An infinitely long wire is charged uniformly with charge density \( \lambda \) and placed in air, the electric field at distance r from wire will be :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{\lambda}{2\pi\varepsilon_0 r} \)
By symmetry the field of an infinitely long line charge points radially outward and has the same magnitude at all points at distance \( r \). Take a cylindrical Gaussian surface of radius \( r \) and length \( l \) coaxial with the wire. The flat ends give no flux, because the field is parallel to them. Applying Gauss's law to the curved surface:
\( \displaystyle E \times 2\pi r l = \frac{\lambda l}{\varepsilon_0} \)
\( \displaystyle E = \frac{\lambda}{2\pi\varepsilon_0 r} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{4\pi\varepsilon_0}\,\frac{\lambda}{r} \) — Incorrect. This is half the correct value; \( \lambda/(2\pi\varepsilon_0 r) = \frac{1}{4\pi\varepsilon_0}\,\frac{2\lambda}{r} \).
Option (2) \( \displaystyle\frac{1}{4\pi\varepsilon_0}\,\frac{\lambda}{r^{2}} \) — Incorrect. A \( 1/r^{2} \) field belongs to a point charge, not to a line charge.
Option (3) \( \displaystyle\frac{\lambda}{2\varepsilon_0} \) — Incorrect. A field that does not depend on distance, \( \sigma/2\varepsilon_0 \), belongs to an infinite plane sheet; here \( \lambda \) is a charge per unit length and the field falls as \( 1/r \).
NCERT (2026-27):
Applications of Gauss’s Law – “… The surface includes charge equal to λ l. Gauss’s law then gives
\( \displaystyle E \times 2\pi r l = \frac{\lambda l}{\varepsilon_0} \)
\( \displaystyle \text{i.e.,}\ \ E = \frac{\lambda}{2\pi\varepsilon_0\,r} \)
Vectorially, E at any point is given by
\( \displaystyle \mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0\,r}\,\hat{\mathbf{n}} \qquad (1.32) \)
where \( \hat{\mathbf{n}} \) is the radial unit vector in the plane normal to the wire passing through the point. E is directed outward if λ is positive and inward if λ is negative.”