What is the Potential Above a Charged Cylindrical Shell?

[Melawrghk]

Homework Statement

A very long insulating cylindrical shell of radius 6.90 cm carries charge of linear density 8.40 microC/m spread uniformly over its outer surface. Find the potential of a point located 4.4cm above the surface of the shell

Homework Equations

The Attempt at a Solution

I used Gaussian surface to find the electric field created by the shell outside (Q = linear charge density):
QL/$$\epsilon_{0}$$ = E*2*pi*d*L,
where L is length, d is distance to point from the axis of the cylinder.
The Ls cancel out, rearranging, I got:
E=$$\frac{Q}{2pi*d*\epsilon_{0}}$$

I then integrated that to get the potential:
V=$$\frac{Q}{2pi*\epsilon_{0}} \int {(1/d)dd}$$
The integral's upper limit is d-r, and lower limit is r.
I got the equation below as a result:
V=$$\frac{Q}{2pi*d*\epsilon_{0}} ln(\frac{d-r}{r})$$

Substituting in the numbers I got V=-6.8*10^4, but that's wrong. Is it my method? Should I slice the cylinder into rings and do V from each one, then integrate the whole thing?

Thanks in advance

 

[gabbagabbahey]

I then integrated that to get the potential:
V=$$\frac{Q}{2pi*d*\epsilon_{0}} \int {(1/d)dd}$$

What's that extra 1/d doing out front of the integral?

The integral's upper limit is d-r, and lower limit is r.

Why is that?

 

[Melawrghk]

What's that extra 1/d doing out front of the integral?
Why is that?

First one: typo, sorry :)

Second one: now that I think about, I'm not sure. I was thinking of using distance from the cylinder's axis and then r would be the smallest so that you're still on the outside (well, surface) and d-r would be anything beyond that. Should those limits be d and 0 instead?..

EDIT: I got it! Instead of using d as distance to center of the cylinder, I used d as distance to surface and for gaussian surface I had:
E=Q/(epsilon*2*pi*(d+r))
And integrated taht with upper limit being 'd' and lower being 0. And it worked! :D Thanks so much.