Question
Download Solution PDFAccording to Gauss’s law, the electric field due to an infinitely long thin charged wire varies as:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e. \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\)
But we know that Electrical flux through a closed surface is \(\oint \vec E \cdot \overrightarrow {ds} \)
\(\therefore\oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\)
Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.
EXPLANATION:
Electric field due to line charge:
Electric field due to an infinitely long straight conductor is
\(E=\frac{\lambda }{2\pi {{\epsilon }_{o}}r}~\)
Where λ = linear charge density, r = radius of the cylinder, and εo = permittivity of free space.
From the above equation, it is clear that the electric field of an infinitely long straight wire is proportional to 1/r. Hence option 1 is correct.
Last updated on Jun 19, 2025
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