Question

State Gauss's theorem in electrostatics. Using this theorem, derive an expression for the electric field due to an infinitely long straight wire of linear charge density λ.

Answer 1

Gauss’s theorem: The flux of an electric field through any closed surface is `1/(epsilon_0)` times the total charge enclosed by the closed surface.

`phi = q/(epsilon_0)`   ...(1)

By definition, the total electric flux through the closed surface is given by

`phi = "∮"vecE*vec(ds)`   ...(2)

∴ From (1) and (2), Gauss’s theorem may be expressed as follows

`phi = "∮"vecE*vec(ds) = q/epsilon_0`

∴ The surface integral of electric field over a closed surface is equal to `1/epsilon_0` times the total charge enclosed by the surface.

Application of Gauss’s theorem:

To find electric field due to a line charge, let us consider an infinitely long line charge placed along XX’ axis with linear charge density λ. Our aim is to find electric field intensity at a point P distant r from the line charge. We draw a cylindrical surface of radius r and length l coaxial with the line charge. The net flux through the cylindrical gaussian surface, i.e.

`phi = "∮"vecE*vec(ds)`

= `int_(LCF)vecE*vec(ds) + int_(CS)vecE*vec(ds) + int_(RCF)vecE*vec(ds)`

= `int_(LCF)Eds cos90"°" + int_(CS)Eds cos0"°" + int_(RCF)Eds cos90"°"`

`phi = int_(CS)Eds cos0"°" = E*2pirl`   ...(1)

The charge enclosed by the gaussian surface is q = λl   ...(2)

Using Gauss’s theorem from equations (1) and (2)

`E*(2pirl)= (lambdal)/epsilon_0 ⇒ E = lambda/(2piepsilon_0r)`