Finding the electric field of a uniformly charged cylinder using Gauss' law

1. Jul 27, 2012

Miscing

1. The problem statement, all variables and given/known data

A very long cylinder has charge density L/m and radius a. Find the electric field at a distance r, with r > a.

3. The attempt at a solution

I construct a gaussian cylinder around it with radius r, and then from Gauss' law the field at any point on it's side surface is E= Q/Ae, where A is the area of the side of the cylinder, e = vacuum permittivity. So, Q = L*length, A = 2*pi*r. This gets me the right answer in the text book (L/(2*pi*e)), but there's a sticking point in my approach - this assumes all of the flux is concentrated through the sides, whereas there should also be some through the top and bottom of the cylinder, right? The best explanation I can think of is that because the cylinder is "very long", SA of the sides >>> SA of the top and bottom and so flux through these is negligible. This reasoning is a bit too heuristic for me though and it would be great if someone can give a better/more quantitative explanation.

2. Jul 27, 2012

pgardn

I think thats why they said a very long cylinder. Have you seen the Gauss law problems with the infinitely long wire of charge, and the infinitely large plates of charge? thats so you dont have to deal with the ends and edges as that would be very difficult using gauss.

If you have used the integrative technique to find the strength of an E-field at some point outside a line of charge of known length, L, then you know it can be a pain. And then how about a round plate of charge with known radius R, thats a bit of a pain and its only a bit of a pain if you want to know the efield strength at some point that lies right along the axis of the center of the plate. It would be an enormous pain if the point was not in line with the axis of the plate. Now combine that with the fact that you would have more charge contributing to the E.field of some cylinder of known L as well... a HUGE math pain. Better to just measure it.

Its Gauss, its supposed to be a short cut for objects with some sort of spherical symmetry usually.

Last edited: Jul 27, 2012
3. Jul 29, 2012

Miscing

Awesome, thanks. I did try to set up the integral first and it quickly turned into a cluster**** haha