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:
$$V=\int_{-L/2}^{L/2} \rho \frac{\sqrt{R^2 +z^2}dz}{2\epsilon_0}$$