Question

For an infinite line charge with linear charge density λ, the electric field at distance r from the wire is:

पर्याय

• \[E = \frac{\lambda}{2\pi\varepsilon_0 r}\]

• \[E = \frac{\lambda}{4\pi\varepsilon_0}\]

• \[E = \frac{\lambda}{4\pi\varepsilon_0 r^2}\]

• \[E = \frac{\lambda}{\varepsilon_0 r}\]

Answer 1

For an infinite line charge, using a co-axial cylindrical Gaussian surface of radius r and length l, the flux through the curved surface is E × 2πrl. Since the enclosed charge is Qenc = λl, applying Gauss's Law gives E = λ/(2πε₀r). The field decreases as 1/r and is directed radially outward for positive λ.