question_answer

The magnitude of electric field at distance r from an infinitely thin rod having a linear charge density \[\lambda ,\]is (use Gauss's law)

A)  \[E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}\]

B)  \[E=\frac{2\lambda }{\pi {{\varepsilon }_{0}}r}\]

C)  \[E=\frac{2\lambda }{4\pi {{\varepsilon }_{0}}r}\]

D)  \[E=\frac{4\lambda }{\pi {{\varepsilon }_{0}}r}\]

Correct Answer: A

Solution :

Gauss theorem states that the total electric flux \[(\phi )\]through any closed surface is equal to \[\frac{1}{{{\varepsilon }_{0}}}\]times the 'net' charge q enclosed by the surface. \[{{\phi }_{z}}=\int{{\vec{E}}}.\overrightarrow{dA}=\frac{Q}{{{\varepsilon }_{0}}}\] Taking cylindrical Gaussian surface of radius r, height h curved surface \[=2\pi rh\]. Electric flux through it \[=E\times 2\pi rh\] Charge enclosed \[=\lambda h\] \[\therefore \] From Gauss theorem, Charge = \[={{\varepsilon }_{0}}\times \] (Electric flux) \[\Rightarrow \] \[{{\varepsilon }_{0}}E\times 2\pi rh=\lambda h\] \[\Rightarrow \] \[E=\frac{1}{2\pi {{\varepsilon }_{0}}}.\frac{\lambda }{r}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}\]