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 = E x . 4πr 2 Similarly for other circle, outward normal is in direction of –ve x- axis i.e. - ^ i . So, electric flux = E.dS = -E x . 4πr 2 (-1) = E x 4πr 2 So, total flux = 2. E x 4πr 2 Now, using guass law -> φ = q/ε => φ = Electric flux = 2E x 4πr 2 which is equal to q/ ε So, q = 8 π r 2 ε E x Coluomb Please approve it if you like.

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