 # Doubt from electrostatics

My approach is this...
First construct a prism like 3 D shape such that flux through other surfaces are zero (they are radial)

Then find charge enclosed by that prism like thing by gauss law.

Then find the other dimension of the cut to find total area of the cut.Then find the perpendicular area of the cut through which we find solid angle.

For that solid angle we know charge enclosed .. and for 360 it is q.
By this we find length value.
If anyone find me conceptually wrong pls do correct me

I will calculate and will say if I have got the ans....

As shown in the figures, consider two conical regions with central angles 53^\circ and 37^\circ. We can calculate the flux due to the charge that goes through these two conical regions (i.e the flux through the surface of the cylinder that lies above these cones) using Gauss Law and Solid Angle.
For the cone in figure A:
\Omega_1 = 2\pi (1- \cos \theta_1) = \cfrac{4\pi}{5}
For the cone in figure B:
\Omega_2 = 2 \pi (1- \cos \theta_2) = \cfrac{2\pi}{5}
The corresponding fluxes will be
\phi_1 = \cfrac{q}{4 \pi \epsilon_0 }\times \cfrac{4\pi}{5} = \cfrac{4q}{20 \epsilon_0}
\phi_2 =\cfrac{q}{4 \pi \epsilon_0 }\times \cfrac{2\pi}{5} = \cfrac{2q}{20 \epsilon_0}
Now, we can calculate the flux through the thin strip, which contains the window as
\phi = \phi_1 - \phi_2 = \cfrac{q}{10 \epsilon_0}
Let \phi_w represent the flux through the window alone.
Then,
\phi_w \times \text{No. of windows in the strip} = \text{Total flux through the strip}
\phi_w \times \cfrac{2\pi r}{l} = \phi
\boxed{ l = 2 \ cm }

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Pls ignore my before post ..thanks

sir how you developed such skills in physics i have seen ur before post to but now ur physics seems next level how you developed such crystal clear concept... @Viram_2019 sir..

Here you can apply concept of solid angle to get the answer easily as shown by @Viram_2019
Also here you can proceed by integration ( lengthier method but still usable )
For integration , simply consider it to be a complete cylinder and integrate the flux from height r tan 37 to r tan 57
And then the value you get is for 2pieR length and as charge is placed symmetrically , therefore flux through window = flux calculated * L/2pieR