help in here, this exercise is killing meI am.?

Determine the distribution of the electric field originated by a hollow, infinite cylinder of radius R, and surface charge density σ.

Assume cylinder has length L, where L → ∞. Total charge on cylinder surface is Q = σA = σ (2πRL), where 2πRL is cylinder area. According to Gauss' Law, Q = ε E A'. ε is permittivity of medium (presumably, air, but not specified), E is electric field strength, and A' is the area of a Gaussian (cylindrical, in this case) surface built around original cylinder. That is, imagine the original cylinder totally enclosed by an imaginary cylindrical surface, whose radius is r. Clearly, if this cylindrical surface totally encloses original cylinder, its surface area will necessarily be infinite, as well. Disregard this fact for the moment being.

Since A' = 2πrL, by Gauss' Law,

ε E (2πrL) = 2πRLσ

ε E r = σ R

E = σ/ε · R/r

The above expression gives the magnitude of the vector field E, as a function of distance r. r is the distance from any arbitrary point to the axis of the charged cylinder. It gives field strength for r ≥ R. In the region where r < R (inside the cylinder), E = 0, everywhere*. Note that resulting field strength is independent of L. Field points radially outward (perpendicularly to cylinder surface), assuming positive σ.

A device to avoid dealing with infinite length is to work with differentials of area, rather than area itself. Thus,

dQ = σ da = σ (2πR) dL,

ε E · da = ε E · 2πr dL = 2π ε (E · r) dL = 2πRσ dL.

As both E (by symmetry) and r (by definition) have the same radially outward direction, E · r = E r cos 0° = E r. Thus,

ε E r = σ R,

and

E = σ/ε · R/r,

as before.

*Since there's no charge inside the hollow cylinder, charge enclosed by any Gaussian surface inside the cylinder will be naught. If Q = 0, then E = 0, according to Gauss' Law.

Look at symetries in the system: no need of z (height of the cyllinder) and theta (symetry by rotation), and depends only of r the distance to the cylinder.

Take Gauss theorem and a cylinder of radius r and heigth h centered the same as your cylinder as Gauss surface.

if sigma * h * 2pi R is the electric charge inside the gauss surface then :

2 pi R h sigma / epsilon0 = E . 2 pi r h

giving E = R sigma / (epsilon0 r),

note thtat E decreases with r distance to the cyinder (r>R).

Inside the cylinder, as sigma is not into the gaussian surface, E=0.

do your own homework.....

look in the back of the book!!!