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Question
For an infinite line charge with linear charge density λ, the electric field at distance r from the wire is:
Options
\[E = \frac{\lambda}{2\pi\varepsilon_0 r}\]
\[E = \frac{\lambda}{4\pi\varepsilon_0}\]
\[E = \frac{\lambda}{4\pi\varepsilon_0 r^2}\]
\[E = \frac{\lambda}{\varepsilon_0 r}\]
MCQ
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Solution
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.
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