Potential on the axis of a uniformly charged cylinder

[acrimius]

Homework Statement

Posting here because it was over a previous homework assignment and I don't understand the solution that was given out. For reference, the problem is 2.27 is Griffiths' Introduction to Electrodynamics. It reads "Find the potential on the axis of a uniformly charged solid cylinder, a distance z from the center. The length of the cylinder is L, its radius is R, and the charge density is ρ. Use your result to calculate the electric field and this point. (Assume z > L/2.)

Homework Equations

[/B]
V(r) = - (1/4πε_o) * ∫ (ρ(r')/(√(r-r'))) dτ'

The Attempt at a Solution

Initially when I had tried it, I just went straight for this integral over the volume, setting r' = s' s_hat + z' z_hat and r = z z_hat, and for a cylinder dτ' = s'ds'dφ'dz'. I set up the integral, it got messy, etc. It was very wrong essentially, marked "See Solutions" on my paper. In the solutions, the problem started with finding the potential of a singular disk, using the surface charge density, then after finding the potential for a singular disk, integrating again over the volume to find the complete potential of the cylinder at this point. My questions are:

1) Do you have to compute the potential from a single disk first? Or did I just compute/setup my integral incorrectly, and the solution is just a different way of doing this?

2) Normally when we talk about the electric field through a cylinder, we say that it points radially outward, and not through the faces. So why then, if this point is on the axis meaning above one of the faces, are we finding a nonzero potential and nonzero electric field?

 

[kuruman]

1) Do you have to compute the potential from a single disk first? Or did I just compute/setup my integral incorrectly, and the solution is just a different way of doing this?

You don't have to compute the potential from a charged disk first. Think about it. If you set up the triple integral correctly and you do the angular integral first, and then the radial integral you will be left with a single integral over z which is adding disks. So finding the disk potential first is the same as doing the triple integral in two steps.

Normally when we talk about the electric field through a cylinder, we say that it points radially outward

It looks like you are confusing the infinite cylinder (or wire) with the finite cylinder. The latter is the case here.