The electric field due to an INFINITELY long straight uniformly charged wire (linear charge density $\lambda$) at a perpendicular distance $r$ from the wire is

1. Use Gauss's law with a CYLINDRICAL Gaussian surface of radius $r$ and length $L$, coaxial with the wire. 2. By symmetry, $\vec{E}$ is radial. Flux through the two END caps is ZERO (field parallel to end face's normal? No — field is RADIAL, so perpendicular to the cylinder's axial direction, hence parallel to end caps. Flux through end caps actually zero since $\vec{E}\cdot\hat{n}_\text{cap} = 0$.). 3. Flux through the curved side: $\Phi = E\,(2\pi r L)$, since $\vec{E}$ is uniform and normal to that face. 4. Enclosed charge: $Q_\text{enc} = \lambda L$. 5. Gauss's law: $E\,(2\pi r L) = \lambda L / \varepsilon_0$, so $E = \dfrac{\lambda}{2\pi\varepsilon_0\,r}$. 6. Key features: field falls off as $1/r$ (not $1/r^2$ as for a point charge) — a hallmark of cylindrical symmetry. 7. Option B is the point-charge formula (wrong symmetry). Option C has wrong dependence. Option D drops the factor of 2 in the denominator. _Source: NCERT Class 12 Physics Part 1, Ch 1, §1.14 (Gauss's law applications — infinite line charge), p. 31._