Electric Field due to a uniform Infinite Cylindrical Charge

Let us consider that electric charge is distributed uniformly within an infinite cylinder of radius R. Let p be the charge density. We have to find electric field E at any point distant r from the axis lying (i) inside (ii) on the surface (iii) outside the cylindrical charge distribution.

Case (i). When the point lies outside the charge distribution

Let  P1 be a point a distance r (> R) from the axis of the cylinder. Draw a coaxial cylinder of radius r and length l such that P1 lies on the surface of the cylinder. Fro symmetry, the electric field E is everywhere normal to the curved surface and has the same magnitude at all points on it. The electric flux due to plane faces is zero. So the total electric flux is due to the curved surface alone.

The electric flux due to curved surface = Ñ„ E . dS = E ( 2 π rl )
The net charged enclosed by the Gaussian surface = q = (π R2 l) p
. : by Gauss’ law, E (2 π rl ) = π R² lp / ε₀

or    E = pR² / 2 ε₀ r

Case (ii). When the point lies on the surface of charge distribution (r = R)

Let P2 be the point on the surface of charge distribution.

By Gauss’ law, E (2 π Rl ) = π R² lp / ε₀

. :    E = Rp/ 2 ε₀

Case (iii). When the point lies inside the charge distribution (r < R)

Let P3 be the point at a distance r (< R) from the axis of the cylinder. Consider a coaxial cylindrical surface of radius r and length l such that P3 lies on the curved surface of this cylinder.

The charge q’ inside this Gaussian surface = π R² lp
By Gauss’ law, E (2 π Rl) = π R² lp/ ε₀    

. :     E = pr/ 2 ε₀     E

Field due to a uniformly charged Hollow Cylinder

Consider a uniformly charged hollow cylinder of radius R. In this case, charge is uniformly distributed over cylindrical surface. Let λ be the charge per unit length. P is a point at a distance r ( > R) from the axis of the cylinder. Draw a coaxial cylindrical Gaussian surface of radius r and length l. The electric flux due to the top and bottom circular caps is zero.

The electric flux due to curved surface = Ñ„ E . dS = E (2 π rl)
The net charge enclosed by the Gaussian surface = q = λ l
By Gauss’s law, E (2 π rl) = λl/ ε₀
. :        E = λ/ 2 π ε₀ r

The direction of the electric field is radially outwards.
E is independent of the radius R of the charged cylinder.
Let σ be the surface density of charge on the cylinder. If P is infinitely close to the cylinder, then λ = 2 π R σ.
. :        E = σ/ ε₀
If we construct a Gaussian surface inside the hollow cylinder, it will enclose no charge. Therefore, the electric field inside a charge hollow cylinder is zero.