To find the electric field of an infinitely long, straight wire with…

The correct answer is Option (2) → There is no flux through the two flat end faces of the cylinder.

The field of a long, uniformly charged wire is radial and, by symmetry, depends only on the distance r from the wire. The cylindrical Gaussian surface is chosen to match this symmetry.

Option (1) The flux through the curved surface is \( E\times\pi r^{2} \). — Incorrect. \( \pi r^{2} \) is the area of one flat end face. The curved surface has an area \( 2\pi rl \). The field is normal to it at every point, so the flux through it is \( E\times2\pi rl \).

Option (2) There is no flux through the two flat end faces of the cylinder. — Correct. The radial field lies along the plane of each end face, at right angles to its outward normal. So \( \vec{E}\cdot\Delta\vec{S} \) is zero on both ends. All the flux passes through the curved part.

Option (3) The charge enclosed by the surface is \( 2\pi r\lambda \). — Incorrect. The cylinder encloses a length l of the wire, so the charge inside it is \( \lambda l \). The radius of the surface does not decide how much of the wire lies inside.

Option (4) The magnitude of the field changes from point to point on the curved surface. — Incorrect. Every point of the curved surface is at the same distance r from the wire, so the field has the same magnitude all over it. This is what allows the flux to be written as \( E\times2\pi rl \).

NCERT (2026-27):
Field due to an Infinitely Long Straight Uniformly Charged Wire – “To calculate the field, imagine a cylindrical Gaussian surface, … Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. The surface area of the curved part is 2πrl, where l is the length of the cylinder.”