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# Ques attached.

Shivam Chopra
45 Points
6 years ago
As the cylinder is kept horizontally, so, outward normal for the curved surface ( i.e. whose surface area is 2πrl ) is perpendicular to the x- axis & thus electric flux through it wil be zero.
Now, talking of 2 circles, for circle at x>0, outward normal is parallel to x-axis & is in the direction of x-axis i.e. ^i .
So, electric flux = E.dS = Ex . 4πr2
Similarly for other circle, outward normal is in direction of –ve x- axis i.e. -^i .
So, electric flux = E.dS = -Ex . 4πr2 (-1) = Ex4πr2
So, total flux = 2. Ex4πr2
Now, using guass law -> φ = q/ε
=>  φ = Electric flux = 2Ex4πr2   which is equal to  q/ ε
So, q = 8 π r2 ε Ex   Coluomb
Please approve it if you like.