Consider a cylindrical surface of radius R and length l in a uniform electric field E. Compute the electric flux if the axis of the cylinder is parallel to the field direction.

Explanation

We can divide the entire surface into three parts, right and left plane faces and curved portion of its surface. Hence, the surface integral consists of the sum of the three terms:
`phi_(E) = oint E * dA = int_(l eft end) E*dA + int_(right end) E*dA + int_(curved)E*dA`

All the area elements on the left end and electric field E are
at and angle of `180^@`.
`(phi_E)_l eft end = oint_(l eft end )E*dA = oint_(l eft end) E dA cos 180^@`
`=-E oint_(right end) dA = E pi R^2`
Note that E is constant over the entire plane surface of left end, therefore, we take it out from the integral.
Similarly, all the area elements on the right end are parallel
to electric field E, i.e., angle is `0^@`.
`(phi_E)_(right end ) = oint_(right end)E*dA = oint_(right end) E dA cos 0^@`
`=+Eoint_(right end) dA = E pi R^2`
Finally, at every point on the curved surface the area vectors
are perpendicular to the direction of the electric field. Thus,
`(phi_E)_(curved) = oint_(curved surface) E*dA = oint_(curved surface) E dA (cos90^@) = 0`
Total flux `= (phi_E)_(right end) + (phi_E)_(l eft end) + (phi_E)_(curved surface)`
`= (+EpiR^2) + (-EpiR^2) +0 = 0`
Hence, we see that in a uniform field the flux through a closed surface is zero. This is true for any shape of closed surface.