and (3), we get

\(E \times 2\pi rl= \frac {\lambda l}{\varepsilon_0}\)

\(E = \frac {\lambda }{2\pi\varepsilon_0}\)

\(E = \frac 1{2\pi \varepsilon_0} \frac \lambda r\)

The direction of 'E' is perpendicular to the wire and directed away from the wire.

Answer 2

Consider an infinitely long thin straight wive with uniform linear charge q density λ. Let P be a point at ⊥r distance r from the wire.

To calculate the E.F \(\vec E\) at P, imagine a cylindrical Gaussian surface.

∴ The surface area of the curved part S = 2πrl

Total charge enclosed by the Gaussian surface q = λl 

Electric flix through the end Surfaces of the cylinder is Φ = 0 

Electric flux through the curved Surfaces of the cylinder is Φ₂ = Ecosθ.s 

Φ₂ = E x 1 x 2πrl

The total electric flux Φ = Φ₁ + Φ₂ 

Φ = 0 + E₂πrl, Φ₂ = 2πrlE ………(1) 

A/C to Gauss law,

from (1) and (2)