Electric Field at axis of Cylinder!

Consider a Disc of radius $R$ of thickness $dr$ at a distance $r$ from the center $O$ of the cylinder.

Charge on the disc,

$dq = \frac{Q}{\pi R^2 L} (\pi R^2 dr) = \frac{Q}{L} dr$

Electric field due to disc along it's axis

$E_x = \frac{\sigma}{2 \epsilon_0} [1- \frac{x}{\sqrt{x^2 + R^2}}]$

Do the necessary Integration and get the answer!

Final answer $\large \Rightarrow E = \frac{Q}{2 \pi R^2 L \epsilon_0}[ L+(\frac{L^2}{4} +R^2)^{0.5} - (\frac{9L^2}{4} +R^2)^{0.5} ]$

We begin by considering the integral across all infinitesimal contributions to the overall electric field:

$E \ = \ \displaystyle\int \ dE \ = \ \displaystyle\int \frac{k dq}{\ell^2} \ = \ \displaystyle\int \frac{k \cos \theta dq}{z^2 \ + \ r^2}$

Where $\theta$ is the angle that the radius extending from point $P$ to some point in the cylinder makes with the horizontal, since the vertical components of force cancel each other out. We also know that $dV \ = \ r dr d\theta dx$ and $\frac{dq}{dV} \ = \ \frac{Q}{\pi R^2 L}$ . This means that we will have:

$E \ = \ \frac{kQ}{\pi R^2 L} \displaystyle\int \displaystyle\int \displaystyle\int \frac{\cos \theta r}{z^2 \ + \ r^2} dr d\theta dz$

Since $\cos \theta \ = \ \frac{z}{\ell}$ , we have:

$E \ = \ \frac{kQ}{\pi R^2 L} \displaystyle\int \displaystyle\int \displaystyle\int \frac{z r}{(z^2 \ + \ r^2)^{3/2}} dr d\theta dz \ \Rightarrow \ E(P) \ = \ \frac{2 kQ}{ R^2 L} \displaystyle\int_{L/2}^{3L/2} \displaystyle\int_{0}^{R} \frac{z r}{(z^2 \ + \ r^2)^{3/2}} dr dz$

Performing the u-substitution of $u \ = \ r^2$ and $\frac{du}{2} \ = \ r dr$ , we get:

$E(P) \ = \ \frac{kQ}{ R^2 L} \displaystyle\int_{L/2}^{3L/2} \displaystyle\int_{0}^{R^2} \frac{z }{(z^2 \ + \ u)^{3/2}} du dz \ = \ \frac{-2kQ}{ R^2 L} \displaystyle\int_{L/2}^{3L/2} \Big[ z \frac{1}{\sqrt{z^2 \ + \ u}} \Big] \biggr\rvert_{0}^{R^2} dz \ = \ \frac{2kQ}{ R^2 L} \displaystyle\int_{L/2}^{3L/2} \Big[ 1 \ - \ z \frac{1}{\sqrt{z^2 \ + \ R^2}} \Big] dz$

Now, we perform another u-substitution, with $u \ = \ z^2$ and $\frac{du}{2} \ = \ z dz$ , getting:

$E(P) \ = \ \frac{2 kQ}{ R^2 L} \Big[ L \ - \ \frac{1}{2} \displaystyle\int_{L^2 / 4}^{9L^2 / 4} \frac{1}{\sqrt{u \ + \ R^2}} \Big] du \ = \ \frac{2 kQ}{ R^2 L} \Big[ L \ - \ \sqrt{\frac{L^2}{4} \ + \ R^2} \ + \ \sqrt{\frac{9L^2}{4} \ + \ R^2} \Big] \ = \ \frac{Q}{2 \pi R^2 L \epsilon_0} \Big[ L \ - \ \sqrt{\frac{L^2}{4} \ + \ R^2} \ + \ \sqrt{\frac{9L^2}{4} \ + \ R^2} \Big]$

Which leaves us with:

$2 \ + \ 2 \ + \ 2 \ + \ 4 \ + \ 2 \ + \ \frac{1}{2} \ + \ 9 \ + \ 2 \ + \ 4 \ + \ 2 \ + \ \frac{1}{2} \ = \ 30$