How to Find the Electric Potential of a Cylinder on the Z-Axis?

[bfusco]

Homework Statement

An insulating solid cylinder of radius R, length L carries a uniformly distributed electric charge with density \rho. Chose the z-axis along the axis of the cylinder, z=0 in the middle of the cylinder. the cylinder can be boken down into curcular tabs (disks) of thickness dl and surface charge \sigma, the combined slabs integrated over dl make up the cylinder.
(a)Find the potential on the z axis due to a disk; express \sigma in terms of \rho.
(b) find the potential on the z-axis V(z) for the entire cylinder.
(c)Calculate the electric field on the z-axis.

The Attempt at a Solution

(a) i drew a disk of radius R, and called the point where I am calculating the potential at a point P. The disk is the sum of rings (of radius r) from 0 to R, the line from the center of the disk to the point P is z and the line connecting radius r to point P is r'.
The charge distribution \sigma =dq/dA which turns into dq=\sigma 2\pi rdr

Potential is:
V=k\int \frac{dq}{r'}

Plugging the dq into the potential you get:
V=k \int \frac{\sigma 2\pi rdr}{\sqrt{r^2 + z^2}}

Which reduces to:
V=\frac{\sigma *\sqrt{R^2 +z^2}}{2 \epsilon_0}

Where \sigma=\rho dl

Which gives:
V=\rho \frac{ \sqrt{R^2 + z^2} dl}{2\epsilon_0}

(b) I know i have to sum the potentials of all the disks to make the cylinder, but idk how to do that.

is it:
V=\int_{-L/2}^{L/2} \rho \frac{\sqrt{R^2 + z^2}dl}{2\epsilon_0}
?

(c) when i get the answer to (b) i can just take the (-)gradient of it to get E

 

[bfusco]

For part (b), is it:
V=\int_{-L/2}^{L/2} \rho \frac{\sqrt{R^2 +z^2}dz}{2\epsilon_0}