Question

Derive an expression for electric field intensity due to an infinitely long straight charged wire.

Answer 1

Consider an infinitely long, uniformly charged wire with a constant linear charge density (λ) in a permittivity medium (ε).

Let R be the radius of cross section of wire.

We know, λ = `q/l`

Total charge q = λ l

Let P be a location located ‘r’ from the wire’s axis. Draw a Gaussian cylindrical surface with radius r and length l, as indicated in the picture.

Let E be the electric intensity at a point ‘P’ that is orientated radially outward and constant for all places at the same distance from the axis. 

By Gauss law,

T·N·E·I = Total charge

Φ ε.Ecos θ ds = λ l

The angle between the direction of `vec(E)` and normal to the surface of ds is zero.

∴ ε.E Φ ds = λ l     ...(∵ θ = 0 and cos 0 = 1)

where,

Φ ds = Area of curved surface of Gaussian cylinder

= 2 π r l

∴ ε.E 2πrl = λl

E = `λ/(2piεr)`   ...(1)

This is equation for a dielectric medium.

We know ε = kε₀

∴ E = `λ/(2pikε_0r)`   ...(2)

For air medium k = 1

∴ E = `λ/(2piε_0r)`    ...(3)