Just refreshing on Gauss's Law. For a line of charge, we choose a cylindrical Gaussian surface. But if the E field is radially outwards, why is there no flux through the ends of the cylinder and only through the sides? I know that the field is only assumed to be perpendicular to the sides of the cylinder, but isn't there field through the ends as well? Or is the flux through the ends zero because the line of charge is infinitely long and there are field lines in equal and opposite direction at the ends of the cylinder so they cancel out, hence there is no flux?