A cylinder of length 2a and radius ‘a’ has the x-axis as its axis. Its two ends (plane surfaces) are at x = a and x = 3a respectively. Point charges +q and –q are located at x = 2a and x = 0 respectively on the axis of cylinder. The electric flux through the curved surface of cylinder is (nearly)

flux through the base of a cone of semi-solid angle $\theta$is   $\mathrm{\varphi }=\frac{\mathrm{q}}{2{\mathrm{ϵ}}_0}(1-\cos \mathrm{\theta })$.

Flux through the curved surface due to charge inside the cylinder, ${\mathrm{\varphi }}_1=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}-2\left\{\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_{\mathrm{o}}}\left(1-\cos 45°\right)\right\}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}\cos 45°=\frac{\mathrm{q}}{\sqrt{2} {\mathrm{\epsilon }}_{\mathrm{o}}}$

Flux through the curved surface due to charge outside the cylinder,

 $$\begin{gathered} {\varphi }_2=\frac{-q}{2{\epsilon }_o}\left(1-\cos 45°\right)-\left\{\frac{-q}{2{\epsilon }_o}\left(1-\cos \left\{{\tan }^{-1}\left(\frac{1}{3}\right)\right\}\right)\right\} \\ =\frac{-q}{2{\epsilon }_o}\left(\cos \left\{{\tan }^{-1}\left(\frac{1}{3}\right)\right\}-\cos 45°\right) \\ =\frac{-q}{2{\epsilon }_o}\left(\frac{3}{\sqrt{10}}-\frac{1}{\sqrt{2}}\right) \end{gathered}$$

$\Rightarrow {\mathrm{\varphi }}_{\mathrm{c}}={\mathrm{\varphi }}_1+{\mathrm{\varphi }}_2=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}\left(\frac{1}{\sqrt{2}}-\frac{3}{2\sqrt{10}}+\frac{1}{2\sqrt{2}}\right)=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}\left[\frac{3}{2\sqrt{2}}\left(1-\frac{1}{\sqrt{5}}\right)\right]\simeq 0.6\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$